A rental car company, Affordable Autos, charges $30 per day plus $0.51 per mile driven. A second car rental company, Cheap Cars, charges $25 per day plus $0.57 per mile driven. For a short distance, Cheap Cars offers the better deal. At what point (after how many miles in a single day) does the Affordable Autos rental company offer the better deal?
Best Method to Solve a Linear System
Any of the methods (graphing, substitution, linear combination) learned in this unit can be used to solve a linear system of equations. Sometimes, however it is more efficient to use one method over another based on how the equations are presented. For example
 If both equations are presented in slope intercept form \begin{align*}(y=mx+b)\end{align*}, then either graphing or substitution would be most efficient.
 If one equation is given in slope intercept form or solved for \begin{align*}x\end{align*}, then substitution might be easiest.
 If both equations are given in standard form \begin{align*}(Ax+By=C)\end{align*}, then linear combinations is usually most efficient.
Solve the following systems
\begin{align*}y &= x+5\\ y &= \frac{1}{2}x+2\end{align*}
Since both equations are in slope intercept form we could easily graph these lines. The question is whether or not the intersection of the two lines will lie on the “grid” (whole numbers). If not, it is very difficult to determine an answer from a graph. One way to get around this difficulty is to use technology to graph the lines and find their intersection.
The first equation has a \begin{align*}y\end{align*}intercept of 5 and slope of 1. It is shown here graphed in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*}.
The second equation has a \begin{align*}y\end{align*}intercept of 2 and a slope of \begin{align*}\frac{1}{2}\end{align*}. It is shown here graphed in \begin{align*}{\color{red}\mathbf{red}}\end{align*}.
The two lines clearly intersect at (2, 3).
Alternate Method: Substitution may be the preferred method for students who would rather solve equations algebraically. Since both of these equations are equal to \begin{align*}y\end{align*}, we can let the right hand sides be equal to each other and solve for \begin{align*}x\end{align*}:
\begin{align*}& \quad x+5 = \frac{1}{2}x+2\\ & \ 2\left(x+5=\frac{1}{2}x+2\right) \ \leftarrow \text{Multiplying the equation by 2 eliminates the fraction}.\\ & 2x+10=x+4\\ & \qquad \quad \ \ 6=3x\\ & \qquad \quad \ \ x=2\end{align*}
Now solve for \begin{align*}y\end{align*}:
\begin{align*}y &= (2)+5\\ y &=3\end{align*}
Solution: (2, 3)
Check your answer:
\begin{align*}3 &= 2+5\\ 3 &= \frac{1}{2}2+2 \rightarrow 1 + 2\end{align*}
Solve the system:
\begin{align*}15x+y &=24\\ y &= 4x+2\end{align*}
This time one of our equations is already solved for \begin{align*}y\end{align*}. It is easiest here to use this expression to substitute into the other equation and solve:
\begin{align*}15x+(4x+2) &=24\\ 15x4x+2 &=24\\ 11x &= 22\\ x &= 2\end{align*}
Now solve for \begin{align*}y\end{align*}:
\begin{align*}y &= 4(2)+2\\ y &= 8+2\\ y &= 6\end{align*}
Solution: (2, 6)
Check your answer:
\begin{align*}15(2)+(6) &= 306=24\\ 6 =4(2)+2 & =8+2=6 \end{align*}
Solve the system:
\begin{align*}6x+11y &= 86\\ 9x13y &= 115\end{align*}
Both equations in this example are in standard form so the easiest method to use here is linear combinations. Since the LCM of 6 and 9 is 18, we will multiply the first equation by 3 and the second equation by 2 to eliminate \begin{align*}x\end{align*} first:
\begin{align*}& 3(6x+11y=86) \quad \Rightarrow \quad  \cancel{18x}+33y=258\\ & \ \ 2(9x13y=115) \qquad \quad \ \underline{\cancel{18x}26y=230 \; \; \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 7y=28\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad y=4\end{align*}
Now solve for \begin{align*}x\end{align*}:
\begin{align*}6x+11(4)&=86\\ 6x+44 &=86\\ 6x &=42\\ x &= 7 \end{align*}
Solution: (7, 4)
Check your answer:
\begin{align*}6(7)+11(4) &= 42+44=86\\ 9(7)13(4) &= 6352=115 \end{align*}
Examples
Example 1
Earlier, you were asked at what point does the rental company offer the better deal.
Set up equations to represent the total cost (for one day’s rental) for each company:
Affordable Autos \begin{align*}\Rightarrow y=0.51x+30\end{align*}
Cheap cars \begin{align*}\Rightarrow y=0.57x+25\end{align*}
It is easiest to use substitution here. Substituting \begin{align*}y=0.51x+30\end{align*} into the second equation, we get:
\begin{align*}0.51x+30 = 0.57x+25\\ 5 = 0.06x\\ x = 83.333\end{align*}
Therefore, Affordable Autos has a better deal if we want to drive more than 83 and onethird miles during our oneday rental.
Solve the following systems using the most efficient method:
Example 2
\begin{align*}y &= 3x+2\\ y &= 2x3\end{align*}
This one could be solved by graphing, graphing with technology or substitution. This time we will use substitution. Since both equations are solved for \begin{align*}y\end{align*}, we can set them equal and solve for \begin{align*}x\end{align*}:
\begin{align*}3x+2 &= 2x3\\ 5 &= 5x\\ x &= 1\end{align*}
Now solve for \begin{align*}y\end{align*}:
\begin{align*}y &= 3(1)+2\\ y &= 3+2\\ y &=1 \end{align*}
Solution: (1, 1)
Example 3
\begin{align*}4x + 5y &= 5\\
x &= 2y11\end{align*}
Since the second equation here is solved for \begin{align*}x\end{align*}, it makes sense to use substitution:
\begin{align*}4(2y11)+5y &= 5\\ 8y44+5y &= 5\\ 13y &= 39\\ y &= 3\end{align*}
Now solve for \begin{align*}x\end{align*}:
\begin{align*}x &= 2(3)11\\ x &= 611\\ x &= 5\end{align*}
Solution: (5, 3)
Example 4
\begin{align*}4x5y &= 24\\
15x+7y &= 4\end{align*}
This time, both equations are in standard form so it makes the most sense to use linear combinations. We can eliminate \begin{align*}y\end{align*} by multiplying the first equation by 7 and the second equation by 5:
\begin{align*}& \quad \ 7(4x5y=24) \quad \Rightarrow \quad 28x\cancel{35y}=168\\ & 5(15x+7y=4) \qquad \quad \ \underline{75x+\cancel{35y}=20 \; \; \;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 47x = 188\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ x=4\end{align*}
Now find \begin{align*}y\end{align*}:
\begin{align*}4(4)5y &= 24\\ 165y &= 24\\ 5y &= 40\\ y &= 8 \end{align*}
Solution: (4, 8)
Review
Solve the following systems using linear combinations.
 .

 \begin{align*}5x2y &= 1\\ 8x+4y &= 56\end{align*}
 .

 \begin{align*}3x+y &= 16\\ 4xy &= 21\end{align*}
 .

 \begin{align*}7x+2y &= 4\\ y &= 4x+1\end{align*}
 .

 \begin{align*}6x+5y &= 25\\ x &= 2y+24\end{align*}
 .

 \begin{align*}8x+10y &= 1\\ 2x6y &= 2\end{align*}
 .

 \begin{align*}3x+y &= 18\\ 7x+3y &= 10\end{align*}
 .

 \begin{align*}2x+15y &= 3\\ 3x5y &= 6\end{align*}
 .

 \begin{align*}15xy &= 19\\ 13x+2y &= 48\end{align*}
 .

 \begin{align*}x &= 9y2\\ 2x15y &= 6\end{align*}
 .

 \begin{align*}3x4y &= 1\\ 2x+3y &= 1\end{align*}
 .

 \begin{align*}xy &= 2\\ 3x2y &= 7\end{align*}
 .

 \begin{align*}3x+12y &= 18\\ y &= \frac{1}{4}x\frac{3}{2}\end{align*}
 .

 \begin{align*}2x8y &= 2\\ x &= \frac{1}{2}y+10\end{align*}
 .

 \begin{align*}14x+y &= 3\\ 21x3y &= 3\end{align*}
 .

 \begin{align*}y &= \frac{4}{5}x+7\\ 8x10y &= 2\end{align*}
Solve the following word problem by creating and solving a system of linear equations.
 Jack and James each buy some small fish for their new aquariums. Jack buys 10 clownfish and 7 goldfish for $28.25. James buys 5 clownfish and 6 goldfish for $17.25. How much does each type of fish cost?
 The sum of two numbers is 35. The larger number is one less than three times the smaller number. What are the two numbers?
 Rachel offers to go to the coffee shop to buy cappuccinos and lattes for her coworkers. She buys a total of nine drinks for $35.75. If cappuccinos cost $3.75 each and the lattes cost $4.25 each, how many of each drink did she buy?
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 3.9.