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# 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

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

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

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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Continue studying heart of algebra subscore with linear inequalities topic.