Question

practice 4.13
problems 7–8: ali graphed this system:
\

$$\begin{cases} y = -\\frac{1}{4}x - 1 \\\\ y = \\frac{1}{4}x - 3 \\end{cases}$$

ali marked its solutions with points p and r.

1. which statement describes ali’s solutions?

a. the solutions are correct.
b. the y intercepts are marked instead of the intersection point of the two lines.
c. the y intercepts are marked instead of the x–intercepts
d. only the y–intercepts are marked instead of the x–and y–intercepts.

1. what is the solution to the system of equations?

Explanation:

Question 7

To determine the correct statement about Ali's solutions, we analyze the system of equations and the marked points (P and R). The solution to a system of linear equations is the intersection point of the two lines. Let's find the y - intercepts of each equation:

• For \(y =-\frac{1}{4}x - 1\), when \(x = 0\), \(y=-1\)? Wait, no, looking at the graph, point P is at (0, - 1)? Wait, no, the first equation \(y =-\frac{1}{4}x-1\): when \(x = 0\), \(y=-1\)? Wait, the second equation \(y=\frac{1}{4}x - 3\): when \(x = 0\), \(y=-3\). But in the graph, point R is at (0, - 3)? Wait, no, let's re - examine. The system of equations is \(

$$\begin{cases}y =-\frac{1}{4}x-1\\y=\frac{1}{4}x - 3\end{cases}$$

\). The solution of the system is found by setting the two equations equal: \(-\frac{1}{4}x-1=\frac{1}{4}x - 3\). Solving for \(x\): \(-\frac{1}{4}x-\frac{1}{4}x=-3 + 1\), \(-\frac{2}{4}x=-2\), \(-\frac{1}{2}x=-2\), \(x = 4\). Then \(y=-\frac{1}{4}(4)-1=-1 - 1=-2\). So the solution is (4, - 2). But Ali marked points P and R. Let's check the y - intercepts: for \(y =-\frac{1}{4}x-1\), y - intercept is (0, - 1) (point P?); for \(y=\frac{1}{4}x - 3\), y - intercept is (0, - 3) (point R?). So instead of marking the intersection point of the two lines, Ali marked the y - intercepts of the two lines. So option B says "The y intercepts are marked instead of the intersection point of the two lines" which matches.

Step 1: Set the two equations equal

To find the solution of the system \(

$$\begin{cases}y =-\frac{1}{4}x-1\\y=\frac{1}{4}x - 3\end{cases}$$

\), we set \(-\frac{1}{4}x-1=\frac{1}{4}x - 3\).

Step 2: Solve for \(x\)

First, move the \(x\) terms to one side: \(-\frac{1}{4}x-\frac{1}{4}x=-3 + 1\).
Simplify the left - hand side: \(-\frac{2}{4}x=-2\), or \(-\frac{1}{2}x=-2\).
Multiply both sides by - 2: \(x = 4\).

Step 3: Solve for \(y\)

Substitute \(x = 4\) into one of the equations, say \(y =-\frac{1}{4}x-1\).
\(y=-\frac{1}{4}(4)-1=-1 - 1=-2\).

Answer:

B. The y intercepts are marked instead of the intersection point of the two lines.

Question 8