# Systems Using Substitution

## Solve for one variable, substitute the value in the other equation

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Solving Linear Systems by Substitution

Kathleen’s rectangular garden has a perimeter of 280 yards. The length is three times the width. If she wants to buy fencing panels to surround it, then Kathleen must determine the garden's dimensions.

How can Kathleen use a system of equations to calculate the length and width of her rectangular garden?

In this concept, you will learn to solve linear systems by substitution.

### Substitution Method

There is an ordered pair \begin{align*}(x,y)\end{align*} that makes both equations in the following system of linear equations true. This means that ‘\begin{align*}x\end{align*}’ in the first equation equals ‘\begin{align*}x\end{align*} in the second equation and ‘\begin{align*}y\end{align*}’ in the first equation equals ‘\begin{align*}y\end{align*}’ in the second equation.

\begin{align*}\begin{array}{rcl} 3y &=& x-2 \\ y-x &=& 4 \end{array}\end{align*}

The solution to a system of two linear equations with two variables \begin{align*}x\end{align*} and \begin{align*}y\end{align*} is a value for \begin{align*}x\end{align*} and a value for \begin{align*}y\end{align*} that will make both equations true. One way to find these values for the variables is by using the substitution method. The substitution method involves solving one of the equations for one of the variables and replacing that variable in the other equation with its equivalent expression. The result will be an equation with one variable which can be solved using algebra.

Let’s solve the given system of linear equations using the substitution method.

\begin{align*}\begin{array}{rcl} 3y &=& x-2 \\ y-x &=& 4 \end{array}\end{align*}

First, solve the second equation for the variable \begin{align*}y\end{align*} by adding \begin{align*}x\end{align*} to both sides of the equation.

\begin{align*}\begin{array}{rcl} y-x &=& 4 \\ y-x+x &=& 4+x \end{array}\end{align*}

Next, simplify both sides of the equation.

\begin{align*}\begin{array}{rcl} y-x+x &=& 4+x \\ y &=& 4+x \end{array}\end{align*}

Next, substitute this expression for \begin{align*}y\end{align*} into the first equation.

\begin{align*}\begin{array}{rcl} y &=& 4+x \\ 3y &=& x-2 \\ 3(4+x) &=& x-2 \end{array}\end{align*}

Next, do the multiplication on the left side of the equation to clear parentheses.

\begin{align*}\begin{array}{rcl} 3(4+x) &=& x-2 \\ 12+3x &=& x- 2 \end{array}\end{align*}

Next, subtract 12 from both sides of the equation and simplify.

\begin{align*}\begin{array}{rcl} 12+3x &=& x-2 \\ 12-12+3x &=& x-2-12 \\ 3x &=& x-14 \end{array}\end{align*}

Next, subtract \begin{align*}x\end{align*} from both sides of the equation and simplify to isolate the variable on one side.

\begin{align*}\begin{array}{rcl} 3x &=& x-14 \\ 3x-x &=& x-x-14 \\ 2x &=& -14 \end{array}\end{align*}

Then, divide both sides of the equation by 2 to solve for the variable \begin{align*}x.\end{align*}

\begin{align*}\begin{array}{rcl} 2x &=& -14 \\ \overset{1} {\frac{\cancel{2}x}{\cancel{2}}} &=& -\overset{7} {\frac{\cancel{14}}{\cancel{2}}} \\ x &=& -7 \end{array}\end{align*}

Next, substitute the value for \begin{align*}x\end{align*} into the expression for \begin{align*}y\end{align*} and simplify the right side of the equation.

\begin{align*}\begin{array}{rcl} y &=& 4+x \\ y &=& 4+(-7) \\ y &=& 4-7 \\ y &=& -3 \end{array}\end{align*}

The answer is \begin{align*}x=-7\end{align*} and \begin{align*}y=-3\end{align*}.

The solution for the system of linear equations is:

\begin{align*}\dbinom{x}{y} = \dbinom{\text{-}7}{\text{-}3}\end{align*}

### Examples

#### Example 1

Earlier, you were given a problem about Kathleen and her rectangular garden. She needs to figure out the dimensions of the garden. How can Kathleen use a system of equations to help her figure this out?

First, write a system of linear equations to represent the information given in the story.

Let \begin{align*}l\end{align*} represent length and let \begin{align*}w\end{align*} represent the width.

The length of the garden is three times the width and the perimeter is 280 yards.

\begin{align*}\begin{array}{rcl} l &=& 3w \\ 2l+2w &=& 280 \end{array}\end{align*}

Next, substitute the expression for \begin{align*}l\end{align*} into the second equation.

\begin{align*}\begin{array}{rcl} l &=& 3w \\ 2(3w)+2w &=& 280 \end{array}\end{align*}

Next, perform the multiplication on the left side of the equation to clear parentheses and simplify.

\begin{align*}\begin{array}{rcl} 2(3w)+2w &=& 280 \\ 6w+2w &=& 280 \\ 8w &=& 280 \end{array}\end{align*}

Next, divide both sides of the equation by 8 to solve for \begin{align*}w\end{align*}.

\begin{align*}\begin{array}{rcl} 8w &=& 280 \\ \overset{1} {\frac{\cancel{8}w}{\cancel{8}}} &=& \overset{35} {\frac{\cancel{280}}{\cancel{8}}} \\ w &=& 35 \end{array}\end{align*}

The width of the garden is 35 yards.

Next, substitute the value for \begin{align*}w\end{align*} into the expression for the variable \begin{align*}l\end{align*} and simplify the right side of the equation.

\begin{align*}\begin{array}{rcl} l &=& 3w \\ l &=& 3(35) \\ l &=& 105 \end{array}\end{align*}

The length of the garden is 105 yards.

The solution for the system of linear equations is \begin{align*}\dbinom{w}{l} = \dbinom{35}{105}\end{align*} which means the dimensions of Kathleen’s garden are 35 yards by 105 yards.

#### Example 2

Solve the following system of linear equations using the substitution method.

\begin{align*} \begin{array}{rcl} 2x+3y &=& 6 \\ x+y &=& 6 \end{array}\end{align*}

First, solve the second equation for the variable \begin{align*}x\end{align*} by subtracting \begin{align*}y\end{align*} from both sides of the equation.

\begin{align*}\begin{array}{rcl} x+y &=& 6 \\ x+y-y &=& 6-y \end{array}\end{align*}

Next, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} x+y-y &=& 6-y \\ x&=& 6-y \end{array}\end{align*}

Next, substitute this expression for \begin{align*}x\end{align*} into the first equation.

\begin{align*}\begin{array}{rcl} x &=& 6-y \\ 2x+3y &=& 6 \\ 2(6-y)+3y &=& 6 \end{array}\end{align*}

Next, do the multiplication on the left side of the equation to clear parentheses.

\begin{align*}\begin{array}{rcl} 2(6-y)+3y &=& 6 \\ 12-2y+3y &=& 6 \end{array}\end{align*}

Next, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 12-2y+3y &=& 6 \\ 12+y &=& 6 \end{array}\end{align*}

Next, subtract 12 from both sides of the equation to solve for the variable \begin{align*}y.\end{align*}

\begin{align*}\begin{array}{rcl} 12+y &=& 6 \\ 12-12+y &=& 6-12 \end{array}\end{align*}

Next, simplify both sides of the equation.

\begin{align*}\begin{array}{rcl} 12-12+y &=& 6-12 \\ y &=& -6 \end{array}\end{align*}

Next, substitute the value for \begin{align*}y\end{align*} into the expression for \begin{align*}x\end{align*} and simplify the right side of the equation.

\begin{align*}\begin{array}{rcl} x &=& 6-y \\ x &=& 6-(-6) \\ x &=& 6+6 \\ x &=& 12 \end{array}\end{align*}

The answer is \begin{align*}x=12\end{align*} and \begin{align*}y=-6\end{align*}.

The solution for the system of linear equations is:

\begin{align*}\dbinom{x}{y} = \dbinom{12}{-6}\end{align*}

#### Example 3

Solve the following system of linear equations using the substitution method.

\begin{align*}\begin{array}{rcl} 2y &=& x+4 \\ y &=& 3x \end{array}\end{align*}

First, substitute the expression for \begin{align*}y\end{align*} into the first equation.

\begin{align*}\begin{array}{rcl} y &=& 3x \\ 2y &=& x+4 \\ 2(3x) &=& x+4 \end{array}\end{align*}

Next, perform the multiplication on the left side of the equation to clear parentheses.

\begin{align*}\begin{array}{rcl} 2(3x) &=& x+4 \\ 6x &=& x+4 \end{array}\end{align*}

Next, subtract \begin{align*}x\end{align*} from both sides of the equation and simplify to isolate the variable on one side.

\begin{align*}\begin{array}{rcl} 6x &=& x+4 \\ 6x-x &=& x-x+4 \end{array}\end{align*}

Next, simplify both sides of the equation.

\begin{align*}\begin{array}{rcl} 6x-x &=& x-x+4 \\ 5x &=& 4 \end{array}\end{align*}

Next, divide both sides of the equation by 5 to solve for the variable \begin{align*}x.\end{align*}

\begin{align*}\begin{array}{rcl} 5x &=& 4 \\ \overset{1} {\frac{\cancel{5}}{\cancel{5}}x} &=& \frac{4}{5} \\ x &=& \frac{4}{5} \end{array}\end{align*}

Next, substitute the value for \begin{align*}x\end{align*} into the expression for \begin{align*}y\end{align*} and simplify the right side of the equation.

\begin{align*}\begin{array}{rcl} y &=& 3x \\ y &=& 3 \left(\frac{4}{5} \right) \\ y &=& \frac{12}{5} \\ \end{array}\end{align*}
The answer is \begin{align*}x = \frac{4}{5}\end{align*} and \begin{align*}y=\frac{12}{5}\end{align*}.

The solution for the system of linear equations is:

\begin{align*}\dbinom{x}{y} = \dbinom{\frac{4}{5}}{\frac{12}{5}}\end{align*}

#### Example 4

Solve the following system of linear equations using the substitution method.

\begin{align*}\begin{array}{rcl} x+y &=& 4 \\ 2x-3y &=& 18 \end{array}\end{align*}

First, solve the first equation for the variable ‘\begin{align*}x\end{align*}’ by subtracting ‘\begin{align*}y\end{align*}’ from both sides of the equation.

\begin{align*}\begin{array}{rcl} x+y &=& 4 \\ x+y-y &=& 4-y \end{array}\end{align*}

Next, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} x+y-y &=& 4-y \\ x &=& 4-y \end{array}\end{align*}

Next, substitute this expression for ‘\begin{align*}x\end{align*}’ into the first equation.

\begin{align*}\begin{array}{rcl} x &=& 4-y \\ 2x-3y &=& 18 \\ 2(4-y)-3y &=& 18 \end{array}\end{align*}

Next, do the multiplication on the left side of the equation to clear parentheses.

\begin{align*}\begin{array}{rcl} 2(4-y)-3y &=& 18 \\ 8-2y-3y &=& 18 \end{array}\end{align*}

Next, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 8-2y-3y &=& 18 \\ 8-5y &=& 18 \end{array}\end{align*}

Next, subtract 8 from both sides of the equation to isolate the variable ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} 8-5y &=& 18 \\ 8-8-5y &=& 18-8 \end{array}\end{align*}

Next, simplify both sides of the equation.

\begin{align*}\begin{array}{rcl} 8-8-5y &=& 18-8 \\ -5y &=& 10 \end{array}\end{align*}

Next, divide both sides of the equation by -5 to solve for the variable ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} -5y &=& 10 \\ \overset{1} {\frac{\cancel{-5}}{\cancel{-5}}y} &=& \overset{-2} {\frac{\cancel{10}}{\cancel{-5}}} \\ y &=& -2 \end{array}\end{align*}

Next, substitute the value for ‘\begin{align*}y\end{align*}’ into the expression for ‘\begin{align*}x\end{align*}’ and simplify the right side of the equation.

\begin{align*}\begin{array}{rcl} x &=& 4-y \\ x &=& 4-(-2) \\ x &=& 4+2 \\ x &=& 6 \end{array}\end{align*}
The answer is \begin{align*}x=6\end{align*} and \begin{align*}y=-2\end{align*}.

The solution for the system of linear equations is:

\begin{align*}\dbinom{x}{y} = \dbinom{6}{-2}\end{align*}

#### Example 5

Solve the following system of linear equations using the substitution method.

\begin{align*}\begin{array}{rcl} 3y &=& x-22 \\ y &=& 4x \end{array}\end{align*}

First, substitute the expression for ‘\begin{align*}y\end{align*}’ into the first equation.

\begin{align*}\begin{array}{rcl} y &=& 4x \\ 3y &=& x-22 \\ 3(4x) &=& x-22 \end{array}\end{align*}

Next, perform the multiplication on the left side of the equation to clear parentheses.

\begin{align*}\begin{array}{rcl} 3(4x) &=& x-22 \\ 12x &=& x-22 \end{array}\end{align*}

Next, subtract ‘\begin{align*}x\end{align*}’ from both sides of the equation and simplify to isolate the variable on one side.

\begin{align*}\begin{array}{rcl} 12x &=& x-22 \\ 12x -x &=& x-x-22 \end{array}\end{align*}

Next, simplify both sides of the equation.

\begin{align*}\begin{array}{rcl} 12x-x &=& x-x-22 \\ 11x &=& -22 \end{array}\end{align*}

Next, divide both sides of the equation by 11 to solve for the variable ‘\begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 11x &=& -22 \\ \overset{1} {\frac{\cancel{11}}{\cancel{11}}x} &=& \overset{-2} {\frac{\cancel{-22}}{\cancel{11}}} \\ x &=& -2 \end{array}\end{align*}

Next, substitute the value for ‘\begin{align*}x\end{align*}’ into the expression for ‘\begin{align*}y\end{align*}’ and simplify the right side of the equation.

\begin{align*}\begin{array}{rcl} y &=& 4x \\ y &=& 4(-2) \\ y &=& -8 \end{array}\end{align*}
The answer is \begin{align*}x=-2\end{align*} and \begin{align*}y=-8\end{align*}.

The solution for the system of linear equations is:

\begin{align*}\dbinom{x}{y} = \dbinom{-2}{-8}\end{align*}

### Review

Solve each linear system by substitution.

1. \begin{align*}\begin{array}{rcl} y &=& 8 \\ 2y+2x &=& 2 \end{array} \end{align*}

2. \begin{align*}\begin{array}{rcl} x &=& 16 \\ 2y+2x &=& 2 \end{array}\end{align*}

3. \begin{align*}\begin{array}{rcl} 2x &=& 8 \\ 2y+x &=& 10 \end{array}\end{align*}

4. \begin{align*}\begin{array}{rcl} 3y &=& 9 \\ y+2x &=& 11 \end{array}\end{align*}

5. \begin{align*}\begin{array}{rcl} y-8 &=& 8 \\ 2y+x &=& 20 \end{array}\end{align*}

6. \begin{align*}\begin{array}{rcl} 4y &=& 8 \\ y-2x &=& 8 \end{array}\end{align*}

7. \begin{align*}\begin{array}{rcl} x-2 &=& 4 \\ 4x+y &=& 12 \end{array}\end{align*}

8. \begin{align*}\begin{array}{rcl} y &=& x+8 \\ y+2x &=& 11 \end{array}\end{align*}

9. \begin{align*}\begin{array}{rcl} 2y+8 &=& 12 \\ y+2x &=& 20 \end{array}\end{align*}

10. \begin{align*}\begin{array}{rcl} y-3 &=& 6 \\ 3y+3x &=& 9 \end{array}\end{align*}

11. \begin{align*}\begin{array}{rcl} 4y-1 &=& 11 \\ y-4x &=& 5 \end{array}\end{align*}

12. \begin{align*}\begin{array}{rcl} 2y-8 &=& 8 \\ 2y+2x &=& 2 \end{array}\end{align*}

13. \begin{align*}\begin{array}{rcl} 4x+y &=& -2 \\ -2x-3y &=& 1 \end{array}\end{align*}

14. \begin{align*}\begin{array}{rcl} y &=& 2x \\ 6x-y &=& 8 \end{array} \end{align*}

15. \begin{align*}\begin{array}{rcl} x+4y &=& -6 \\ 2x+10y &=& -6 \end{array}\end{align*}

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### Vocabulary Language: English

TermDefinition
Consistent A system of equations is consistent if it has at least one solution.
distributive property The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
linear equation A linear equation is an equation between two variables that produces a straight line when graphed.
substitute In algebra, to substitute means to replace a variable or term with a specific value.
system of equations A system of equations is a set of two or more equations.