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.

## Summary of systems of two equations with two unknowns

We can solve systems of two equations using three main methods: elimination, substitution, and graphical methods. 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 A

*x*+B

*y*=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.

## 10 Examples of systems of equations with answers

**EXAMPLE 1**

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

##### Solution

**Step 1:** We have nothing to simplify.

**Step 2:** Solving the first equation for

*x*, we have:

$latex x+2y=10$

$latex x=10-2y$

**Step 3:** We substitute the expression $latex x=10-2y$ into 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 $latex y=3$ into 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: $latex \begin{cases}x-y=3 \\ 2x+y=12 \end{cases}$

##### Solution

**Step 1:** We have nothing to simplify, and both equations are already in the form A

*x*+B

*y=*C.

**Step 2:** We already have opposite coefficients in the variable

*y.*

**Step 3:** Adding the equations, we have:

$latex x-y=3$

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

___________________

$latex 3x=15$

**Step 4:** Solving for

*x*, we have:

$latex 3x=15$

$latex x=5$

**Step 5:** We substitute $latex x=5$ into 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: $latex \begin{cases}-2x-y=1 \\ 3x+4y=6 \end{cases}$

##### Solution

**Step 1:** We have nothing to simplify.

**Step 2:** Solving the first equation for

*y*, we have:

$latex -2x-y=1$

$latex -y=1+2x$

$latex y=-1-2x$

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

$latex 3x+4y=6$

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

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

**Step 4:** Solving for

*x*, we have:

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

$latex -5x=10$

$latex x=-2$

**Step 5: **We substitute $latex x=-2$ into 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: $latex \begin{cases}y=2x+7 \\ -6x-2y=-4 \end{cases}$

##### Solution

**Step 1:** We write the equations in the form A

*x*+B

*y=*C:

$latex \begin{cases}-2x+y=7 \\ -6x-2y=-4 \end{cases}$

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

*y*:

$latex \begin{cases}-4x+2y=14 \\ -6x-2y=-4 \end{cases}$

**Step 3:** Adding the equations:

$latex -4x+2y=14$

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

___________________

$latex -10x=10$

**Step 4:** Solving for

*x*, we have:

$latex -10x=10$

$latex x=-1$

**Step 5:** We substitute $latex x=-1$ into 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: $latex \begin{cases}2(2x-4)+y=3 \\ -x+2y=4 \end{cases}$

##### Solution

**Step 1:** We simplify the first equation:

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

**Step 2:** Solving the first equation for

*y*, we have:

$latex 4x-8+y=3$

$latex y=-4x+11$

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

$latex -x+2y=4$

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

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

**Step 4:** Solving for

*x*, we have:

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

$latex -9x=-18$

$latex x=2$

**Step 5: **We substitute $latex x=2$ into 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: $latex \begin{cases}2x=3y-14 \\ 2y=x+8 \end{cases}$

##### Solution

**Step 1:** We write the equations in the form A

*x*+B

*y=*C:

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

**Step 2:** Multiplying the second equation by 2 to obtain opposite coefficients in the

*x*, we have:

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

**Step 3:** Adding the equations, we have:

$latex 2x-3y=-14$

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

___________________

$latex y=2$

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

*y*:

$latex y=2$

**Step 5:** Substituting $latex y=2$ into the first equation, we have:

$latex 2x=3y-14$

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

$latex 2x=-8$

$latex x=-4$

**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 $latex x=1,~~y=2$.

## Systems of equations 2×2 – Practice problems

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

Write the answer in the input box x=?, y=?.

## See also

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