Question

ali and sid graphed this system:

$y = -\frac{1}{4}x - 1$

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

they marked its solutions with points $p$ and $r$.

which statement describes ali and sids solutions?

• their solutions are correct.
• they marked the $y$-intercepts instead of the intersection point of the two lines.
• they marked the $y$-intercepts instead of the $x$-intercepts.
• they marked only the $y$-intercepts instead of the $x$- and $y$-intercepts.

Explanation:

1. First, recall that a system of linear equations \(y = m_1x + b_1\) and \(y = m_2x + b_2\) has a solution at the point where the two lines intersect, which is found by solving the system. Also, the \(y\)-intercept of a line \(y=mx + b\) is at \(x = 0\), so the \(y\)-intercepts are \((0, b_1)\) and \((0, b_2)\).
2. For the first line \(y=-\frac{1}{4}x - 1\), the \(y\)-intercept is when \(x = 0\), so \(y=-1\), giving the point \((0, -1)\). For the second line \(y=\frac{1}{4}x - 3\), the \(y\)-intercept is when \(x = 0\), so \(y=-3\), giving the point \((0, -3)\).
3. Now, solve the system of equations \(

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

\). Set the two equations equal: \(-\frac{1}{4}x - 1=\frac{1}{4}x - 3\). Add \(\frac{1}{4}x\) to both sides: \(-1=\frac{2}{4}x - 3\) or \(-1=\frac{1}{2}x - 3\). Add 3 to both sides: \(2=\frac{1}{2}x\). Multiply both sides by 2: \(x = 4\). Substitute \(x = 4\) into \(y=\frac{1}{4}x - 3\): \(y=\frac{1}{4}(4)-3=1 - 3=-2\). So the solution (intersection point) is \((4, -2)\).

1. Ali and Sid marked points \(P\) and \(R\). If we consider the \(y\)-intercepts, they are \((0, -1)\) and \((0, -3)\). So instead of marking the intersection point (the solution of the system), they marked the \(y\)-intercepts of the two lines.

Answer:

B. They marked the \(y\)-intercepts instead of the intersection point of the two lines.