Example: Determining Whether an Ordered Pair Is a Solution to a System of Equations

Determine whether the ordered pair [latex]\left(5,1\right)[/latex] is a solution to the given system of equations.

[latex]\begin{array}{l}x+3y=8\hfill \\ 2x - 9=y\hfill \end{array}[/latex]

Answer: Substitute the ordered pair [latex]\left(5,1\right)[/latex] into both equations.

[latex]\begin{array}{ll}\left(5\right)+3\left(1\right)=8\hfill & \hfill \\ \text{ }8=8\hfill & \text{True}\hfill \\ 2\left(5\right)-9=\left(1\right)\hfill & \hfill \\ \text{ }\text{1=1}\hfill & \text{True}\hfill \end{array}[/latex] The ordered pair [latex]\left(5,1\right)[/latex] satisfies both equations, so it is the solution to the system.

Analysis of the Solution

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.  

Example: Solving a System of Equations in Two Variables by Graphing

Solve the following system of equations by graphing. Identify the type of system.

[latex]\begin{array}{c}2x+y=-8\\ x-y=-1\end{array}[/latex]

Answer: Solve the first equation for [latex]y[/latex].

[latex]\begin{array}{c}2x+y=-8\\ y=-2x - 8\end{array}[/latex]

Solve the second equation for [latex]y[/latex].

[latex]\begin{array}{c}x-y=-1\\ y=x+1\end{array}[/latex]

Graph both equations on the same set of axes: The lines appear to intersect at the point [latex]\left(-3,-2\right)[/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.

[latex]\begin{array}{ll}2\left(-3\right)+\left(-2\right)=-8\hfill & \hfill \\ \text{ }-8=-8\hfill & \text{True}\hfill \\ \text{ }\left(-3\right)-\left(-2\right)=-1\hfill & \hfill \\ \text{ }-1=-1\hfill & \text{True}\hfill \end{array}[/latex]

The solution to the system is the ordered pair [latex]\left(-3,-2\right)[/latex], so the system is independent.

Example: Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution. [latex-display]\begin{array}{l}\text{ }-x+y=-5\hfill \\ \text{ }2x - 5y=1\hfill \end{array}[/latex-display]

Answer: First, we will solve the first equation for [latex]y[/latex]. [latex-display]\begin{array}{l}-x+y=-5\hfill \\ \text{ }y=x - 5\hfill \end{array}[/latex-display] Now we can substitute the expression [latex]x - 5[/latex] for [latex]y[/latex] in the second equation. [latex-display]\begin{array}{l}\text{ }2x - 5y=1\hfill \\ 2x - 5\left(x - 5\right)=1\hfill \\ \text{ }2x - 5x+25=1\hfill \\ \text{ }-3x=-24\hfill \\ \text{ }x=8\hfill \end{array}[/latex-display] Now, we substitute [latex]x=8[/latex] into the first equation and solve for [latex]y[/latex]. [latex-display]\begin{array}{l}-\left(8\right)+y=-5\hfill \\ \text{ }y=3\hfill \end{array}[/latex-display] Our solution is [latex]\left(8,3\right)[/latex]. Check the solution by substituting [latex]\left(8,3\right)[/latex] into both equations. [latex-display]\begin{array}{llll}-x+y=-5\hfill & \hfill & \hfill & \hfill \\ -\left(8\right)+\left(3\right)=-5\hfill & \hfill & \hfill & \text{True}\hfill \\ 2x - 5y=1\hfill & \hfill & \hfill & \hfill \\ 2\left(8\right)-5\left(3\right)=1\hfill & \hfill & \hfill & \text{True}\hfill \end{array}[/latex-display]

Example: Using the Addition Method When Multiplication of Both Equations Is Required

Solve the given system of equations in two variables by addition. [latex-display]\begin{array}{c}2x+3y=-16\\ 5x - 10y=30\end{array}[/latex-display]

Answer: One equation has [latex]2x[/latex] and the other has [latex]5x[/latex]. The least common multiple is [latex]10x[/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate [latex]x[/latex] by multiplying the first equation by [latex]-5[/latex] and the second equation by [latex]2[/latex]. [latex-display]\begin{array}{l} -5\left(2x+3y\right)=-5\left(-16\right)\hfill \\ \text{ }-10x - 15y=80\hfill \\ \text{ }2\left(5x - 10y\right)=2\left(30\right)\hfill \\ \text{ }10x - 20y=60\hfill \end{array}[/latex-display] Then, we add the two equations together. [latex-display]\begin{array}\ −10x−15y=80 \\ 10x−20y=60 \\ \text{______________} \\ \text{ }−35y=140 \\ y=−4 \end{array}[/latex-display] Substitute [latex]y=-4[/latex] into the original first equation. [latex-display]\begin{array}{c}2x+3\left(-4\right)=-16\\ 2x - 12=-16\\ 2x=-4\\ x=-2\end{array}[/latex-display] The solution is [latex]\left(-2,-4\right)[/latex]. Check it in the other equation. [latex-display]\begin{array}{r}\hfill \text{ }5x - 10y=30\\ \hfill 5\left(-2\right)-10\left(-4\right)=30\\ \hfill \text{ }-10+40=30\\ \hfill \text{ }30=30\end{array}[/latex-display]

Example: Using the Addition Method in Systems of Equations Containing Fractions

Solve the given system of equations in two variables by addition. [latex-display]\begin{array}{l}\frac{x}{3}+\frac{y}{6}=3\hfill \\ \frac{x}{2}-\frac{y}{4}=\text{ }1\hfill \end{array}[/latex-display]

Answer: First clear each equation of fractions by multiplying both sides of the equation by the least common denominator. [latex-display]\begin{array}{l}6\left(\frac{x}{3}+\frac{y}{6}\right)=6\left(3\right)\hfill \\ \text{ }2x+y=18\hfill \\ 4\left(\frac{x}{2}-\frac{y}{4}\right)=4\left(1\right)\hfill \\ \text{ }2x-y=4\hfill \end{array}[/latex-display] Now multiply the second equation by [latex]-1[/latex] so that we can eliminate the x-variable. [latex-display]\begin{array}{l}-1\left(2x-y\right)=-1\left(4\right)\hfill \\ \text{ }-2x+y=-4\hfill \end{array}[/latex-display] Add the two equations to eliminate the x-variable and solve the resulting equation. [latex-display]\begin{array}\ \hfill 2x+y=18 \\ \hfill−2x+y=−4 \\ \text{_____________} \\ \hfill 2y=14 \\ \hfill y=7 \end{array}[/latex-display] Substitute [latex]y=7[/latex] into the first equation. [latex-display]\begin{array}{l}2x+\left(7\right)=18\hfill \\ \text{ }2x=11\hfill \\ \text{ }x=\frac{11}{2}\hfill \\ \text{ }=7.5\hfill \end{array}[/latex-display] The solution is [latex]\left(\frac{11}{2},7\right)[/latex]. Check it in the other equation. [latex-display]\begin{array}{c}\frac{x}{2}-\frac{y}{4}=1\\ \frac{\frac{11}{2}}{2}-\frac{7}{4}=1\\ \frac{11}{4}-\frac{7}{4}=1\\ \frac{4}{4}=1\end{array}[/latex-display]

Example: Finding a Solution to a Dependent System of Linear Equations

Find a solution to the system of equations using the addition method.

[latex]\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}[/latex]

Answer: With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating [latex]x[/latex]. If we multiply both sides of the first equation by [latex]-3[/latex], then we will be able to eliminate the [latex]x[/latex] -variable.

[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \left(-3\right)\left(x+3y\right)=\left(-3\right)\left(2\right)\hfill \\ \text{ }-3x - 9y=-6\hfill \end{array}[/latex]

Now add the equations.

[latex]\begin{array} \hfill−3x−9y=−6 \\ \hfill+3x+9y=6 \\ \hfill \text{_____________} \\ \hfill 0=0 \end{array}[/latex]

We can see that there will be an infinite number of solutions that satisfy both equations.

Analysis of the Solution

If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.

[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \text{ }3y=-x+2\hfill \\ \text{ }y=-\frac{1}{3}x+\frac{2}{3}\hfill \\ 3x+9y=6\hfill \\ \text{ }9y=-3x+6\hfill \\ \text{ }y=-\frac{3}{9}x+\frac{6}{9}\hfill \\ \text{ }y=-\frac{1}{3}x+\frac{2}{3}\hfill \end{array}[/latex]

Look at the graph below. Notice the results are the same. The general solution to the system is [latex]\left(x, -\frac{1}{3}x+\frac{2}{3}\right)[/latex].  

Example: Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile.[footnote]Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/[/footnote] When will Keep on Trucking, Inc. be the better choice for Jamal?

Answer: The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

Input	d, distance driven in miles
Outputs	K(d): cost, in dollars, for renting from Keep on TruckingM(d) cost, in dollars, for renting from Move It Your Way
Initial Value	Up-front fee: K(0) = 20 and M(0) = 16
Rate of Change	K(d) = $0.59/mile and P(d) = $0.63/mile

A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations

[latex]\begin{array}{l}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{array}[/latex]

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\left(d\right)<M\left(d\right)[/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\left(d\right)[/latex] function is smaller. These graphs are sketched above, with K(d) in blue. To find the intersection, we set the equations equal and solve:

[latex]\begin{array}{l}K\left(d\right)=M\left(d\right)\hfill \\ 0.59d+20=0.63d+16\hfill \\ 4=0.04d\hfill \\ 100=d\hfill \\ d=100\hfill \end{array}[/latex]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\left(d\right)[/latex] is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[/latex].

Example: Solve a Chemical Mixture Problem

A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?

Answer: We will use the following table to help us solve this mixture problem:

	Amount	Part	Total
Start
Add
Final

We start with 70 mL of solution, and the unknown amount can be x. The part is the percentages, or concentration of solution 0.5 for start, 0.8 for add.

	Amount	Part	Total
Start	70mL	0.5
Add	x	0.8
Final	[latex]70+x[/latex]	0.6

Add amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.

	Amount	Part	Total
Start	70mL	0.5	35
Add	x	0.8	[latex]0.8x[/latex]
Final	[latex]70+x[/latex]	0.6	[latex]42+0.6x[/latex]

Multiply amount by part to get total. be sure to distribute on the last row:[latex](70 + x)0.6[/latex]. If we add the start and add entries in the Total column, we get the final equation that represents the total amount and it's concentration. [latex-display]\begin{array}{c}35+0.8x = 42+0.6x\\0.2x=7\\\frac{0.2}{0.2}x=\frac{7}{0.2}\\x=35\end{array}[/latex-display] 35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane. The same process can be used if the starting and final amount have a price attached to them, rather than a percentage.