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Systems Using Substitution

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

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Substitution Method for Systems of Equations

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

Substitution Method

A \begin{align*}2 \times 2\end{align*} system of linear equations can be solved algebraically by the substitution method. In order to use this method, follow these steps:

Step 1: Solve one of the equations for one of the variables.

Step 2: Substitute that expression into the remaining equation. The result will be a linear equation, with one variable, that can be solved.

Step 3: Solve the remaining equation.

Step 4: Substitute the solution into the other equation to determine the value of the other variable.

Step 5: The solution to the system is the intersection point of the two equations and it represents the coordinates of the ordered pair.

Let's solve the following systems of linear equations by using the substitution method:

  1.    

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

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 with a coefficient that is either +1 or –1. In the above system the first equation has ‘\begin{align*}y\end{align*}’ with a coefficient of 1.

\begin{align*}& 3x+y = 1\\ & 3x {\color{red}-3x}+y = 1 {\color{red}-3x}\\ & \boxed{y = 1-3x}\end{align*}

Substitute \begin{align*}(1-3x)\end{align*} into the second equation for ‘\begin{align*}y\end{align*}’.

\begin{align*}2x+5y &= 18\\ 2x+5({\color{red}1-3x}) &= 18\end{align*}

Apply the distributive property and solve the equation.

\begin{align*}& 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}\end{align*}

Substitute –1 for \begin{align*}x\end{align*} into the equation \begin{align*}\boxed{y=1-3x}\end{align*}.

\begin{align*}y &= 1-3x\\ y &= 1-3({\color{red}-1})\end{align*}

\begin{align*}& \ y = 1 {\color{red}+3}\\ & \boxed{y = 4}\end{align*}

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:

\begin{align*}\boxed{\text{Line1}} : 3x+y=1\end{align*}

\begin{align*}\boxed{\text{Line2}} : 2x+5y=18\end{align*}

Line 1 intersects Line 2 at (–1, 4).

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

  1.     

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

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 for the variable ‘\begin{align*}x\end{align*}’.

\begin{align*}& 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}\end{align*}

Substitute \begin{align*}(-4-2y)\end{align*} into the first equation for ‘\begin{align*}x\end{align*}’.

\begin{align*}8x-3y &= 6\\ 8 ({\color{red}-4-2y})-3y &= 6\end{align*}

Apply the distributive property and solve the equation.

\begin{align*}& {\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}\end{align*}

Substitute –2 for \begin{align*}y\end{align*} into the equation \begin{align*}\boxed{x=-4-2y}\end{align*}

\begin{align*}x &= -4-2y\\ x &= -4-2 ({\color{red}-2})\end{align*}

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

Examples

Example 1

Earlier, you were asked to solve the following system of equations by substitution:\begin{align*}\begin{Bmatrix} 2x+3y=13 \\ y=4x-5 \end{Bmatrix}\end{align*}

The second equation is solved in terms of the variable ‘\begin{align*}y\end{align*}’. The expression \begin{align*}(4x+5)\end{align*} can be used to replace ‘\begin{align*}y\end{align*}’ in the first equation.

\begin{align*}2x+3y &= 13\\ 2x+3({\color{red}4x-5}) &= 13\end{align*}

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

\begin{align*}2x+{\color{red}12x-15}=13\end{align*}

Combine like terms to simplify the equation.

\begin{align*}{\color{red}14x}-15=13\end{align*}

Solve the equation.

\begin{align*}& 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}\end{align*}

To determine the value of ‘\begin{align*}y\end{align*}’, substitute this value into the equation \begin{align*}y=4x-5\end{align*}.

\begin{align*}y &= 4x-5\\ y &= 4({\color{red}2})-5\end{align*}

\begin{align*}& \ y = {\color{red}8}-5\\ & \boxed{y = 3}\end{align*}

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

Example 2

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

Both equations are equal to the variable ‘\begin{align*}x\end{align*}’. Set \begin{align*}(2y+1)=(4y-3)\end{align*}

\begin{align*}2y+1=4y-3\end{align*}

Solve the equation.

\begin{align*}& 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}\end{align*}

Substitute this value for ‘\begin{align*}y\end{align*}’ into one of the original equations.

\begin{align*}x &= 2y+1\\ x &= 2({\color{red}2})+2\end{align*}

\begin{align*}& \ x = {\color{red}4}+1\\ & \boxed{x = 5}\\ & \boxed{l_1 \cap l_2 @ (5,2)}\end{align*}

Example 3

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

 

The first equation has the variable ‘\begin{align*}y\end{align*}’ with a coefficient of 1. Solve the equation in terms of ‘\begin{align*}y\end{align*}’.

\begin{align*}& 2x+y = 3\\ & 2x {\color{red}-2x}+y = 3 {\color{red}-2x}\\ & \boxed{y = 3-2x}\end{align*}

Substitute \begin{align*}(3-2x)\end{align*} into the second equation for ‘\begin{align*}y\end{align*}’.

\begin{align*}3x+2y &= 12\\ 3x+2 ({\color{red}3-2x}) &= 12\end{align*}

Apply the distributive property and solve the equation.

\begin{align*}& 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}\end{align*}

Substitute this value for ‘\begin{align*}x\end{align*}’ into the equation \begin{align*}y=3-2x\end{align*}.

\begin{align*}y &= 3-2x\\ y &= 3-2({\color{red}-6})\end{align*}

\begin{align*}& \ y = 3 {\color{red}+12}\\ & \boxed{y = 15}\\ & \boxed{l_1 \cap l_2 @ (-6,15)}\end{align*}

Example 4

Solve the following system of linear equations by substitution: \begin{align*}\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}\end{align*}

\begin{align*}\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}\end{align*}

Begin by multiplying each equation by the LCM of the denominators to simplify the system.

\begin{align*}\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2}\end{align*} The LCM for the denominators is 20.

\begin{align*}{\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\end{align*}

\begin{align*}-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}\end{align*} The LCM for the denominators is 12.

\begin{align*}-{\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\end{align*}

The two equations that need to be solved by substitution are: \begin{align*}\begin{Bmatrix} 8m+15n=50 \\ -8m+6n=9 \end{Bmatrix}\end{align*}

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 ‘\begin{align*}m\end{align*}’.

\begin{align*}& 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}\end{align*}

Substitute this value for ‘\begin{align*}m\end{align*}’ into the second equation.

\begin{align*}-8m+6n &= 9\\ -8 \left({\color{red}\frac{25}{4}-\frac{15}{8}n}\right)+6n &= 9\end{align*}

Apply the distributive property and solve the equation.

\begin{align*}& {\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}}\end{align*}

Substitute this value into the equation that has been solved in terms of ‘\begin{align*}m\end{align*}’ 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.

\begin{align*}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}}\end{align*}

A common denominator is required to subtract the fractions.

\begin{align*}& \overset{ \quad \ {\color{red}42}}{4 \overline{ ) {168}}}\\ & \underline{- 16 {\color{blue}\downarrow}}\\ & \quad \ \ 8\\ & \underline{- \;\;\;8}\\ & \quad \ \ 0\end{align*}

Multiply \begin{align*}\frac{25}{4}\end{align*} by \begin{align*}{\color{red}\frac{42}{42}}\end{align*} :

\begin{align*}& 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)}\end{align*}

Example 5

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 $0.50 every month. His other plan has a base fee of $5.00 plus $1 per month.

Write an equation to represent each plan.

Solve the system of equations.

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:

\begin{align*}y=0.50x+25\end{align*}

\begin{align*}y=1x+5\end{align*}

where ‘\begin{align*}y\end{align*}’ represents the cost and ‘\begin{align*}x\end{align*}’ represents the number of months. Both equations are equal to ‘\begin{align*}y\end{align*}’. Therefore, the expression for \begin{align*}y\end{align*} can be substituted for the \begin{align*}y\end{align*} in the other equation.

\begin{align*}\begin{Bmatrix} y=0.50x+25 \\ y=1x+5 \end{Bmatrix}\end{align*}

\begin{align*}& 0.50x+25 = 1x+5 \\ & 0.50x+25 {\color{red}-25} = 1x+5 {\color{red}-25}\\ & 0.50x = 1x {\color{red}-20}\\ & 0.50x {\color{red}-1x} = 1x {\color{red}-1x}-20\\ & {\color{red}-0.50x} = -20\\ & \frac{-0.50x}{{\color{red}-0.50}} = \frac{-20}{{\color{red}-0.50}}\\ & \frac{\cancel{-0.50}x}{\cancel{-0.50}} = \frac{\overset{{\color{red}40}}{\cancel{-20}}}{\cancel{-0.50}}\\ & \boxed{x = 40 \ months}\end{align*}

Since the equations were equal, the value for ‘\begin{align*}x\end{align*}’ can be substituted into either of the original equations. The result will be the same.

\begin{align*}y &= 1x+5\\ y &= 1({\color{red}40})+5\end{align*}

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

Review

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

.

\begin{align*}\begin{Bmatrix} y=3x \\ 5x-2y=1 \end{Bmatrix}\end{align*}

.

\begin{align*}\begin{Bmatrix} y=3x+1 \\ 2x-y=2 \end{Bmatrix}\end{align*}

.

\begin{align*}\begin{Bmatrix} x=2y \\ x=3y-3 \end{Bmatrix}\end{align*}

.

\begin{align*}\begin{Bmatrix} x-y=6 \\ 6x-y=40 \end{Bmatrix}\end{align*}

.

\begin{align*}\begin{Bmatrix} x+y=6 \\ x+3(y+2)=10 \end{Bmatrix}\end{align*}

.

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

.

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

.

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

.

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

.

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

.

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

.

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

.

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

.

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

.

\begin{align*}\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}\end{align*}

Review (Answers)

To see the Review answers, open this PDF file and look for section 5.2. 

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Vocabulary

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.

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