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# 5.2: Substitution Method for Systems of Equations

Difficulty Level: At Grade Created by: CK-12

When you solve a system of consistent and independent equations by graphing, a single ordered pair is the solution. The ordered pair satisfies both equations and the point is the intersection of the graphs of the linear equations. The coordinates of this point of intersection are not always integers. How can you algebraically solve a system of equations like the one below?

$\begin{Bmatrix}2x+3y=13 \\y=4x-5\end{Bmatrix}$

### Guidance

A $2 \times 2$ system of linear equations can be solved algebraically by the substitution method. In order to use this method, follow these steps:

1. Solve one of the equations for one of the variables.
2. Substitute that expression into the remaining equation. The result will be a linear equation, with one variable, that can be solved.
3. Solve the remaining equation.
4. Substitute the solution into the other equation to determine the value of the other variable.
5. The solution to the system is the intersection point of the two equations and it represents the coordinates of the ordered pair.

#### Example A

Solve the following system of linear equations by substitution:

$\begin{Bmatrix}3x+y=1 \\2x+5y=18\end{Bmatrix}$

Solution:

To begin, solve one of the equations in terms of one of the variables. This step is simplified if one of the equations has one variable whose coefficient is either +1 or -1. In the above system the first equation has ‘ $y$ ’ that has a coefficient of 1.

$& 3x+y = 1\\& 3x {\color{red}-3x}+y = 1 {\color{red}-3x}\\& \boxed{y = 1-3x}$

Substitute $(1-3x)$ into the second equation for ‘ $y$ ’.

$2x+5y &= 18\\2x+5({\color{red}1-3x}) &= 18$

Apply the distributive property and solve the equation.

$& 2x+{\color{red}5-15x} = 18\\& {\color{red}-13x}+5 = 18\\& -13x+5 {\color{red}-5} = 18 {\color{red}-5}\\& -13x = {\color{red}13}\\& \frac{-13x}{{\color{red}-13}} = \frac{13}{{\color{red}-13}}\\& \frac{\cancel{-13}x}{\cancel{-13}} = \frac{\overset{{\color{red}-1}}{\cancel{-13}}}{\cancel{-13}}\\& \boxed{x = -1}$

Substitute -1 for $x$ into the equation $\boxed{y=1-3x}$ .

$y &= 1-3x\\y &= 1-3({\color{red}-1})$

Evaluate the equation.

$& \ y = 1 {\color{red}+3}\\& \boxed{y = 4}$

The solution is (-1, 4). This represents the intersection point of the lines if the equations were graphed on a Cartesian grid. Another way to write ‘the lines intersect at (-1, 4)’ is:

$\boxed{\text{Line1}} : 3x+y=1$

$\boxed{\text{Line2}} : 2x+5y=18$

Line 1 intersects Line 2 at (-1, 4)

$& \qquad \quad \text{at}\\& \qquad \quad {\color{red}\uparrow}\\& \boxed{l_1 \cap l_2 @ (-1,4)}\\& \quad \ {\color{red}\downarrow}\\& \text{intersects}$

#### Example B

Solve the following system of linear equations by substitution:

$\begin{Bmatrix}8x-3y=6 \\6x+12y=-24\end{Bmatrix}$

Solution:

There is no variable that has a coefficient of +1 or of -1. However, the second equation has coefficients and a constant that are multiples of 6. The second equation will be solved in terms of the variable ‘ $x$ ’.

$& 6x+12y =-24\\& 6x+12y {\color{red}-12y} = -24 {\color{red}-12y}\\& 6x = -24-12y\\& \frac{6x}{{\color{red}6}} = \frac{-24}{{\color{red}6}}-\frac{12y}{{\color{red}6}}\\& \frac{\cancel{6}x}{\cancel{6}} = \frac{\overset{{\color{red}-4}}{\cancel{-24}}}{\cancel{6}}-\frac{\overset{{\color{red}2}}{\cancel{12}y}}{\cancel{6}}\\& \boxed{x = -4-2y}$

Substitute $(-4-2y)$ into the first equation for ‘ $x$ ’.

$8x-3y &= 6\\8 ({\color{red}-4-2y})-3y &= 6$

Apply the distributive property and solve the equation.

$& {\color{red}-32-16y}-3y = 6\\& -32 {\color{red}-19y} = 6\\& -32 {\color{red}+32}-19y = 6 {\color{red}+32}\\& -19y = {\color{red}38}\\& \frac{-19y}{{\color{red}-19}} = \frac{38}{{\color{red}-19}}\\& \frac{\cancel{-19}y}{\cancel{-19}} = \frac{\overset{{\color{red}-2}}{\cancel{38}}}{\cancel{-19}}\\& \boxed{y =- 2}$

Substitute -2 for $y$ into the equation $\boxed{x=-4-2y}$

$x &= -4-2y\\x &= -4-2 ({\color{red}-2})$

Evaluate the equation.

$& x = -4 {\color{red}+4}\\& \boxed{x = 0}\\& \boxed{l_1 \cap l_2 @ (0,-2)}$

#### Example C

Jason, who is a real computer whiz, decided to set up his own server and to sell space on his computer so students could have their own web pages on the Internet. He devised two plans. One plan charges a base fee of $25.00 plus$.50 every month. His other plan has a base fee of $5.00 plus$1 per month.

i) Write an equation to represent each plan.

ii) Solve the system of equations.

Solution:

Both plans deal with the cost of buying space from Jason’s server. The cost involves a base fee and a monthly rate. The equations for the plans are:

$y=.50x+25$ and $y=1x+5$ where ‘ $y$ ’ represents the cost and ‘ $x$ ’ represents the number of months. Both equations are equal to ‘ $y$ ’. Therefore, the equations are equal to each other.

$\begin{Bmatrix}y=.50x+25 \\y=1x+5\end{Bmatrix}$

$& .50x+25 = 1x+5 && \text{Set the equations equal to each other and solve the equation.}\\& .50x+25 {\color{red}-25} = 1x+5 {\color{red}-25}\\& .50x = 1x {\color{red}-20}\\& .50x {\color{red}-1x} = 1x {\color{red}-1x}-20\\& {\color{red}-.50x} = -20\\& \frac{-.50x}{{\color{red}-.50}} = \frac{-20}{{\color{red}-.50}}\\& \frac{\cancel{-.50}x}{\cancel{-.50}} = \frac{\overset{{\color{red}40}}{\cancel{-20}}}{\cancel{-.50}}\\& \boxed{x = 40 \ months}$

Since the equations were equal, the value for ‘ $x$ ’ can be substituted into either of the original equations. The result will be the same.

$y &= 1x+5\\y &= 1({\color{red}40})+5$

Evaluate the equation.

$& y = {\color{red}40}+5\\& \boxed{y = 45 \ dollars}\\& \boxed{l_1 \cap l_2 @ (40,45)}.$

#### Concept Problem Revisited

When graphing is not a feasible method for solving a system, you can solve by substitution:

$\begin{Bmatrix}2x+3y=13 \\y=4x-5\end{Bmatrix}$

The second equation is solved in terms of the variable ‘ $y$ ’. The expression $(4x+5)$ can be used to replace ‘ $y$ ’ in the first equation.

$2x+3y &= 13\\2x+3({\color{red}4x-5}) &= 13$

The equation now has one variable. Apply the distributive property.

$2x+{\color{red}12x-15}=13$

Combine like terms to simplify the equation.

${\color{red}14x}-15=13$

Solve the equation.

$& 14x-15 {\color{red}+15} = 13 {\color{red}+15}\\& 14x = {\color{red}28}\\& \frac{14x}{{\color{red}14}} =\frac{28}{{\color{red}14}}\\& \frac{\cancel{14}x}{\cancel{14}} = \frac{\overset{{\color{red}2}}{\cancel{28}}}{\cancel{14}}\\& \boxed{x = 2}$

To determine the value of ‘ $y$ ’, substitute this value into the equation $y=4x-5$ .

$y &= 4x-5\\y &= 4({\color{red}2})-5$

Evaluate the equation.

$& \ y = {\color{red}8}-5\\& \boxed{y = 3}$

The solution is (2, 3). This represents the intersection point of the lines if the equations were graphed on a Cartesian grid.

### Vocabulary

Substitution Method
The substitution method is a way of solving a system of linear equations algebraically. The substitution method involves solving an equation for a variable and substituting that expression into the other equation.

### Guided Practice

1. Solve the following system of linear equations by substitution: $\begin{Bmatrix}x=2y+1 \\x=4y-3\end{Bmatrix}$

2. Solve the following system of linear equations by substitution: $\begin{Bmatrix}2x+y=3 \\3x+2y=12\end{Bmatrix}$ .

3. Solve the following system of linear equations by substitution: $\begin{Bmatrix}\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2} \\-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}\end{Bmatrix}$

1.

$\begin{Bmatrix}x=2y+1 \\x=4y-3\end{Bmatrix}$
Both equations are equal to the variable ‘ $x$ ’. Set $(2y+1)=(4y-3)$
$2y+1=4y-3$
Solve the equation.
$& 2y+1 {\color{red}-1} = 4y-3 {\color{red}-1}\\& 2y = 4y {\color{red}-4}\\& 2y {\color{red}-4y} = 4y {\color{red}-4y}-4\\& {\color{red}-2y} = -4\\& \frac{-2y}{{\color{red}-2}} = \frac{-4}{{\color{red}-2}}\\& \frac{\cancel{-2}y}{\cancel{-2}} = \frac{\overset{{\color{red}2}}{\cancel{-4}}}{\cancel{-2}}\\& \boxed{y = 2}$
Substitute this value for ‘ $y$ ’ into one of the original equations.
$x &= 2y+1\\x &= 2({\color{red}2})+2$
Evaluate the equation.
$& \ x = {\color{red}4}+1\\& \boxed{x = 5}\\& \boxed{l_1 \cap l_2 @ (5,2)}$

2.

$\begin{Bmatrix}2x+y=3 \\3x+2y=12\end{Bmatrix}$
The first equation has the variable ‘ $y$ ’ with a coefficient of 1. Solve the equation in terms of ‘ $y$ ’.
$& 2x+y = 3\\& 2x {\color{red}-2x}+y = 3 {\color{red}-2x}\\& \boxed{y = 3-2x}$
Substitute $(3-2x)$ into the second equation for ‘ $y$ ’.
$3x+2y &= 12\\3x+2 ({\color{red}3-2x}) &= 12$
Apply the distributive property and solve the equation.
$& 3x+ {\color{red}6-4x} = 12\\& 6 {\color{red}-x} = 12\\& 6 {\color{red}-6}-x = 12 {\color{red}-6}\\& -x = {\color{red}6}\\& \frac{-x}{{\color{red}-1}} = \frac{6}{{\color{red}-1}}\\& \frac{{\cancel{-1}}x}{{\cancel{-1}}} = \frac{\overset{{\color{red}-6}}{\cancel{6}}}{\cancel{-1}}\\& \boxed{x = -6}$
Substitute this value for ‘ $x$ ’ into the equation $y=3-2x$ .
$y &= 3-2x\\y &= 3-2({\color{red}-6})$
Evaluate the equation.
$& \ y = 3 {\color{red}+12}\\& \boxed{y = 15}\\& \boxed{l_1 \cap l_2 @ (-6,15)}$

3.

$\begin{Bmatrix}\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2} \\-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}\end{Bmatrix}$
Begin by multiplying each equation by the LCM of the denominators to simplify the system.
$\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2}$ The LCM for the denominators is 20.
${\color{red}20} \left(\frac{2}{5}\right)m+ {\color{red}20} \left(\frac{3}{4}\right)n &= {\color{red}20} \left(\frac{5}{2}\right)\\\overset{{\color{red}4}}{\cancel{20}} \left(\frac{2}{\cancel{5}}\right)m+\overset{{\color{red}5}}{\cancel{20}} \left(\frac{3}{\cancel{4}}\right)n &= \overset{{\color{red}10}}{\cancel{20}} \left(\frac{5}{\cancel{2}}\right)\\{\color{red}8}m+{\color{red}15}n &= {\color{red}50}\\8m+15n &= 50$
$-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}$ The LCM for the denominators is 12.
$-{\color{red}12} \left(\frac{2}{3}\right)m+{\color{red}12} \left(\frac{1}{2}\right)n &= {\color{red}12} \left(\frac{3}{4}\right)\\\overset{{\color{red}4}}{-\cancel{12}} \left(\frac{2}{\cancel{3}}\right)m+\overset{{\color{red}6}}{\cancel{12}} \left(\frac{1}{\cancel{2}}\right)n &= \overset{{\color{red}3}}{\cancel{12}}\left(\frac{3}{\cancel{4}}\right)\\{\color{red}-8}m+{\color{red}6}n &= {\color{red}9}\\-8m+6n &= 9$
The two equations that need to be solved by substitution are: $\begin{Bmatrix}8m+15n=50 \\-8m+6n=9\end{Bmatrix}$
Neither of the equations have a variable with a coefficient of 1 nor does one equation have coefficients that are multiples of a given coefficient. Solve the first equation in terms of ‘ $m$ ’.
$& 8m+15n = 50\\& 8m+15n {\color{red}-15n} = 50 {\color{red}-15n}\\& 8m = 50-15n\\& \frac{8m}{{\color{red}8}} = \frac{50}{{\color{red}8}}-\frac{15n}{{\color{red}8}}\\& \frac{\cancel{8}m}{\cancel{8}} = \frac{50}{8}-\frac{15n}{8}\\& m = {\color{red}\frac{25}{4}}-\frac{15}{8}n\\& \boxed{m = \frac{25}{4}-\frac{15}{8}n}$
Substitute this value for ‘ $m$ ’ into the second equation.
$-8m+6n &= 9\\-8 \left({\color{red}\frac{25}{4}-\frac{15}{8}n}\right)+6n &= 9$
Apply the distributive property and solve the equation.
$& {\color{red}-\frac{200}{4}+\frac{120}{8}n}+6n = 9\\& -\frac{\overset{{\color{red}50}}{\cancel{200}}}{\cancel{4}}+\frac{\overset{{\color{red}15}}{\cancel{120}}}{\cancel{8}}n+6n = 9\\& {\color{red}-50+15n}+6n = 9\\& -50 {\color{red}+21n} = 9\\& -50 {\color{red}+50}+21n = 9 {\color{red}+50}\\& 21n = {\color{red}59}\\& \frac{21n}{{\color{red}21}} = \frac{59}{{\color{red}21}}\\& \frac{\cancel{21}n}{\cancel{21}} = \frac{59}{21}\\& \boxed{n = \frac{59}{21}}$
Substitute this value into the equation that has been solved in terms of ‘ $m$ ’ or into one of the original equations or into one of the new equations that resulted from multiplying by the LCM.
Whichever substitution is performed, the same result will occur.
$m &= \frac{25}{4}-\frac{15}{8}n\\m &= \frac{25}{4}-\frac{15}{8} \left({\color{red}\frac{59}{21}}\right)\\m &= \frac{25}{4}-{\color{red}\frac{885}{168}}$
A common denominator is required to subtract the fractions.
$& \overset{ \quad \ {\color{red}42}}{4 \overline{ ) {168}}}\\& \underline{- 16 {\color{blue}\downarrow}}\\& \quad \ \ 8\\& \underline{- \;\;\;8}\\& \quad \ \ 0$
Multiply $\frac{25}{4}$ by ${\color{red}\frac{42}{42}}$ :
$& m = {\color{red}\frac{42}{42}} \left(\frac{25}{4}\right)-\frac{885}{168}\\& m = {\color{red}\frac{1050}{168}}-\frac{885}{168}\\& m= {\color{red}\frac{165}{168}}\\& m = {\color{red}\frac{55}{56}}\\& \boxed{m = \frac{55}{56}}\\& \boxed{l_1 \cap l_2 @ \left(\frac{55}{56}, \frac{59}{21}\right)}$

### Practice

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

$\begin{Bmatrix}2x+y=5 \\3x-4y=2\end{Bmatrix}$

$\begin{Bmatrix}5x-2y=-4 \\4x+y=-11\end{Bmatrix}$

$\begin{Bmatrix}3y-x=-10 \\3x+4y=-22\end{Bmatrix}$

$\begin{Bmatrix}4e+2f=-2 \\2e-3f=1\end{Bmatrix}$

$\begin{Bmatrix}\frac{1}{4}x+y=-\frac{7}{2} \\\frac{1}{2}x-\frac{1}{4}y=1\end{Bmatrix}$

$\begin{Bmatrix}x=-4+y \\x=3y-6\end{Bmatrix}$

$\begin{Bmatrix}3y-2x=-3 \\3x-3y=6\end{Bmatrix}$

$\begin{Bmatrix}2x=5y-12 \\3x+5y=7\end{Bmatrix}$

$\begin{Bmatrix}3y=2x-5 \\2x=y+3\end{Bmatrix}$

$\begin{Bmatrix}\frac{x+y}{3}+\frac{x-y}{2}=\frac{25}{6} \\\frac{x+y-9}{2}=\frac{y-x-6}{3}\end{Bmatrix}$

## Date Created:

Dec 19, 2012

Jun 13, 2014
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