The graph shows the lines and , which represent a system of two linear equations. A new system is formed by these two equations and the equation , where is a constant. Which choice for makes the system of three equations have exactly one solution ?

Solution available with step-by-step explanation

SAT Systems of Two Linear Equations Question 39 | Free SAT Prep | Aniko

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Question 39·Hard·Systems of Two Linear Equations in Two Variables

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The graph shows the lines $r$ and $s$ , which represent a system of two linear equations.

A new system is formed by these two equations and the equation $4x-5y=c$ , where $c$ is a constant.

Which choice for $c$ makes the system of three equations have exactly one solution $(x,y)$ ?

When a graph shows two lines, treat the solution to the system as their intersection. For a third linear equation to keep exactly one solution, it must pass through that same intersection point (otherwise there is no common solution). So: read two points on each line to write both equations, solve the two-equation system to get the intersection point, then substitute that point into the third equation’s left-hand side to determine the needed constant.

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Use two points to build each line

Each line has two labeled points. Find the slope of each line from those points, then write an equation in the form $y=mx+b$ .

Find where the two lines meet

Set the two expressions for $y$ equal to each other and solve for $x$ . Then substitute to get $y$ .

Make the third equation go through that point

The system will have exactly one solution only if the intersection point also satisfies $4x-5y=c$ . Plug the intersection point into $4x-5y$ to get $c$ .

Desmos Guide

Enter the two lines using points from the graph

From the graph, use the two points on line $r$ to compute its slope $m_r$ and then type an equation like $y=m_r x+b_r$ .

Do the same for line $s$ .

Find the intersection

Click the intersection point of the two graphed lines and record its coordinates $(x,y)$ . (Desmos will display the coordinates when you click the intersection.)

Compute $c$ from the intersection coordinates

In a new expression line, type $4x-5y$ but replace $x$ and $y$ with the intersection coordinates you found (you can type it as $4(\text{x-coordinate})-5(\text{y-coordinate})$ ).

Use the computed value to match one of the answer choices.

Step-by-step Explanation

Write an equation for line $r$ from the graph

From the graph, line $r$ passes through the points $(-6,8)$ and $(3,2)$ .

Slope:

\frac{2-8}{3-(-6)}=\frac{-6}{9}=-\frac{2}{3}

Use $(3,2)$ in $y=mx+b$ :

2=-\frac{2}{3}(3)+b=-2+b \Rightarrow b=4

So $r$ is:

y=-\frac{2}{3}x+4

Write an equation for line $s$ from the graph

From the graph, line $s$ passes through the points $(-2,-2)$ and $(4,1)$ .

Slope:

\frac{1-(-2)}{4-(-2)}=\frac{3}{6}=\frac{1}{2}

Use $(4,1)$ in $y=mx+b$ :

1=\frac{1}{2}(4)+b=2+b \Rightarrow b=-1

So $s$ is:

y=\frac{1}{2}x-1

Find the intersection point of $r$ and $s$

Set the expressions for $y$ equal:

-\frac{2}{3}x+4=\frac{1}{2}x-1

Add $\frac{2}{3}x$ to both sides and add $1$ to both sides:

5=\left(\frac{1}{2}+\frac{2}{3}\right)x=\left(\frac{3}{6}+\frac{4}{6}\right)x=\frac{7}{6}x

x=5\cdot\frac{6}{7}=\frac{30}{7}

Substitute into $y=\frac{1}{2}x-1$ :

y=\frac{1}{2}\cdot\frac{30}{7}-1=\frac{15}{7}-\frac{7}{7}=\frac{8}{7}

The intersection is $\left(\frac{30}{7},\frac{8}{7}\right)$ . This is the only solution to the original two-equation system.

Choose $c$ so the third line passes through the intersection

For the three-equation system to have exactly one solution, the intersection point must also satisfy $4x-5y=c$ .

Compute $c$ using $\left(\frac{30}{7},\frac{8}{7}\right)$ :

\begin{aligned} c&=4\left(\frac{30}{7}\right)-5\left(\frac{8}{7}\right) \\ &=\frac{120}{7}-\frac{40}{7} \\ &=\frac{80}{7} \end{aligned}

So the correct choice is $\frac{80}{7}$ .

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Step-by-step Explanation

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Step-by-step Explanation