Heart of algebra

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Linear equations on the SAT test

Linear equations on the SAT test

SAT Subscore: Heart of Algebra

Linear equations formula sheet

linear equation is an algebraic equation in which each term has an exponent of one. It is called linear because it can be graphed as a straight line in the xy-plane. 

On the SAT test linear equations are part of heart of algebra subscore questions. Since this is a fundamental topic it appears in many SAT questions. 

Let’s look at the different types of questions and examples for each one of them.

Calculating an output of an expression SAT questions

In these questions you need to calculate the value of an expression. In some questions you are given a word problem and in other questions you are given an equation that you have to use it order to solve the expression.

The skills required are
Translating a word problem into a mathematical expression.
Solving equation.
Performing calculations on an equation to match a given expression.

Before making any calculations, it is necessary to know the rules for calculations with negative numbers:
Negative * negative = positive. For example: -5*-4=20
Positive * negative = negative. For example: 5*-4=-20
Negative : negative = positive. For example: -8:-4=2
Positive : negative = negative. For example: 8:-4=-2
Negative : positive = negative. For example: -8:4=-2

Calculating an output of an expression given a word problem

Consider this example:

The length of the rectangle is 5 cm, the width of the rectangle is 20 percent smaller than its length.

What is the area of the rectangle?

We need to write an expression and calculate it. Rectangle’s area is the equals to its length multiplied by its width. If the length of a rectangle is 5 cm then the width of the rectangle is 5* 80%=4 cm. The area of ​​the rectangle is 5*4=20 cm.

Calculating and expression given an equation

In these questions we are be asked to calculate an expression given an equation that we don’t need to solve. All we need in order to get the answer is to change the equation to match the expression. 

Consider this example:

2x+10=3x+25

What is 6x+50 equal to?

Remember that if you need to divide or multiple you need to do it for each component of the expression. In this case you can add parentheses to the expression. We see that the expression 6X+50 is 2 times larger than the expression 3x+25 therefore we can multiply the given equation by 2 and get the answer.

Before multiplying each side of the equation by 2 we add parentheses to the both sides of the equation:
(2x+10)*2=(3x+25)*2
4x+20=6x+50
So the answer is that 6x+50 is equal to 4x+20

Sometimes we will be asked to calculate an expression given an equation that we need to solve first. 

Consider this example:

We know that 3x+10=25, calculate 4x+12.

Step 1: Solving the equation:
3x+10=25
3x=25-10
3x=15
x=5

Step 2: Calculating the expression:
4x+12=4*5+12=20+12=32

Solving a linear equation SAT questions

In these questions you are given an equation and you are asked to solve it.

The skills required are solving an equation with one variable and making calculations including fractions, decimal fractions and absolute values.
In addition there are questions with two variables where you need to find the value of one variable from a word problem before solving for the second variable.

The rules for maintaining equality

Before starting to solve the equation, it is necessary to know the rules for maintaining equality:
1. You can add or subtract the same value from each side of the equation.
2. You can divide or multiply each side of the equation by a same value.

Consider this example:

2x+5=14

What is the solution to the given equation?

2x+5=14
2x+5-5=14-5
2x=9
2x:2=9:2
x=4.5

 

The steps for solving a linear equation

Distributing formula for removing parentheses:

(a+b)x=ax+bx
(a-b)x=ax-bx

For example: (2+x)5=10+5x

Combining like terms formula:

ax+bx=(a+b)x
ax-bx=(a-b)x

For example: 8x-2x=(8-2)x=6x

The goal of solving an equation is finding the x variable value. In order to do so we need to isolate x on the one side of the equation so that the other side of the equation will contain the constant (the answer).

a. Removing parentheses and combining like terms in each side of the equation.
Distributing formula for removing parentheses: (a±b)x=ax±bx, for example (2+x)5=10+5x
Combining like terms formula: ax±bx=(a±b)x, for example 8x-2x=(8-2)x=6x
b. Isolating the variable term on one side of the equation by using addition or subtraction.
c. Solving the equation (finding x=) by using multiplication or division.

Consider this example:

5x-10+x=2x+9(5+5)

What is the solution to the given equation?

Step 1: Solving the equation
a. Removing parentheses and combining like term in each side of the equation:
6x-10=2x+90

b. Subtracting 2x and adding 10 to each side of the equation:
6x-10-2x+10=2x+90-2x+10
4x=100

c. Dividing each side of the equation by 4:
4x:4=100:4
x=25

Step 2: checking the answer:
5*25-10+25=2*25+9(5+5)
125-10+25=50+9*10
140=140

Solving a linear equation with decimal fractions

Consider this example:

(2x-6)*0.5=(3x-10)*0.4-0.6

What is the solution to the given equation?

Step 1: Solving the equation:
To solve the equation we need to open parentheses on both of its sides.
2*0.5x-6*0.5=3*0.4X-10*0.4-0.6
x-3=1.2x-4-0.6
-0.2x=-4-0.6+3
-0.2x=-1.6
x=1.6/0.2
x=8

Step 2: checking the answer:
(2*8-6)*0.5=10*0.5=5
(3*8-10)*0.4-0.6=14*0.4-0.6=5.6-0.6=5

Solving a linear equation with fractions

Addition and subtraction formula:

 ab±cb=(a±c)b

For example: 14+514=512

Multiplication formula:

ab*cd=a*cb*d

For example: 14*2=1*24*1=24 =12

Division formula:

ab : cd=ab*dc=a*db*c

For example: 14 : 2=14*12=1*14*2=1

To add or subtract fractions they must have the same denominator (the bottom value). We need to add or subtract the numerators and leave the denominator.
Addition and subtraction formula: ab±cb=(a±c)b.   For example: 14+514=512

To multiply fractions, you have to multiply the nominators and multiply the denominators.
Multiplication formula: ab*cd=a*cb*d. For example: 14*2=1*24*1=24 =12

To divide fractions, you must flip the second fraction and then multiply it with the first fraction.
Division formula: ab : cd=ab*dc=a*db*c. For example: 14 : 2=14*12=1*14*2=1

If the denominators are different you can first reach a common denominator in order to cancel the denominators and then you can continue to solve the equation. This is done by multiplying by the smallest common multiple of the denominators of the fractions.

Consider this example:

312x-214=2x+514

What is the solution to the given equation?

Step 1: Solving the equation: The common denominator for the numbers 2 and 4 is 4 so we need to multiply the equation by 4. 4*(312x-21⁄4)=4*(2x+51⁄4) 4*3x +4*12x-4*2-4*1⁄4=8x+4*5+ 4*1⁄4 12x+2x-8-1=8x+20+1 14x-9=8x+21 6x=30 x=5 Step 2: Checking the solution: 312x-21⁄4=2x+51⁄4 312*5-21⁄4=2*5+514 15+2+12-21⁄4=1514 17+12-21⁄4=1514 1514=1514

Solving equations with absolute values

Pay attention that absolute value calculations are rare on the SAT.

The absolute value of a negative number is always positive. Therefore, we need to split an absolute value equation into 2 equations.

Consider this example:

What is the solution for the equation |6x-10|=2?

Step 1: Solving the equation
Since we don’t know the sign of |6x-10|=2, we need to write 2 equations.

Equation 1: (6x-10)=2
6x=12
x=2

Equation 2: (6x-10)=-2
6x=8
x=86=126=113

Step 2: Checking the answers
|6*2-10|=2
2=2

|6*86-10|=2
|486-10|=2
|8-10|=2
|-2|=2

Solving an equation with two variables given an input of one variable

In these questions you are given a word problem and an equation with two variables. You need to find the value of one variable inside the word problem and then plug it into the equation. After that you will have an equation with one variable that you can solve.

Consider this example:

0.6x+0.4y=86

The equation above represents a weighted average of 86  in two subjects, defined x as a grade in course 1 and y as a grade in course 2. What was the second grade?

The first grade should be placed in the equation instead of x and then we will be able to solve an equation with 1 variable y.

Step 1: Plugging the input into the equation:
0.6*90+0.4y=86

Step 2: Solving the equation with 1 variable:
54+0.4y=86
0.4y=32
y=80
The second grade was 80.

Step 3: Checking the answer:
In order to test the answer we will place it in the equation: 0.6*90+0.4*80=54+32=86.

Linear equations that don't have one solution SAT questions

Previous examples included common equations that had one single solution.

In these questions you are asked to solve an equation that has no solution or an infinite number of solutions.

The skills required are solving an equation with one variable and identifying the conditions for no solution or an infinite number of solutions to the equation.

Linear equations with no solution

Equations with no solution appear when the x variable is eliminated from the equation so what is left is only 2 constants a=b. Since a is a different from b this is a false statement and the equation has no solution. 

Consider this example:

What is the solution for the equation 6x+12=3(2x+2)?

6x+12=3(2x+2)
6x+12=6x+6
12-6=6x-6x
6=0 is a false statement therefore the equation has no solution.

Finding the condition for no solution:

In this type of questions we are given an equation with a as a constant and asked for which value of a there is no solution to the equation. In order to find the solution, we need to solve the equation. Another possibility is to place the answers into the equation and see if we get a false statement.

Consider this example:

For which value of a the equation 6ax+12=3(2x+2) has no solution? a=0,1,2?

Option 1: Solving the equation
6ax+12=3(2x+2)
6ax+12=6x+6
6ax-6x=-6
6x(a-1)=-6
x(a-1)=-1
x=-1/(a-1)
a≠1 since division by zero is not allowed therefore there is no solution when a=1.

Option 2: Placing given answers into the equation
For a=0 we will get
12=6x+6
6x=6
x=1
We can check this answer by plugging x=1 and a=0 into the equation to see if the quality holds:
6*0*1+12=3(2*1+2)
0+12=3*4
12=12

For a=1 we will get
6x+12=6x+4
12=4 is a false statement therefore there is no solution when a=1.

For a=2 we will get
12x+12=6x+6
6x=-6
x=-1
We can check this answer by plugging x=-1 and a=2 into the equation to see if the quality holds:
6*2*-1+12=3(2*-1+2)
-12+12=3*(-2+2)
0=3*0
0=0

Linear equations with an infinite number of solutions

Equation with an infinite number of solutions appears when the constants are eliminated from the equation so what is left is only 2 x variables in the form x=x.  This is a true statement for every x therefore the equation has an infinite number of solutions. 

Consider this example:

What is the solution for the equation 6x+12=3(2x+4)?

6x+12=3(2x+4)
6x+12=6x+12
6x=6x
x=x
x=x is a true statement for every x therefore the equation has an infinite number of solutions.

We can also continue to solve:
x=x
x-x=0
0=0 is also a true statement for every x

Creating and solving an equation from a word problem SAT questions

In these questions you are given a word problem from which you need to write an equation and solve it.  

The skills required are defining x variable in a word problem, writing an equation from a word problem and solving an equation with one variable.

Consider this example:

The inventory of a store contained bags with balls received from the supplier so that in each bag were packed 10 balls. The store worker decided to pack the balls in smaller packages in order to sell them to customers. He took a certain number of bags and emptied the balls from them. He then added to the balls he took out another 30 balls left in the store from previous orders and then he divided these balls into 75 bags so that each bag contained 2 balls. The worker noticed that he forgot to write down how many bags he had taken from stock. What was the amount?

In this example we will need to write an equation in order to reach a solution.

Step 1: X definition: Let’s look at the question and mark in x the number of bags the employee took from stock.
The worker took x bags containing 10 balls in each bag and added them 30 balls: 10x+30
Then he divided these the balls into 75 bags so that each bag contained 2 balls: 75*2

Step 2: writing the expressions in each side of equation: We need to find 2 equal expressions.
Since we know that the worker sorted the same amount of balls, we know what to write on each side of the equation:
10x+30=75*2.

Step 3: Solving the equation:
10x+30=75*2
10x=150-30
10x=120
x=12
The solution is x=12.

Solving a word problem using two equations

What if the question is not about is the x variable value? In this case after calculating x value, we will write another equation using y variable (y will be defined as what is asked in the question).

Steps for writing a second equation: 

Step 1: Looking at the relevant data parts again, this time ignoring the data we needed to calculate x.
Step 2: Identifying parts of the data which will give us an equality.
Step 1+2: Writing the second equation consisting y variable and solving it.

Let’s add to the above example: There were 50 bags with balls in the store’s inventory.
I addition, let’s assume that the question is not to find the amount off bags that the worker took from the store’s inventory. Instead, the question is how many bags were left in the inventory?

The inventory of a store contained bags with balls received from the supplier so that in each bag were packed 10 balls. There were 50 bags with balls in the store’s inventory. The store worker decided to pack the balls in smaller packages in order to sell them to customers. He took a certain number of bags (we solved x=12)  and emptied the balls from them. He then added to the balls he took out another 30 balls left in the store from previous orders and then he divided these balls into 75 bags so that each bag contained 2 balls. How many bags were left in the inventory?

The first equation was 10x+30=75*2. We solved it so x=12.

Writing a second equation: We will use the letter y for the new variable. The y variable will be according to the question: y=the number off bags that were in the inventory. The equation will be: y=50-x. We had already solved x=12, so now we know that y=38.

The question could also be even more complicated: How many bags were left in the inventory? In this case the second equation should be y=(50-12)*10 and the answer will be 380 balls.

Linear equations formula sheet

Systems of linear equations on the SAT Test

Systems of Linear Equations on the SAT Test

SAT Subscore: Heart of Algebra

System of linear equations is a set of two or more linear equations. In SAT questions we usually see systems of equations containing 2 equations with 2 variables x and y. If we are given a word problem, we first need to define the variables and then write 2 equations with these variables.

On the SAT test linear equations are part of heart of algebra subscore questions. 

Let’s look at the different types of questions and examples for each one of them.

Creating a system of linear equations SAT questions

In these questions you need to create a system of linear equations from a word problem.

The skills required:
Defining variables.
Writing two equations with two variables.

Why do we need two equations instead of one?

This type of questions contain a word problem with 2 variables. This means that we will have to write 2 linear equations with two variables in each equation.

Each one of the equations separately has an infinite number of solutions, therefore in order to reach one solution we need to solve two equations as a system.

For example: Given the equation 2x+3y=5 we can plug any x into the equation and find its y value. For x=1 the y value is y=1. For x=2 the y value is y=1/3….
Only combining 2 equations and solving them together can have a single solution.

Note that if the system has 1 solution, it will be the intersection point of the graphs of the equations.

Consider this example:

The supermarket sells small egg packs containing 14 eggs and large egg packs containing 30 eggs. The order for the eggs was $ 1,300 and 70 packages were placed on the shelf. Write a system of equations that can be used to determine the amounts of packages that were ordered of each type.

Step 1: Variables definition– Define x as the number of small packs and y as the number of large packs.

Step 2: Writing the equations:

14x+30y=1300

x+y=70

Methods for solving a system of linear equations SAT questions

In these questions you need to solve a system of linear equations. Some question may require first to translate a word problem into a system of equations.

The skills required:
Defining variables.
Writing two equations with two variables.
Solving two equations with either substitution or elimination.

In order to solve the equations, we need to reduce 2 equations with 2 variables to 1 equation with 1 variable. This can be done in two methods: substitution and elimination. 

Solving two equations by substitution

This can be done by substituting an expression for a variable (plugging an expression instead of a variable). This will leave us with one equation with one variable that we can solve.

Step 1: Isolate one of the variables in one equation so you have an equation containing x=expression or y=expression. Look which variable in which equation is easiest to isolate.

Step 2: Substitute the expression from step 1 for a variable into the other equation. Now you stay with one equation that has only one variable that can be solved.

Step 3: Solve the equation for the remaining value and plug it into one of the equations to calculate the other variable.

Consider this example:

Solve the equations using substitution
2x+3y=5
x=y

We plug y instead of x into 2x+3y=5 and get an equation without x variable: 2y+3y=5.
Now we can solve the equation.
5y=5
y=1
x=y=1

 

Solving two equations by elimination

This can be done by adding or subtracting the two equations in a way that will cancel one of the variables.  This will leave us with one equation with one variable that we can solve.

Step 1: Identify which variable has the same coefficients. For example: 4x and -4x, -6y and 6y).

Step 2: If there is no variable with the same coefficient multiply or divide one of the equations in order to reach for the same coefficient.

Step 3: Add or subtract the equations to eliminate the variable with the same coefficient.

Step 4: Solve what is left- one equation with one variable and then plug the answer to one of the equations to calculate the other variable. 

Step 3- Adding or subtracting- which side of the equations to use?

We know that we can add or decrease the same amount from both sides of the equation because the quality will be maintained. Since both sides of the equation are equal, we can add or decrease 2 sides of one equation from the other.

For example:
Equation 1: 5=3+2
Equation 2: 3=2+1.

Adding the left side of the first equation to the left side of the second equation and adding the right side of the first equation with the right side of the second equation will result in 5+3=3+2+2+1, 8=8.

Adding the left side of the first equation to the right side of the second equation and adding the right side of the first equation with the left side of the second equation will result in 5+2+1=3+2+3, 8=8.

This can be also done with subtraction, for example:
Subtracting the left side of the first equation from the left side of the second equation and subtracting the right side of the first equation from the right side of the second equation will result in 3-5=2+1-3-2, -2=-2.

We need the adding or the subtracting to eliminate one of the variables so we will choose the action accordingly.

Consider this example:

Solve the equations using elimination
2x+3y=5
-2x=y

Step 1: Solving the equations:
We have 2x in one equation and -2x in the other, therefore the action that will eliminate x variable should be adding the left sides of the equations and adding the right sides of the equations.
2x+3y-2x=5+y
3y=5+y
2y=5
y=2.5

Calculating x:
2x+3*2.5=5
2x+3*2+3*0.5=5
2x+6+1.5=5
2x+7.5=5
2x=-2.5
x=-1.25

Step 2: Checking the answer:
Plug x=1.25 and y=2.5 into one of the equations
-2*-1.25=2.5
2.5=2.5

Step 3- Adding or subtracting- what to do if the coefficients of the variables are not the same?

Sometimes the coefficients of the x and y variables in the two equations are not the same, therefore adding or subtructing the equations will not eliminate a variable.
In this case first multiply of divide one of the equations to reach the coefficient that is identical to the coefficient in the other equation.

Consider this example:

Solve the equations using elimination
2x+3y=5
-x=y

Solution 1:
We can multiply the second equation by 2 and then add the equations to eliminate x variable:
-x=y
-2x=2y
2x+3y=5

Adding the left and the right sides of the equations:
-2x+2x+3y=2y+5
3y=2y+5
y=5
-x=5
x=-5

Solution 2:
We can also multiply the second equation by 3 and then subtract the equations to eliminate y variable:
-x=y
-3x=3y
2x+3y=5

Subtracting the right side of first equation from the left side of the second equation and subtracting the left side of first equation from the right side of the second equation will give us
2x+3y-3y=5–3x
2x=5+3x
x=-5
-x=y
–5=y
y=5

Solving a system of linear equations SAT questions

In these questions you need to solve a system of linear equations. Some question may require first to translate a word problem into a system of equations.

The skills required:
Defining variables.
Writing two equations with two variables.
Solving two equations with either substitution or elimination.

Solving a system of linear equations- one solution

Most systems of linear equations have one solution.
Consider this example:

The sum of two numbers is 20. One number is 200 percent larger than the other number.

What are the numbers?

Step 1: Variables definition– Define x as the first number and y as the second number.

Step 2: Writing the equations:
x+y=20
y=x+200%*x

Step 3: Solving the equations: Since there is y variable on the left side of the first equation and the left side of the second equation we can subtract one side from the other.
(x+y)-y=20-(x+200%*x)
x+y-y=20-x-200%*x
x=20-x-2x
x+3x=20
4x=20
x=20/4
x=5 meaning that the first number is 5.
Now we can calculate the second number: x+y=20, y=15

Step 5: checking the answer:
5+15=20
15=5+200%*5

Solving a system of equations with no solution

Solving a system of equations with no solution will lead us to a false statement. Consider this example:
Solve at the system of equations 2x+3y=5 6x+9y=7

We can multiply each side of 2x+3y=5 by 3 and get 6x+9y=15.
Then we can subtract each side of 6x+9y=15 from the corresponding side of 6x+9y=7.
The answer will be 0=8, this is a false statement therefore there is no solution.

Solving a system of equations with infinite number of solutions

Solving a system of equations with infinite number of solutions will lead us to a statement that is always true.

Consider this example:

Solve at the system of equations
x+2y=4
4x+8y=16

We can multiply the first equation by 4 and get an equation that is identical to the second equation. Since we can’t solve one equation with 2 variables the system will have an infinite number of solutions. For example: if x=0 then y=2, if x=2 then y=1.

We can also subtract each side of 4x+8y=16 from the corresponding side of 4x+8y=16 and get 0=0, this is a statement that is true for every x and y. 

Determining the number of solutions for a system of linear equations SAT questions

In these questions you need to determine if the system of equations has one solution, no solution or an infinite number of solutions.

The skills required:
Writing equation in a slope intercept form.
Understanding the concepts slope and intercept and their meaning.
Performing a comparison between 2 equations.

In order to know the number of the solutions we don’t have to solve the equations. We just need to write both equations in a slope intercept form y=mx+b and see if the slopes m and intercepts b of the equations are the same.

  1. If the equations have the same slope m and a different intercept b the system has no solution.
  2. If the equations have the same slope m and the same intercept b the system has an infinite number of solutions.
  3. If the equations have a different slope m then the system has one solution.

Note that we can also graph the lines of the equations and see if they have an intersection point.

Consider this example:

-2x+y=6
4x+y=18
How many solutions does the equations system have?

We need to write the equations in a slope intercept form:

-2x+y=6
y=2x+6 (m=2, b=6)

4x+y=18
y=-4x+18 (m=-4, b=18)

We can see the equations have different slopes therefore there is one solution for the system of the equations.

Note that we can solve the equations and calculate that x=2 and y=10. 

Graphic presentation of a system of linear equations SAT questions

In these questions you need to identify a graph of a given system of equations.

The skills required:
Writing a system of linear equations in a slope intercept form.
Graphing 2 linear equations in the xy plane.

As said above, we can draw each equation as a line in the xy plane. The solution of the system is the intersection point of the lines. Let’s look at the graphic presentation of the systems that have one solution, no solution or an infinite number of solutions. 

Graphing a system of equations with no solution

If the graphs of the equations have different intercepts and same slopes, they are parallel. Parallel lines will never intersect.

Consider this example:

Draw the graphs of a system of equations
2x+3y=5
6x+9y=7

In order to draw the equations, we need to write them in a slope intercept form.
2x+3y=5
3y=-2x+5
y=-23x+53

6x+9y=7
9y=-6x+7
y=-69x+79
y=-23x+79

The equations have the same slope m=-23 and different intercepts b1=7⁄9 b2=5⁄3 therefore their graphs will never meet.

The graphic presentation of this system:

linear equations systems wuth zero solutions

Graphing a system of equations with an infinite number of solutions

If the graphs of the equations have the same intersect and the same slope, we will see the same graph twice, therefore the lines will overlap. Every point on the graph is a solution therefore the system has an infinite number of solutions.

Draw the graphs of a system of equations
x+2y=4
4x+8y=16

We can multiply each side of x+2y=4 by 4 and get 4x+8y=16. We left with only one equation 4x+8y=16 therefore there is an infinite number of solutions. For example: if x=0 then y=2, if x=2 then y=1.

The graphic presentation of this system:

linear system graph

Graphing a system of equations with one solution

If the lines have different slopes they will always intersect once, therefore in this case there is one solution to the system of equations. 

The graphs below show the linear system of equations y=3x-3 and y=-2x+5. The table near the graphs shows some selected points from the graphs.

The solution of the equation system x=1.6 y=1.8 (1.6,1.8) is the point where the two graphs intersect.

linear relationship plot

The graphs below show the linear system of equations that were solved above y=-4x+18 and y=2x+6.

The table near the graphs shows some selected points from the graphs.

The solution of the equation system x=2 y=10 (2,10) is the point where the two graphs intersect.

two lines grafps

Linear Functions on the SAT Test

Linear Functions on the SAT Test

SAT Subscore: Heart of Algebra

linear function is an equation that represents the relationship between two variables, most commonly called x and y. It is called linear because it can be graphed as a straight line in the xy-plane. 

On the SAT test linear functions are part of heart of algebra questions. Since this is a fundamental topic it appears in many SAT questions. 

Let’s look at the different types of questions and examples for each one of them.

Creating a linear function SAT questions

In these questions you are given a word problem and you are asked to write the function that describes the problem without solving it.
The skill required is translating a word problem into a mathematical equation.

Consider this example:

The full price of the item is X dollars. During the holiday sale all the products in the store are sold at a twenty percent discount. In addition, there is a second discount of 10 percent on the item.

What is the function P that represents the price after the discounts in terms of x?
Case 1: The additional discounts are calculated on the price after the holiday discount.
Case 2: The additional discounts are calculated on the original price.

Case 1: The price of the item after the discount is: P= x*(1-0.2)*(1-0.1)=x*0.8*0.9=x*0.72=0.72x
Case 2: P=x*(1-0.2-0.1)= x*0.7=0.7x

Note that the difference between the two answers is 10 percent from 20 percent: 0.1 times 0.2= 0.02

Finding the value of an output of a given function SAT questions

In these questions you are given the equation of the function and you are asked to solve this function given the values of the variables.
The skills required are the ability to plug data instead of variables and making a simple calculation.
Solution steps: Data identification, plugging the data into the function, solving the expression.

Consider this example:

The plant uses function c defined by c(f,n,v)=f/n+v to calculate the cost c(f,n,v), in dollars, of producing a product. F is defined as a fixed cost of the plant; v is defined as a variable production cost per unit and n is defined as the quantity of the products.
If the fixed cost is $ 50,000 and the variable cost is $ 1, what is the manufacturing cost per product in the production of 100,000 products?

In order to answer this question, we must plug the data into the function and solve it.
Step 1: writing the data:
c(f,n,v)=f/n+v
f=50,000
v=1
n=100,000

Step 2: Plugging the data into the function:
c(f=50000, n=100000, v=1)= 50,000/100,000+1

Step 3: Solving the expression:
50,000/100,000+1=0.5+1=1.5

The answer is that the manufacturing cost of 1 unit is 1.5 dollars.

Finding the input that corresponds to a given output of a function SAT questions

In these questions you know the equation of the function and you are asked to calculate the input of this function given the value of the output.
The skills required are the ability to plug data instead of variables and solving the equation by isolating its input variable.
Solution steps: Plugging the data into the function, solving the equation.

The camp manager uses the function f(x) defined by f(x)=430+55*x to calculate the total number of the children f(x) in the summer camp. More children join the camp every day on x buses.
If there were 760 children in the camp today, how many buses arrived at the camp?

In this question we are given the output of the function which is 760 children. This allows us to write an equation from the function and find the variable x.
Step 1: Plugging the output into the function:
760=430+55x

Step 2: Solving the equation:
55x=330
x=6
The answer is that 6 buses arrived today.

Step 3: checking the answer:
In order to test the answer we will place it in the equation: 430+55*6=430+300+30=760.

Interpretation of a linear function SAT questions

In these questions you need to analyze the linear function in its slope-intercept form: y=mx+b.

The skills required:
Understanding the structure of the function.
Rewriting the function in a slope-intercept form.
Performing calculations to find the slope m and the intercept b.
Finding the equation of the function.
Presentation and analysis of the graph of a linear function.

What is a slope-intercept form of a linear function?

Slope intercept form of a linear function:
y=mx+b

For example: y=4x-3

Standard form of a linear function:
ax+ty=c

For example: 3x+4y=20

A linear function is a linear equation with 2 variables. We often see it the standard form as ax+ty=c (where a ,t and c are constants). For example: The price of 3 pens and 4 pencils is 10 dollars 3x+4y=20.

It is important to know that a linear function can be written in a slope intercept form y=mx+b where x and y are variables and m and b are constants. This form is essential to understanding how variable x affects variable y and is used to draw the function on a xy-plane.

The lines in the plot below represent 3 linear functions with different slopes and intercepts. 

slope intercept graph

The constant m represents the slope of the function:
If the slope m is positive then the line trends upward from left to right (see the blue line).
If the slope m is negative then the line trends downward from left to right (see the orange line).
If the slope m is equal to zero then the line is parallel to the x axis (see the red line).

The constant b represents the intercept of the function with y axis. This is the point where the line crosses the y axis. In order to find b, we can plug x=0 into the equation and find y. 

We can also calculate the intercept of the line with the x axis by plugging y=0 into the equation and finding x.

How to identify the slope m and the intercept b in the function equation? 

As said earlier we often see the standard form of a linear function as ax+ty=c (where a ,t and c are constants). Since the equation isn’t written as y=mx+b we need to rewrite it in slope-intercept form. 

Consider this example:

What is the slope and the intercept of the function 6x-3y+12=0?

6x-3y+12=0

6x+12=3y

y=2x+4

 m=2 is the slope and b=4 is the intercept.

Writing the equation of the linear function

In order to write an equation of a linear function we need to know its slope and intercept and plug them in the formula of the function: y=mx+b. Consider this example:

The slope of a linear function is 5, the intercept is -3. Write the equation of the linear function.

We are given that m=5 and b=-3 therefore the equation in a form of y=mx+b is y=5x-3.

How can we calculate the slope m?

Linear slope formula:
m =(Ya-Yb)/(Xa-Xb)

For example: For A(-2,-9) and B(0,-3) the slope is m=(-9–3)/(-2-0)=-6/-2=3

We can calculate the slope m by using the x and y values ​​of any two points A and B that are on the graph of the function.
The slope m is the difference between the y values and divided by the difference in the x values ​​ of the 2 points. Since the function is a straight line the slope is fixed and does not change at any point selected in the function.
The formula for calculating a slope is: m =(Ya-Yb)/(Xa-Xb)
Note that if you write the y value of point A first, then you should also write the x value of point A first.
Consider this example:

There are 2 points on the graph of a linear function A(-2,-9), B(0,-3). What is the slope m of the function?

m=(Ya-Yb)/(Xa-Xb)=(-9–3)/(-2-0)=-6/-2=3. The answer is m=3.

Note that we can write first the values of point B and subtract from them the values of point A, the slope m will be the same: m=(-3–9)/(0–2)=6/2=3.

Finding the equation of a linear function SAT questions

In these questions you need to write the equation of the linear function y=mx+b given values of its points or its slope b.

The skills required:
Understanding the structure of the function.
Performing calculations to find the slope m and the intercept b.
Finding the equation of the function.
Solving an equation with one variable.
Solving an equation with two variables.

As we know the equation of a linear formula is y=mx+b. Without any other data we have 2 unknown variables (x and y) and 2 unknown constants (m and b). In order to write the equation, we need to find the value of m and b.

There are 2 ways we can find the equation:
1. If we are given x and y values of one point on the line and the value of the slope m. We can write an equation with only one variable which in b and solve it.
2. If we are given x and y values of two points on the line. We can write 2 equations with 2 variables m and b and solve them. Another option is to calculate the slope m and then write an equation with only one variable b and solve it. 

Finding the equation of a linear function given the value of one point and the value of the slope m:

We know the slope so we only need to calculate the intercept b by plugging the m value and the point values into the function formula y=mx+b. Consider this example:

In a linear function: the slope m=3, there is a point on the function which values are A(-2,-9). What is the equation of the function? 

The linear function formula is y=mx+b
-9=3*-2+b
-9=-6+b
b=-3
The function equation is therefore y=3x-3

Finding the equation of a linear function given x and y values of two points:
Option 1 for solution– Using the slope formula in the calculation:
Step 1: Calculating the slope m using the slope formula.
Step 2: Calculating the intercept b by placing the m value and the point values into the function formula y=mx+b
Option 2 for solution: Solving with a system of equations:
We can 2 write  and solve 2 equations so that each equation represents a point and the variables are the slope m and the intercept b.

Consider this example:

There are 2 points on the graph of a linear function A(-2,-9), B(0,-3). What is the equation of the function?

Option 1 for solution:
Step 1: We had already calculated the slope m in the previous question as
m=(Ya-Yb)/(Xa-Xb)=(-9–3)/(-2-0)=-6/-2=3
Step 2: We had already calculated the intercept b in the previous question as
-9=3*-2+b
-9=-6+b
b=-3

Option 2 for solution:
Solving the equation system y=3x-3, y=-2x+5
3x-3=-2x+5
5x=8
x=8/5=1+3/5=1+0.6=1.6
y=3x-3=3*1.6-3=3+3*0.6-3=3*0.6=1.8

The solution is x=1.6 y=1.8 (1.6,1.8).

Graphic presentation of a linear function SAT questions

In these questions you need to identify a graph of a given linear function or to find the equation of a linear function y=mx+b given the graph of the function.
The skills required:
Understanding the structure of a linear function.
Writing a linear function in a slope intercept form.
Graphing a linear function in the xy plane.
Finding the equation of the function:
Performing calculations to find the slope m and the intercept b.
Solving an equation with one variable.
Solving an equation with two variables.

The graph of a linear function is a line in the xy plane. We can graph linear equation by plugging different numbers into the equation. Note that you need to plug x=0 x in order to find the intercept with the y axis represented as m.

How can we draw a linear graph? 
In order to draw the graph, we need to mark 2 points and draw a line between them. To make sure you have found the right points it is best to mark a third point. If there is no mistake all the points will be on the line.
Which points to choose? Mark the points of intersection with the x-axis (y=0) and intersection with the y-axis (x=0) and then select a third point that is located between these two points.

Finding the equation of a linear function given the graph of the function:

Sometimes no values ​​are given and instead we see the graph of the function. How can we write the equation of the function using only it’s graph?

Step 1: Finding two points values:
For making an easy and accurate calculation we need to look for 2 points whose (x,y) values are not a fraction.

Step 2: Calculating the slope m value by plugging the points we found from the graph into the slope formula m=(Ya-Yb)/(Xa-Xb).

3. Step 3: Finding the function formula by plugging one of the points and the slope m we calculated into the function formula y=mx+b.

Consider this example:

What is the equation of the function in the graph?

linear function

Step 1: Finding two points values:
We can see that the y intercept has the value of 14 so we know one point (0,14). This point is also the intercept b from the line formula y=mx+b.
The x intercept is between 3 and 4 therefore we don’t know its exact value and we can’t take it as a point.
We can easily see the point (2,6) on the graph so we can use it.

Step 2: Calculating the slope m value by plugging the points (2,6) and (0,14) into the slope formula:
m=(Ya-Yb)/(Xa-Xb)=(6-14)/(2-0)=-8/2=-4

3. Step 3: Finding the function formula by plugging one of the points (2,6) or (0,14) and the slope m=-4 into the function formula y=mx+b:
6=-4*2+n
n=14

14=n+-4*0
n=14

Relationships between two linear functions SAT questions

In these questions you need to find the slope of a linear function given the slope of another function or to find the equation of a linear function given an equation of another function.

The skills required:
Understanding the structure of a linear function.
Writing a linear function in a slope intercept form.
Finding the slope of one line by using the slope of a parallel line.
Finding the slope of one line by using the slope of a perpendicular line.
Finding the equation of the function:
Performing calculations to find the intercept b
Performing calculations to find the x value or y value of a point.
Solving an equation with one variable.

Finding the slope of a linear function given the slope of another function:

Parallel lines slope formula:
m1=m2

For example: if m1=-2 then m2=-2

Perpendicular lines slope formula:
m1*m2=-1

For example: if m1=-2 then m2=1/2

There are two cases when we can find the slope of one line by using the slope of another line: the lines are parallel or the lines are perpendicular.

Parallel lines have the same slope m, so if we know the slope of one line then the slope of the parallel line will be the same.

Perpendicular lines slopes form 90° angles at their cutting point.
The product of the slopes of vertical lines is equal to minus 1 meaning that

m1*m2=-1  (m1 is the slope of the first line and m2 is the slope of the second line).

Consider this example:

The equation of function a is 3y+9x-36=0. Function b is parallel to function a and function c is perpendicular to function a.

What are the slopes of the functions a ,b and c? 

Step 1: We will write the function a in the form y=mx+b:
3y+9x-36=0
3y=-9x+36
y=-3x+12

Step 2: Finding the slope of the parallel function b:
The slope of function b equals to the slope of function a, therefore the slope of function b is -3.

Step 3: Finding the slope of the perpendicular function c:
We need to plug the slope of function a into the formula m1*m2=-1.
m1=-3
-3*m2=-1
m2=-1/-3
m2=1/3
The slope of function c is 1/3.

The graph shown below presents the functions.

parallel and perpendicular functions

Finding the equation of a linear function given an equation of another function:

The equation of function a is y=-3x+12. Function b is perpendicular to function a and intersects it in a point x=4 so that the angle between the functions is equal to 90 degrees.

Write the equation of function b.

Step 1: Finding the slope of function b: 
From the equation of function a we know that its slope is m=-3.
We are given that function b is perpendicular to function a, therefore we need to plug the slope of function a into the formula m1*m2=-1.
-3*m2=-1
m2=-1/-3
m2=1/3
The slope of function b is 1/3.

Step 2: Finding the y value of the intersection point: 
We know that the functions intersect at x=3. We can plug x=3 into function a and find the value of y.
y=-3x+12
y=-3*4+12
y=0

The intersection point is (-3,0).

Step 3: Finding the y-intercept b of function b:
We know the value of a point on function b and the value its slope, therefore we can find the value its intercept b.
y=mx+b
0=13*4+b  
b=-4/3

The formula of function b is y=13x-43

Step 4: Checking the answer: 
Multiply the slopes to see if the result is -1.
(13)*(-3)=(1*-3)/3=-3/3=-1

Plug the point (4,0) into each equation to see if you get y=0.
y=-3x+12=-3*4+12=0
y=13x+43=(13)*4-43=4343=0

Inequalities on the SAT Test

Linear Inequalities on the SAT Test

SAT Subscore: Heart of Algebra

Linear inequalities on SAT test

Linear inequality is linear equation in which the equal sign is replaced by one of the symbols of inequality. The solution of a linear equation is a range of values, rather than one specific value.

A system of linear inequalities is similar to a system of linear equations, the difference is in the sign. We use the system of inequalities when we a have a word problem with number of constrains instead of one (we write an inequality for each constrain). 

On the SAT test linear inequalities are part of heart of algebra questions.

Let’s look at the different types of questions and examples for each one of them.

Creating and solving a linear inequality SAT questions

In these questions you need to create a linear inequality from a word problem and solve it.

The skills required:
Translating a word problem into an inequality.
Defining x variable.
Solving inequality with one variable.

Statements representing the inequalities in the context of the question

First of all, you need to recognize that the question is about inequality, look for the statements representing the inequalities in the context of the question:
Statements for greater than sign ≥: at least, more than, no less than, greater than, minimum value.
Statements for less than sign ≤: less than, at most, no more than, smaller than, maximum value, the greatest value.

Rules for solving a linear inequality

The direction of the inequality sign:
 When we divide by a negative number, we must change the inequality sign direction. This happens when the x variable sign is negative, therefore we need to divide by a negative number to isolate x.

Range of solutions instead of one:
In contrast to a linear equation where we can find a single solution there is a range of solutions for a linear inequality. If the solution is x>5 the value of the solution can be 6, 7,8……

Rounding the answer:
If the answer is a decimal value and it needs to be a whole number, you need to round it.
Round up example: x≥4.5, when x represents the minimum number of students in the class. The answer is x≥5.
Round down example: x≤38.6, when x represents the maximum number of students in the bus. The answer is x≤38.

Solving a linear inequality without changing the direction of the inequality sign

Solving linear inequalities is similar to solving of linear equations except for treating the inequality sign. When we divide by a negative number, we must change the inequality sign direction.

Consider this example:

In the previous working day, the machine produced 152 products. Every hour the machine produces 10 more products. Twenty percent of the products are disqualified (turning out to be defective) after a quality control check. The minimum quantity of products for delivery is 170. If the current working day starts at 9 am, at what time will it be possible to prepare the shipment?

Since the question is about a minimum quantity, we will need to construct an inequality in order to reach a solution. The quantity should be at least 170 so we will write an expression that should be greater than 170 (instead of equal to 170)

Step 1: X definition: We know the initial quantity of the products and their rate of production. The question is how soon will we reach 170 products, therefore the variable x should be the time (we define x as the number of hours).

Step 2: Writing the expressions in each side of the inequality: The expression on one side of the inequality will be the products in stock from yesterday plus the products that the machine produces today and on the other side of the inequality we will write 170.
152+10*(1-0.2)*x>=170

Step 3: Solving inequality:
152+10*0.8x≥170
152+8x≥170
8x≥18
x≥2+2⁄8
x≥2+1⁄4 
x≥21⁄4

Since the working day starts at 9 o’clock and it takes at least 2.25 hours to reach 170 products, the hour when it will be possible to prepare the delivery will be 11 o’clock plus 0.25 hours. That is 60 minutes double 0.25 (quarter) = 15 minutes. The answer is that the time when it’s possible to prepare the shipment is after 11:15.

Step 4: checking the answer:
In order to test the answer we will solve the left side of the inequality: 152+10*0.8*2.25= 152+8*2.25= 152+16+8*0.25=152+16+2=170

Solving a linear inequality that requires a change in the direction of the inequality sign

Pay attention that when we divide by a negative number, we must change the inequality sign direction!

Consider this example:

Solve the inequality -4x+10<18

Step 1: Solving the inequality:
-4x+10<18
-4x<8-18-10
-4x<8
Now we need to divide both sides by a negative number -4 so we will change the sign direction:
-4x/-4>8/-4
x>-2

Step 2: Checking the answer
We can place any number that is bigger than -2 and see if the inequality holds. Let’s take x=0.
-4*0+10<18
10<18 is a true statement

Note that if we don’t change the direction of the sign the answer x<-2 is incorrect.
For x=-3 for example we will get
-4*-3+10<18
22<18 is a false statement.

Finding values that don't satisfy an inequality SAT questions

In these questions you need to find values that are not a solution for a linear inequality.

The skills required:
Solving inequality with one variable.
Identifying values that are not included in the solution of the inequality.

If we are asked to find values that are not a solution for the inequality, we first need to solve the inequality. All the values that are not included in the solution are values that don’t satisfy the inequality.

Consider this example:

Which values are not a solution to the inequality -4x+10<18?

We solved this inequality in the previous example so we know that the solution is x>-2.
Therefore the values that are not a solution are x≤-2.

Checking the solution:
Let’s plug -2 and a number that is smaller than -2 like -3 into the inequality -4x+10<18

For x=-2:
-4*-2+10<18
8+10<18
18<18 is a false statement

For x=-3:
-4*-3+10<18
12+10<18
22<18 is a false statement

Graphic presentation of a linear inequality SAT questions

In these questions you need to identify a graphic presentation of an inequality. The answers to the inequality are represented by the area below or above the line of its equation. 

The skills required:
Writing an inequality in a slope intercept form.
Graphing the line of the inequality in the xy plane.
Marking the area that corresponds to the inequality sign.

Steps for graphing an inequality


Step 1: Write the inequality in a slope intercept form y>mx+b or y<mx+b.
Step 2: Draw the graph of the inequality as a straight line (as if the sign is =).
Step 3: Shade the range of inequality by its sign:
If y<mx+b shade the area below the graph.
If y>mx+b shade the area above the graph.

Consider this example:

Graph the inequality y+4x>14.

Step 1: Writing the inequality in a slope intercept form y=mx+b:
y+4x>14
y>-4x+14

Step 2: Drawing the equality graph: the graph is represented by the line y=-4x+14 in orange.

Step 3: Shading the area above the graph:
For every x, instead of marking the point where y is equal to -4x+14, we need to mark all the points for which y is bigger than -4x+14. That will result in a yellow area above the graph.

Note that the region y<-4x+14 is the white area below the graph.

linear inequality area

Creating a system of linear inequalities SAT questions

In these questions you need to create a system of  linear inequalities from a word problem.

The skills required:
Defining variables.
Finding the statements representing the inequalities in the context of the question.
Writing two inequalities with two variables.

Statements representing the inequalities in the context of the question

First of all, you need to recognize that the question is about inequality, look for the statements representing the inequalities in the context of the question:
Statements for greater than sign ≥: at least, more than, no less than, greater than, minimum value.
Statements for less than sign ≤: less than, at most, no more than, smaller than, maximum value, the greatest value.

Consider this example:

On the first day the tourists rode the bus at a speed of x  km per hour. On the second day they walked at a speed of y km per hour. In total the tourists traveled z km and walked k km. The tour guide wants to cover a total distance of at most 100 km and the time she wants to walk and ride the bus is at most 15 hours. Write a system of inequalities that represents her plan.

Step 1: writing the first inequality: The tour guide wants to cover a total distance of at most 100 km and we know that the tourists traveled z km and walked k km.
Therefore the first inequality is simple: z+x≤100.

Step 2: writing the second inequality: The time that the guide wants to walk and ride the bus is at most 15 hours so we need to write an equation from riding time plus walking time.
Riding time equals to the riding distance z km divided by riding speed x km.
Walking time equals to the walking distance k km divided by walking speed y km.
Therefore the second equation will be z/x+k/y≤15

Finding possible solutions for a system of linear inequalities SAT questions

In these questions you need to decide which values satisfy a given system of linear inequalities. Sometimes you will need to translate a word problem into a system of inequalities as explained before.

The skills required:
Finding the values that meet the constrains of both inequalities by plugging possible solutions into each inequality.

Consider this example:

Daniel needs to send gifts to ten customers. Each customer should receive at least one gift. The price of a book is $7 and the price of a chocolate box of is $5. The shipping cost is $ 1 per gift. Daniel’s budget is $80. Offer 3 options for gift combinations.

Step 1: Variables definition– Define x as the books amount and y as the chocolate boxes amount.

Step 2: Writing the inequalities:
x+y≥10
(7+1)*x+(5+1)*y≤80

Step 3: Writing possible solutions: Since there is a budget limit in this question, we should try to choose first the smallest x and y. We will start with a x=1 and choose the smallest value of y so that the first condition x + Y> 10 will hold in equality y=9. We will plug these x and y values in the second condition and check whether the second condition is met.
1+9≥10
8*1+6*9=62, 62≤80
We see that the conditions are met, therefore x=1, y=9 is a possible solution for the inequalities system.
To get more answers we can continue to increase x and y values as shown in the table below. Once the second condition is not met (the budget gets above 80) we continue to examine options where x=2.

inequality values

Solving a system of linear inequalities SAT questions

In these questions you need to turn a system of linear inequalities to one inequality that can be solved easily. This can be achieved using additional data or by turning one of the inequalities to an equation.
You may also need to translate a word problem to a system of inequalities.

The skills required:
Plugging a given value into an inequality.
Finding x or y maximum or minimum value.
Solving inequality with one variable

Solving a system of linear inequalities using additional data

If we receive x or y value we have only 1 variable so we can find a solution. Since the inequalities remain, we may get a number of solutions.

Steps for solving the system using an additional data:
Step 1: Plugging the additional data into each one of the inequalities.
Step 2: Finding the solution by combining the answers from two inequalities.

Consider this example:

 Daniel needs to send gifts to ten customers. Each customer should receive at least one gift. The price of a book is $7 and the price of a chocolate box of is $5. The shipping cost is $1 per gift. Daniel’s budget is $80. 

If Daniel buys 5 books, what is the possible number of chocolate boxes that he can buy?

Step 1: Plugging the additional data into each one of the inequalities:

We need to substitute x=5 into the first inequality and solve it:
x+y≥10
5+y≥10
y≥5

We can substitute x=5 into the second inequality and solve it:
8x+6y≤80
8*5+6y≤80
6y≤40
y≤40/6
y≤6 4/6
y≤6 2/3
The answer is y<6 since y can’t be a fraction.

Step 2: Combining the answers from two inequalities:
The answer is x=5, 5≤y≤6 meaning we have 2 answers (5,5) or (5,6).

Step 3: Checking the answers:
We can check the answer (5,5): 5+5≥10, 8*5+6*5=70 ≤80.
We can check the answer (5,6): 5+6≥10, 8*5+6*6=76 ≤80.
We can check a bigger answer (5,7) to see that it doesn’t satisfy the second condition: 5+7≥10, 8*5+6*7=82 ≥80.

Finding maximum or minimum value of a variable

There are questions in which you can be asked about x or y maximum or minimum value.

Consider this example:

Daniel needs to send gifts to ten customers. Each customer should receive at least one gift. The price of a book is $ 7 and the price of a chocolate box of is $ 5. The shipping cost is $ 1 per gift. Daniel’s budget is $ 80.

What is the maximum number of books that Daniel can buy?

In this case we need to turn one of the inequalities to an equality and isolate y by writing it in a form of y= . Then we need to plug the value of y into the inequality and solve it for x that we are asked about.

Note that we are asked about the value of x so we need to leave the x variable and isolate y variable.

Also note that we are also limited by the budget therefore we need to solve the second inequality with the x variable and turn the first inequality x+y≥10 to y=10-x.

Step 1: Turning the first inequality to an equation and finding y=:
x+y≥10
y=10-x

Step 2: Substituting 10-x for y in the second inequality:
8x+6y≤80
8x+6(10-x)≤80
8x+60-6x≤80
2x≤20
x≤10

The maximum number of books is 10.

Step 3: Checking the answer:
Plugging x=10 in the second inequality that was limited will give us 80+6y≤80. For y=0 this inequality exists as 80≤80.
Plugging x=11 which is not in the answer range will give us 88+6y≤80. Even for y=0 this inequality doesn’t exist because 88≤80 is false statement.
As we see the answer is x≤10.

Graphic presentation of a system of linear inequalities SAT questions

In these questions you need to identify a graphic presentation of a system of inequalities given its formulas or to write the formulas of the system that correspond to a given graphic presentation.
The answer to the system of the inequalities is represented by the area below or above the lines of both equations.

The skills required:
Writing an inequality in a slope intercept form.
Graphing the line of the inequality in the xy plane.
Finding areas that correspond to the inequality signs of each equation and both equations.
Finding the formulas from the graphic presentation of an inequality system.

Graphic presentation of a system of linear inequalities that has a solution

Drawing the system of the inequalities will make it easy to see their possible solutions.

Steps for graphing a system of inequalities:
Step 1: Write each inequality in a slope intercept form y<mx+b or y>mx+b
Step 2: Draw the graphs of the inequalities as straight lines (as if the signs are =).
Step 3: Shade the ranges of the inequalities by their signs:
If y<mx+b shade the area below the graph.
If y>mx+b shade the area above the graph.
Step 4: The solution is only the area that was shaded twice, because both of the conditions must be met.

 

Consider this example:

Graph the solution of the system of the linear inequalities
8<4x+y
6>-2x+y

Step 1: Writing each inequality in a slope intercept form:
8<4x+y
y>-4x+8

6>-2x+y
y<2x+6

Step 2: Drawing the graphs of the inequalities as straight lines: The graphs are in blue and orange.

Step 3: Shading the ranges of the inequalities by their signs:
For y>-4x+8 we shade the area above the graph, this area is marked in yellow.
For y<2x+6 we shade the area below the graph, this area is marked in grew.

Step 4: Finding the solution: The solution is only the area that was shaded twice, because both of the conditions must be met. The area is marked in dark grew.

graphing two linear inequalities

Graphic presentation of a system of linear inequalities with no solution

If there is no area that is shaded twice then the system of the inequalities that has no solution.

Consider this example:

Graph the solution of the system of the linear inequalities
y>-2⁄3x+5⁄3
y<-2⁄3x+7⁄9

Step 2: Drawing the graphs of the inequalities as straight lines: The graphs are in blue and orange.

Step 3: Shading the ranges of the inequalities by their signs:
For y>-2⁄3x+5⁄3 we shade the area above the graph, this area is marked in gray.
For y<-2⁄3x+7⁄9 we shade the area below the graph, this area is marked in yellow.

Step 4: Finding the solution: The solution is only the area that was shaded twice, because both of the conditions must be met. The is no such area, therefore there is no solution for the inequalities system.  

graphs of linear inequalities no solution

Note that you can also understand why this system of equations has no solution by testing the equations. Plug x=1 in the equations, according to the first equation y should be greater than 1 and according to the second equation y should be less than 1/9. Those cannot exist at the same time.

Writing inequalities represented by a given graphs of linear equations system

In these questions the graphs of the inequality system are given and you need to find the formulas. 

Consider this example:

Write the inequalities represented by the green area below for the system of the linear equations
y=3x-3
y=-2x+5

system of linear inequalities

The equation y=3x-3 is represented by the blue graph. We can see that because of its positive slope (m=3) and  its y intercept (for x=0 y=-3). Therefore, we know that the shaded area is above the graph, this means that the inequality should be y>3x-3.

The equation y=-2x+5 is represented by the orange graph. We can see that because of its negative  slope (m=-2) and its y intercept (for x=0 y=5). Therefore, we know that the shaded area is under the graph, this means that the inequality should be y<-2x+5.