# Systems of Equations – Examples with Answers

Systems of equations are systems that have two or more equations and two or more unknowns. There are several different methods for solving these systems of equations. In this case, we will focus on two methods, the elimination method and the substitution method. Specifically, we will look at systems of two equations with two unknowns.

We will start by exploring a brief summary of how to solve these systems of equations. Then, we will look at some examples with answers.

##### ALGEBRA

Relevant for

Learning to solve systems of two equations with two unknowns.

See examples

##### ALGEBRA

Relevant for

Learning to solve systems of two equations with two unknowns.

See examples

## Summary of systems of two equations with two unknowns

We can solve systems of two equations by using three main methods: the elimination method, the substitution method, and the graphical method. Here, we will focus on the elimination method and the substitution method.

### Solving systems of equations with the substitution method

We can follow the following steps to solve the system by substitution:

Step 1: Simplify the equations: This includes removing parentheses, combining like terms, and removing fractions.

Step 2: Solve an equation for one variable. It does not matter which equation or variable we choose.

Step 3: Substitute the expression obtained in step 2 into the other equation. This will result in a single equation with one variable.

Step 4: Solve the equation obtained in step 3.

Step 5: Substitute the value from step 4 into any of the other equations and solve for the other unknown.

### Solving systems of equations with the elimination method

We use the following steps to solve the system of equations by elimination:

Step 1: Simplify the equations and put them in the form Ax+By=C.

Step 2: Multiply one or both equations by some number so that we get opposite coefficients either for x or for y. We need to eliminate one of the variables when adding the equations. Therefore, we have to get one coefficient to be a and the other –a.

Step 3: Add the equations. By doing this we will eliminate a variable and we will have an equation with one unknown.

Step 4: Solve the equation from step 3 for the remaining variable.

Step 5: Substitute the value from step 4 into any equation and solve for the second variable.

## Systems of equations – Examples with answers

The following examples of systems of equations can be used to fully understand the equation-solving process detailed above. Each of these examples has its respective solution using the method indicated in the question.

### EXAMPLE 1

Solve the system of equations using the substitution method: $latex \left\{ {\begin{array}{*{20}{c}} {x+2y=10} \\ {2x-y=5} \end{array}} \right.$

Step 1: We have nothing to simplify.

Step 2: We can solve the first equation for x:

$latex x+2y=10$

$latex x=10-2y$

Step 3: We substitute the expression $latex x=10-2y$ in the second equation:

$latex 2x-y=5$

$latex 2(10-2y)-y=5$

$latex 20-4y-y=5$

Step 4: Solve for y:

$latex 20-4y-y=5$

$latex -5y=-15$

$latex y=3$

Step 5: We substitute $latex y=3$ in the first equation:

$latex x+2y=10$

$latex x+2(3)=10$

$latex x=4$

### EXAMPLE 2

Solve the system of equations using the elimination method: $latex \left\{ {\begin{array}{*{20}{c}} {x-y=3} \\ {2x+y=12} \end{array}} \right.$

Step 1: We have nothing to simplify and both equations are already in the form Ax+By=C.

Step 2: We already have opposite coefficients in the variable y.

Step 3: We add the equations:

$latex x-y=3$

$latex + \hspace{1cm} 2x+y=12$

___________________

$latex 3x=15$

Step 4: We solve for x:

$latex 3x=15$

$latex x=5$

Step 5: We substitute $latex x=5$ in the second equation:

$latex 2x+y=12$

$latex 2(5)+y=12$

$latex 10+y=12$

$latex y=2$

### EXAMPLE 3

Solve the following using the substitution method: $latex \left\{ {\begin{array}{*{20}{c}} {-2x-y=1} \\ {3x+4y=6} \end{array}} \right.$

Step 1: We have nothing to simplify.

Step 2: We solve the first equation for y:

$latex -2x-y=1$

$latex -y=1+2x$

$latex y=-1-2x$

Step 3: We substitute the expression $latex y=-1-2x$ in the second equation:

$latex 3x+4y=6$

$latex 3x+4(-1-2x)=6$

$latex 3x-4-8x=6$

Step 4: Solve for x:

$latex 3x-4-8x=6$

$latex -5x=10$

$latex x=-2$

Step 5: We substitute $latex x=-2$ in the first equation:

$latex -2x-y=1$

$latex -2(-2)-y=1$

$latex 4-y=1$

$latex -y=-3$

$latex y=3$

### EXAMPLE 4

Solve the system of equations using the elimination method: $latex \left\{ {\begin{array}{*{20}{c}} {y=2x+7} \\ {-6x-2y=-4} \end{array}} \right.$

Step 1: We write the equations in the form Ax+By=C:

$latex \left\{ {\begin{array}{*{20}{c}} {-2x+y=7} \\ {-6x-2y=-4} \end{array}} \right.$

Step 2: We multiply the first equation by 2 to obtain opposite coefficients in y:

$latex \left\{ {\begin{array}{*{20}{c}} {-4x+2y=14} \\ {-6x-2y=-4} \end{array}} \right.$

Step 3: We add the equations:

$latex -4x+2y=14$

$latex + \hspace{1cm} -6x-2y=-4$

___________________

$latex -10x=10$

Step 4: We resolve for x:

$latex -10x=10$

$latex x=-1$

Step 5: We substitute $latex x=-1$ in the first equation:

$latex y=2x+7$

$latex y=2(-1)+7$

$latex y=5$

### EXAMPLE 5

Solve the system of equations using the substitution method: $latex \left\{ {\begin{array}{*{20}{c}} {2(2x-4)+y=3} \\ {-x+2y=4} \end{array}} \right.$

Step 1: We simplify the first equation:

$latex \left\{ {\begin{array}{*{20}{c}} {4x-8+y=3} \\ {-x+2y=4} \end{array}} \right.$

Step 2: We solve the first equation for y:

$latex 4x-8+y=3$

$latex y=-4x+11$

Step 3: We substitute the expression $latex y=-4x+11$ in the second equation:

$latex -x+2y=4$

$latex -x+2(-4x+11)=4$

$latex -x-8x+22=4$

Step 4: We solve for x:

$latex -x-8x+22=4$

$latex -9x=-18$

$latex x=2$

Step 5: We substitute $latex x=2$ in the second equation:

$latex -x+2y=4$

$latex -2+2y=4$

$latex 2y=6$

$latex y=3$

### EXAMPLE 6

Solve the following using the elimination method: $latex \left\{ {\begin{array}{*{20}{c}} {2x=3y-14} \\ {2y=x+8} \end{array}} \right.$

Step 1: We write the equations in the form Ax+By=C:

$latex \left\{ {\begin{array}{*{20}{c}} {2x-3y=-14} \\ {-x+2y=8} \end{array}} \right.$

Step 2: We multiply the second equation by 2 to obtain opposite coefficients in the x:

$latex \left\{ {\begin{array}{*{20}{c}} {2x-3y=-14} \\ {-2x+4y=16} \end{array}} \right.$

Step 3: We add the equations:

$latex 2x-3y=-14$

$latex + \hspace{1cm} -2x+4y=16$

___________________

$latex y=2$

Step 4: We have already obtained the value of y:

$latex y=2$

Step 5: We substitute $latex y=2$ in the first equation:

$latex 2x=3y-14$

$latex 2x=3(2)-14$

$latex 2x=-8$

$latex x=-4$

## Systems of equations – Practice problems

Practice using the substitution method and the elimination method with the following problems. Select an answer and verify that you selected the correct one. You can use the solved examples above to follow the process step by step.

#### Solve with the substitution method: $latex \left\{ {\begin{array}{*{20}{c}} {-2x+3y=7} \\ {3x-y=7} \end{array}} \right.$  