Question

use the graphs of the linear functions on the right. write the linear system whose solution set is {(6, -2)}. express each equation in the system in slope-intercept form. choose the correct answer below. a. y = (1/3)x + 4; y = x - 8 b. y = - (1/3)x + 4; y = x - 8 c. y = (1/3)x - 4; y = x - 8 d. y = - (1/3)x + 4; y = (1/3)x + 4

Explanation:

Step1: Check the first equation

We know the solution is \((6, -2)\). Let's substitute \(x = 6\) and \(y=-2\) into each option's first equation.

• Option A: \(y=\frac{1}{3}x + 4\). Substitute \(x = 6\): \(\frac{1}{3}\times6+4=2 + 4=6

eq - 2\). Eliminate A.

• Option B: \(y =-\frac{1}{3}x+4\). Substitute \(x = 6\): \(-\frac{1}{3}\times6 + 4=-2 + 4 = 2

eq - 2\)? Wait, no, wait. Wait, the solution is \((6,-2)\)? Wait, no, maybe I misread. Wait, the solution set is \(\{(6, - 2)\}\)? Wait, no, looking at the graph, maybe the intersection is \((6,4)\)? Wait, no, the options: let's check the second equation \(y=x - 8\). For \(x = 6\), \(y=6 - 8=-2\). So the second equation in B, C, A is \(y=x - 8\), which gives \(y=-2\) when \(x = 6\). Now check the first equation.
For option B: first equation \(y=-\frac{1}{3}x + 4\). Substitute \(x = 6\): \(y=-\frac{1}{3}\times6+4=-2 + 4 = 2\). But we need \(y=-2\)? Wait, no, maybe the solution is \((6,4)\)? Wait, the problem says solution set \(\{(6,-2)\}\)? Wait, maybe I made a mistake. Wait, let's re - check.
Wait, the second equation \(y=x - 8\): when \(x = 6\), \(y=6 - 8=-2\). Now check the first equation of each option with \(x = 6\) and \(y=-2\).
Option C: first equation \(y=\frac{1}{3}x-4\). Substitute \(x = 6\): \(\frac{1}{3}\times6-4=2 - 4=-2\). Hey, that works. Let's check:
First equation of C: \(y=\frac{1}{3}x - 4\). When \(x = 6\), \(y=\frac{6}{3}-4 = 2 - 4=-2\). Second equation \(y=x - 8\), when \(x = 6\), \(y=6 - 8=-2\). So both equations are satisfied at \((6,-2)\).
Let's check other options:
Option A: first equation \(y=\frac{1}{3}x + 4\), \(x = 6\) gives \(y=2 + 4 = 6
eq - 2\).
Option D: second equation is \(y=\frac{1}{3}x + 4\), not \(y=x - 8\), so eliminate D.
Option B: first equation \(y=-\frac{1}{3}x + 4\), \(x = 6\) gives \(y=-2 + 4 = 2
eq - 2\).
So option C: first equation \(y=\frac{1}{3}x-4\) (when \(x = 6\), \(y = 2-4=-2\)) and second equation \(y=x - 8\) (when \(x = 6\), \(y=6 - 8=-2\)) satisfy the solution \((6,-2)\).

Step2: Confirm the equations

The second equation in options A, B, C is \(y=x - 8\), which is correct for \(x = 6\), \(y=-2\). The first equation in option C is \(y=\frac{1}{3}x-4\), which when \(x = 6\) gives \(y=-2\), matching the solution.

Answer:

C. \(y=\frac{1}{3}x - 4\); \(y=x - 8\)