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# Linear equations on the SAT test

## Studying linear equations

On the SAT test linear equations are part of heart of algebra subscore that includes 4 fundamental topics that appear in many SAT questions. The 4 topics are:

Linear functions– Start studying heart of algebra subscore with linear functions topic.

Linear equations– The current topic.

Systems of linear equations and linear inequalities– Continue to these last 2 topics after you study linear equations topic.

Linear equations topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

### Linear equations- summary

A 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. Since this is a fundamental topic it appears in many SAT questions.

Calculating an output of an expression requires calculating the value of an expression. In some questions we are given a word problem and in other questions we are given an equation that we must use it order to solve the expression.

Solving linear equations- In these questions we are given an equation and we are asked to solve it. Some questions may include fractions or absolute values.

Linear equations that don’t have one solution- In these questions we are asked to solve an equation that has no solution or an infinite number of solutions.

Creating and solving an equation from a word problem- In these questions we are given a word problem from which we need to write an equation and solve it.

The formula sheet for linear equations is listed below.

### Linear equations formula sheet

Linear equations formula sheet is given below, it includes 3 types of basic formulas:

Calculations with negative numbers formulas– these are 5 very basic formulas that include multiplying and dividing positive and negative numbers.

Distributing and combining like terms formulas-one formula for opening parentheses (distributing) and one formula for creating them (combining).

Calculation with fraction formulas– includes three basic formulas for addition/ subtraction, multiplication and division of fractions.

Note that all the formulas are explained in detail on this page.

### Calculating an output of an expression

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

The skills required are:
Translating a word problem into a mathematical expression.
Solving an 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 an 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 we need to divide or multiple, we need to do it for each component of the expression. In this case we 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

In these questions we are given an equation and we 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 we 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. We can add or subtract the same value from each side of the equation.
2. We 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).

Step 1: 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

Step 2: Isolating the variable term on one side of the equation by using addition or subtraction.

Step 3: Solving the equation (finding x=) by using multiplication or division.

Step 4: Checking the answer by plugging the answer from the previous step into the equation.

Consider this example:

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

What is the solution to the given equation?

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

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

Step 3: Dividing each side of the equation by 4
4x:4=100:4
x=25

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?

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

(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

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 we can first reach a common denominator in order to cancel the denominators and then we 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?

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

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?

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

|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 we are given a word problem and an equation with two variables. We need to find the value of one variable inside the word problem and then plug it into the equation. After that we will have an equation with one variable that we can solve.

Consider this example:

0.6x+0.4y=86

The equation above represents a weighted average of 86  in two subjects, x is defined as a grade in course 1 and y is defined as a grade in course 2.

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.

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

Solving the equation with 1 variable:
54+0.4y=86
0.4y=32
y=80

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?

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.

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

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

The skills required are defining the 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).

The 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.

You just finished studying linear equations topic!

Continue studying heart of algebra subscore with systems of linear equations topic.

# Systems of Linear Equations on the SAT Test

### Studying systems of linear equations

On the SAT test systems of linear equations are part of heart of algebra subscore that includes 4 fundamental topics that appear in many SAT questions. The 4 topics are:

Linear functions and linear equations- Start studying heart of algebra subscore with linear functions topic and continue to linear equations topic.

Systems of linear equations– The current topic.

Linear inequalities- Continue to this last topic after you study systems of linear equations topic.

Systems of linear equations topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

### Systems of linear equations- summary

A 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.

Creating a system of linear equations- In these questions we need to create a system of linear equations from a word problem.

Solving a system of linear equations- In these questions we need to solve a system of linear equations. In order to solve the equations, we need to reduce 2 equations with 2 variables to 1 equation with 1 variable. This should be done with either substitution or elimination.

Determining the number of solutions for a system of linear equations- In these questions we are asked to determine if the system of equations has one solution, no solution and an infinite number of solutions.

Graphic presentation of a system of linear equations- In these questions we need to identify a graph of a given system of equations.

### Creating a system of linear equations

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

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

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

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

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

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

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:

#### 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:

#### 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.

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.

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