The graph shows two lines, and . The solution to the system of linear equations represented by the graph is the point where the two lines intersect. Without solving for and , which choice gives the value of in the equation for that solution ?

Solution available with step-by-step explanation

SAT Systems of Two Linear Equations Question 120 | Free SAT Prep | Aniko

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Question 120·Hard·Systems of Two Linear Equations in Two Variables

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The graph shows two lines, $p$ and $q$ . The solution to the system of linear equations represented by the graph is the point $(x,y)$ where the two lines intersect.

Without solving for $x$ and $y$ , which choice gives the value of $k$ in the equation $15x + 2y = k$ for that solution $(x,y)$ ?

When you’re asked for a linear expression (like $15x+2y$ ) at the solution to a 2-variable linear system, you often don’t need $x$ and $y$ explicitly. Put each line into standard form $Ax+By=C$ , then find multipliers so that a linear combination of the left-hand sides becomes the target expression; apply the same multipliers to the constants on the right-hand sides to get the value directly.

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Write each line as an equation

Use the two labeled points on each line to write an equation for line $p$ and an equation for line $q$ .

Convert to standard form

Rewrite both line equations in the form $Ax+By=C$ . The intersection point must satisfy both equations.

Build the expression you want by combining the equations

Try multiplying the two standard-form equations by constants and adding them so that the left-hand side becomes $15x+2y$ . Whatever you do to the left-hand sides, do to the right-hand sides as well.

Desmos Guide

Enter the two lines

Enter the equations of the two lines:

$y=\frac{3}{2}x+7$
$y=-\frac{2}{3}x+\frac{20}{3}$

Find the intersection

Click the intersection point of the two graphs to display its coordinates $(x,y)$ .

Evaluate $15x+2y$

Compute $15x+2y$ using the intersection coordinates by typing (with parentheses)

$15(\text{x-coordinate})+2(\text{y-coordinate})$

and read the resulting value.

Step-by-step Explanation

Write each line in standard form

From the graph, line $p$ passes through $A(-4,1)$ and $B(2,10)$ , so

y=\frac{3}{2}x+7.

Rewrite in standard form:

3x-2y=-14.

Line $q$ passes through $C(-2,8)$ and $D(4,4)$ , so

y=-\frac{2}{3}x+\frac{20}{3}.

Rewrite in standard form:

2x+3y=20.

Use the fact that the intersection satisfies both equations

Let $(x,y)$ be the intersection. Then $(x,y)$ satisfies both

3x-2y=-14 \quad\text{and}\quad 2x+3y=20.

Any valid linear combination of these two equations is also true at $(x,y)$ .

Choose multipliers to create $15x+2y$

Find constants $a$ and $b$ so that

a(3x-2y)+b(2x+3y)=15x+2y.

Match coefficients of $x$ and $y$ :

\begin{aligned} 3a+2b&=15 \\ -2a+3b&=2 \end{aligned}

Solve this system. One efficient way is elimination:

Multiply $3a+2b=15$ by $2$ and $-2a+3b=2$ by $3$ :

\begin{aligned} 6a+4b&=30 \\ -6a+9b&=6 \end{aligned}

Add to get $13b=36$ , so $b=\frac{36}{13}$ . Then

3a+2\left(\frac{36}{13}\right)=15 \implies 3a=\frac{123}{13} \implies a=\frac{41}{13}.

Apply the same combination to the constants to get $k$

Because $3x-2y=-14$ and $2x+3y=20$ at the intersection,

15x+2y=a(-14)+b(20).

Substitute $a=\frac{41}{13}$ and $b=\frac{36}{13}$ :

\begin{aligned} 15x+2y&=\frac{41}{13}(-14)+\frac{36}{13}(20)\\ &=\frac{-574+720}{13}\\ &=\frac{146}{13}. \end{aligned}

So, $k=\frac{146}{13}$ .

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Step-by-step Explanation

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