In the coordinate plane shown, line and line are graphed. For what value of is the y-coordinate of the point on line 3 greater than the y-coordinate of the point on line with the same -coordinate? Which choice gives this value of ?

Solution available with step-by-step explanation

SAT Linear Equations in Two Variables Question 39 | Free SAT Prep | Aniko

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Question 39·Hard·Linear Equations in Two Variables

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In the coordinate plane shown, line $p$ and line $q$ are graphed.

For what value of $x$ is the y-coordinate of the point on line $p$ 3 greater than the y-coordinate of the point on line $q$ with the same $x$ -coordinate?

Which choice gives this value of $x$ ?

When two lines are compared at the same $x$ -value, write each as a function ( $y_p$ and $y_q$ ). Then translate the vertical comparison into an equation like $y_p=y_q+3$ and solve the resulting single linear equation. If the graph is involved, using the labeled points to form each equation is usually faster and more accurate than estimating from the picture.

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Get each line’s equation from the graph

Use the two labeled points on each line to find its slope, then write the equation in slope-intercept form $y=mx+b$ .

Translate “3 greater” into an equation

If the point on $p$ is 3 higher than the point on $q$ for the same $x$ , write an equation relating $y_p$ and $y_q$ .

Solve one linear equation

After substitution, you should get one equation in $x$ with fractions. Clear denominators or combine like terms carefully.

Desmos Guide

Graph the two lines

Enter the two equations in the standard $x$ - $y$ plane:

$y=-\frac{5}{8}x+\frac{11}{2}$

$y=x-\frac{3}{2}$

Graph the shifted comparison line

Because $y_p=y_q+3$ , graph the line that is 3 units above $q$ :

$y=\left(x-\frac{3}{2}\right)+3$

(which simplifies to $y=x+\frac{3}{2}$ ).

Find the matching $x$ -value

Find the intersection point of the graphs from Step 1 (line $p$ ) and Step 2 (the shifted line). The $x$ -coordinate of that intersection is the requested value.

Step-by-step Explanation

Write an equation for line $p$

From the graph, line $p$ passes through $A\,(0,\tfrac{11}{2})$ and $B\,(8,\tfrac{1}{2})$ .

Its slope is

\frac{\tfrac{1}{2}-\tfrac{11}{2}}{8-0}=\frac{-\tfrac{10}{2}}{8}=\frac{-5}{8}.

Since $p$ crosses the $y$ -axis at $\tfrac{11}{2}$ , an equation for line $p$ is

y_p=-\frac{5}{8}x+\frac{11}{2}.

Write an equation for line $q$

From the graph, line $q$ passes through $C\,(0,-\tfrac{3}{2})$ and $D\,(6,\tfrac{9}{2})$ .

Its slope is

\frac{\tfrac{9}{2}-\left(-\tfrac{3}{2}\right)}{6-0}=\frac{\tfrac{12}{2}}{6}=\frac{6}{6}=1.

Since $q$ crosses the $y$ -axis at $-\tfrac{3}{2}$ , an equation for line $q$ is

y_q=x-\frac{3}{2}.

Use the “3 greater” condition and solve

“The y-coordinate on $p$ is 3 greater than the y-coordinate on $q$ (at the same $x$ )” means

y_p=y_q+3.

Substitute the two expressions:

-\frac{5}{8}x+\frac{11}{2}=\left(x-\frac{3}{2}\right)+3.

Simplify the right side and solve:

-\frac{5}{8}x+\frac{11}{2}=x+\frac{3}{2}

\frac{11}{2}-\frac{3}{2}=x+\frac{5}{8}x

4=\frac{13}{8}x.

Find $x$

Multiply both sides by $\tfrac{8}{13}$ :

x=4\cdot\frac{8}{13}=\frac{32}{13}.

So the correct choice is $\frac{32}{13}$ .

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Step-by-step Explanation

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Step-by-step Explanation