Systems of equations are systems of two or more equations with two or more variables. To solve a system of equations, we have to solve the equations simultaneously. Two common methods that we can use to solve systems of two equations with two unknowns are the substitution method and the elimination method.

Here, we will look at a brief summary of how to solve systems of equations. In addition, we will solve some practice problems using the substitution and elimination methods.

## Methods for solving systems of equations

There are several methods we can use to solve systems of equations. However, the most commonly used methods when dealing with 2×2 systems of equations are the substitution method and the elimination method.

### Substitution method

To solve 2×2 systems of equations by substitution, we follow these steps:

**Step 1:** Remove parentheses, combine like terms, and eliminate fractions to simplify the equations.

**Step 2:** Solve any of the equations for any of the variables.

**Step 3:** Substitute the equation obtained in step 2 into the other equation. Thus, we will obtain an equation with a single variable.

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

**Step 5:** Use the value of the variable obtained in step 4 to find the value of the remaining variable.

### Elimination method

To solve 2×2 systems of equations by elimination, we follow these steps:

**Step 1:** Remove parentheses, eliminate fractions, and combine like terms to simplify fractions.

**Step 2:** Write the equations in the form A*x*+B*y*=C.

**Step 3:** Multiply one or both equations by some number, so that we obtain opposite coefficients in one of the variables.

One of the variables has to be eliminated when we add the equations.

**Step 4:** Add the equations. We will obtain a single equation with a single variable.

**Step 5:** Solve the equation from step 4 for the remaining variable.

**Step 6:** Use the value from step 5 in any of the equations to get the value of the second variable.

## 10 system of equations examples with answers

The following examples are solved using the substitution and elimination methods of 2×2 systems of equations. Try to solve the problems yourself before looking at the solution.

**EXAMPLE **1

Solve the system of equations: $latex \begin{cases}x+2y=10 \\ 2x-y=5 \end{cases}$

##### Solution

We are going to solve this system using the substitution method. Thus, we start by solving the first equation for *x*:

$latex x+2y=10$

$latex x=10-2y$

Now, we substitute $latex x=10-2y$ into the second equation, and we have:

$latex 2x-y=5$

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

$latex 20-4y-y=5$

Solving the equation for *y*, we have:

$latex 20-4y-y=5$

$latex -5y=-15$

$latex y=3$

Using the value *y*=3 in the first equation, we have:

$latex x+2y=10$

$latex x+2(3)=10$

$latex x=4$

The solution to the system is $latex x=4,~~y=3$.

**EXAMPLE **2

**EXAMPLE**

Find the solution to the system: $latex \begin{cases}x-y=3 \\ 2x+y=12 \end{cases}$

##### Solution

We are going to solve this system using the elimination method. Thus, we can see that we have opposite coefficients on the variable *y*.

Adding the equations, we have:

$latex x-y=3$

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

___________________

$latex 3x=15$

Solving the equation, we have:

$latex 3x=15$

$latex x=5$

Using the value *x*=5 in the second equation, we have:

$latex 2x+y=12$

$latex 2(5)+y=12$

$latex 10+y=12$

$latex y=2$

The solution to the system of equations is $latex x=5,~~y=2$.

**EXAMPLE **3

**EXAMPLE**

Solve the following system of equations: $latex \begin{cases}-2x-y=1 \\ 3x+4y=6 \end{cases}$

##### Solution

We will solve this by substitution. Thus, we solve the first equation for *y*:

$latex -2x-y=1$

$latex -y=1+2x$

$latex y=-1-2x$

Using the expression $latex y=-1-2x$ in the second equation, we have:

$latex 3x+4y=6$

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

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

Solving the equation, we have:

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

$latex -5x=10$

$latex x=-2$

Using the value *x*=-2 in the first equation, we have:

$latex -2x-y=1$

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

$latex 4-y=1$

$latex -y=-3$

$latex y=3$

The solution to the system is $latex x=-2, ~~y=3$.

**EXAMPLE **4

**EXAMPLE**

Find the solution to the system of equations: $latex \begin{cases}y=2x+7 \\ -6x-2y=-4 \end{cases}$

##### Solution

We will solve this by elimination. Thus, we start by dividing the second equation by 2 to simplify it.

$latex \begin{cases}y=2x+7 \\ 3x+y=2 \end{cases}$

Now, we write both equations in the form A*x*+B*y*=C:

$latex \begin{cases}-2x+y=7 \\ 3x+y=2 \end{cases}$

We can multiply the first equation by -1 to get opposite coefficients on *y*:

$latex \begin{cases}2x-y=-7\\ 3x+y=2 \end{cases}$

Adding the equations, we have:

$latex 2x-y=-7$

$latex + \hspace{1cm} 3x+y=2$

___________________

$latex 5x=-5$

Solving the equation, we have:

$latex 5x=-5$

$latex x=-1$

Using the value *x*=-1 in the first equation, we have:

$latex y=2x+7$

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

$latex y=5$

The solution is $latex x=-1, ~~y=5$.

**EXAMPLE **5

**EXAMPLE**

Solve the system of equations: $latex \begin{cases}2(2x-4)+y=3 \\ -x+2y=4 \end{cases}$

##### Solution

Simplifying the first equation, we have:

$latex \begin{cases}4x-8+y=3 \\ -x+2y=4 \end{cases}$

Now, let’s solve by substitution. Therefore, solving the first equation for *y*, we have:

$latex 4x-8+y=3$

$latex y=-4x+11$

Using $latex y=-4x+11$ in the second equation, we have:

$latex -x+2y=4$

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

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

Solving the equation, we have:

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

$latex -9x=-18$

$latex x=2$

Using the value *x*=2 in the second equation, we have:

$latex -x+2y=4$

$latex -2+2y=4$

$latex 2y=6$

$latex y=3$

The solution to the system is $latex x=2,~~y=3$.

**EXAMPLE **6

**EXAMPLE**

Solve the system of equations: $latex \begin{cases}2x=3y-14 \\ 2y=x+8 \end{cases}$

##### Solution

We will solve this by elimination. Therefore, we write the equations in the form A*x*+B*y*=C:

$latex \begin{cases}2x-3y=-14 \\ -x+2y=8 \end{cases}$

We can multiply the second equation by 2 to get opposite coefficients in *x*:

$latex \begin{cases}2x-3y=-14 \\ -2x+4y=16 \end{cases}$

Adding the equations, we have:

$latex 2x-3y=-14$

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

___________________

$latex y=2$

Using the value *y*=2 in the first equation, we have:

$latex 2x=3y-14$

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

$latex 2x=-8$

$latex x=-4$

The solution is $latex x=-4,~~y=2$.

**EXAMPLE **7

**EXAMPLE**

Solve the system of equations: $latex \begin{cases}2x-3y=7 \\ 2x+3y=1 \end{cases}$

##### Solution

We will solve this by substitution. Therefore, solving the first equation for *x*, we have:

$latex 2x-3y=7$

$latex 2x=3y+7$

$latex x=\frac{3y+7}{2}$

Using the expression $latex x=\frac{3y+7}{2}$ in the second equation, we have:

$latex 2x+3y=1$

$latex 2\left(\frac{3y+7}{2}\right)+3y=1$

$latex 3y+7+3y=1$

Solving the equation for *y*, we have:

$latex 3y+7+3y=1$

$latex 6y=-6$

$latex y=-1$

Using the value *y*=-1 in the second equation, we have:

$latex 2x+3y=1$

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

$latex 2x-3=1$

$latex 2x=4$

$latex x=2$

The solution to the system is $latex x=2,~~y=-1$.

**EXAMPLE **8

**EXAMPLE**

Solve the system of equations: $latex \begin{cases}2x-7y=1 \\ 2x+3y=11 \end{cases}$

##### Solution

We will solve this by elimination. Therefore, we multiply the second equation by -1 to get opposite coefficients in *x*:

$latex \begin{cases}2x-7y=1 \\ -2x-3y=-11 \end{cases}$

Adding the equations, we have:

$latex 2x-7y=1$

$latex + \hspace{1cm} -2x-3y=-11$

___________________

$latex -10y=-10$

Solving for *y*, we have:

$latex y=1$

Using the value *y*=1 in the first equation, we have:

$latex 2x-7y=1$

$latex 2x-7(1)=1$

$latex 2x=8$

$latex x=4$

The solution is $latex x=4,~~y=1$.

**EXAMPLE **9

**EXAMPLE**

Solve the system of equations: $latex \begin{cases}3x-4y=5 \\ 6x-4y=2 \end{cases}$

##### Solution

We can start by simplifying the second equation by dividing it by 2:

$latex \begin{cases}3x-4y=5 \\ 3x-2y=1 \end{cases}$

Now, let’s solve by substitution. Therefore, we solve the second equation for *x*, and we have:

$latex 3x-2y=1$

$latex 3x=2y+1$

$latex x=\frac{2y+1}{3}$

Using $latex x=\frac{2y+1}{3}$ in the first equation, we have:

$latex 3x-4y=5$

$latex 3\left(\frac{2y+1}{3}\right)-4y=5$

$latex 2y+1-4y=5$

Solving the equation for *y*, we have:

$latex 2y+1-4y=5$

$latex -2y=4$

$latex y=-2$

Using the value *y*=-2 in the second equation, we have:

$latex 3x-2y=1$

$latex 3x-2(-2)=1$

$latex 3x+4=1$

$latex 3x=-3$

$latex x=-1$

The solution to the system is $latex x=-1,~~y=-2$.

**EXAMPLE **10

**EXAMPLE**

Find the solution to the system of equations: $latex \begin{cases}3x-y=1 \\ 5x+y=7 \end{cases}$

##### Solution

Let’s solve this by elimination since we have opposite coefficients in *y*:

$latex \begin{cases}3x-y=1 \\ 5x+y=7 \end{cases}$

Adding the equations, we have:

$latex 3x-y=1$

$latex + \hspace{1cm} 5x+y=7$

___________________

$latex 8x=8$

Solving the equation for *x*, we have:

$latex x=1$

Using the value *x*=1 in the second equation, we have:

$latex 5x+y=7$

$latex 5(1)+y=7$

$latex y=2$

The solution is $latex x=1,~~y=2$.

## 5 system of equations practice problems

Solve the following systems of equations using any method.

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