Question

the system of equations $y = -\frac{1}{5}x - 6$ and $y = -2x + 3$ is shown on the graph below.

Explanation:

Assuming the question is to find the solution (intersection point) of the system of linear equations \( y = -\frac{1}{5}x - 6 \) and \( y = -2x + 3 \).

Step1: Set the two equations equal

Since both equal \( y \), set \( -\frac{1}{5}x - 6 = -2x + 3 \)

Step2: Solve for \( x \)

Add \( 2x \) to both sides: \( -\frac{1}{5}x + 2x - 6 = 3 \)
Simplify \( x \) terms: \( \frac{-1 + 10}{5}x - 6 = 3 \) → \( \frac{9}{5}x - 6 = 3 \)
Add 6 to both sides: \( \frac{9}{5}x = 9 \)
Multiply both sides by \( \frac{5}{9} \): \( x = 9 \times \frac{5}{9} = 5 \)

Step3: Find \( y \) using \( x = 5 \) in \( y = -2x + 3 \)

Substitute \( x = 5 \): \( y = -2(5) + 3 = -10 + 3 = -7 \)

Wait, but looking at the graph, maybe I made a mistake. Wait, let's re - check the first equation. The original first equation is \( y = -\frac{1}{5}x - 6 \)? Wait, maybe the first equation was written wrong? Wait, looking at the graph, the red line has a y - intercept around 3? No, wait the problem says \( y = -\frac{1}{5}x - 6 \) and \( y = -2x + 3 \). Wait, maybe the first equation is \( y = -\frac{1}{5}x - 3 \)? No, the problem states \( y = -\frac{1}{5}x - 6 \). Wait, let's re - solve:

\( -\frac{1}{5}x - 6=-2x + 3 \)

Add \( 2x \) to both sides: \( -\frac{1}{5}x+2x-6 = 3\)

\(2x-\frac{1}{5}x=\frac{10x - x}{5}=\frac{9x}{5}\)

So \( \frac{9x}{5}-6 = 3\)

\( \frac{9x}{5}=9\)

\(x = 5\)

Then \( y=-2(5)+3=-7\). But looking at the graph, the red line (probably \( y = - 2x + 3\)) and the blue line ( \( y=-\frac{1}{5}x - 6\)). Wait, maybe the first equation is \( y = -\frac{1}{5}x+3 \)? No, the problem says \( y = -\frac{1}{5}x - 6\).

Wait, maybe the question is to find the intersection point. But according to the calculation, \( x = 5,y=-7 \), but the graph's x - axis goes up to 10 and y - axis down to - 5. There is a discrepancy. Wait, maybe the first equation is \( y=-\frac{1}{5}x + 3\)? Let's try that.

If \( y = -\frac{1}{5}x+3 \) and \( y=-2x + 3 \)

Set equal: \( -\frac{1}{5}x + 3=-2x + 3\)

Subtract 3: \( -\frac{1}{5}x=-2x\)

Add \( 2x \): \( \frac{9}{5}x = 0\) → \( x = 0,y = 3\). But that doesn't match.

Wait, going back to the problem statement. The user provided the system \( y = -\frac{1}{5}x - 6 \) and \( y=-2x + 3 \). Let's re - solve:

\( -\frac{1}{5}x-6=-2x + 3\)

\( -x-30=-10x + 15\) (multiply both sides by 5)

\( -x + 10x=15 + 30\)

\(9x = 45\)

\(x = 5\)

\(y=-2(5)+3=-7\)

But the graph shows the red line (maybe \( y=-2x + 3\)) crossing the x - axis at \( x = 1.5\) (since \( y = 0=-2x+3\) → \( x = 1.5\)) and the blue line ( \( y=-\frac{1}{5}x - 6\)) would have a y - intercept at - 6, which is below the graph. So maybe there was a typo in the equation. But based on the given equations:

The solution to the system \(

$$\begin{cases}y = -\frac{1}{5}x - 6\\y=-2x + 3\end{cases}$$

\) is \( x = 5,y=-7 \). But if we consider the graph, maybe the first equation is \( y = -\frac{1}{5}x+3 \). Let's assume that the first equation was a typo and it's \( y = -\frac{1}{5}x+3 \). Then:

Step1: Set \( -\frac{1}{5}x + 3=-2x + 3\)

Step2: Subtract 3 from both sides: \( -\frac{1}{5}x=-2x\)

Step3: Add \( 2x \) to both sides: \( \frac{9}{5}x = 0\) → \( x = 0\)

Step4: Substitute \( x = 0\) into \( y=-2x + 3\), \( y = 3\)

But since the problem states the first equation as \( y = -\frac{1}{5}x - 6\), the solution via algebra is \( (5,-7) \). However, this doesn't match the graph. There might be an error in the problem statement. But following the given equations:

Answer:

The solution to the system is \( (5,-7) \) (If we follow the given equations. If there is a typo in the first equation, the answer may vary).