9. \begin{cases} y = x^2 + 3x + 1 \\ y = -x + 1 \end{cases} 10. \begin{cases} y = x^2 + 1 \\ y = -2x \end{cases} 11. graph of a parabola (blue) and a line (red) on a coordinate plane with x-axis from -6 to positive, y-axis with markings, grid lines

Question

Accepted Answer

Let's solve problem 9 first. The system of equations is:

$$
\begin{cases}
y = x^2 + 3x + 1 \
y = -x + 1
\end{cases}
$$

# Explanation:
## Step 1: Set the two equations equal to each other
Since both expressions equal $ y $, we can set them equal:
$ x^2 + 3x + 1 = -x + 1 $

## Step 2: Simplify the equation
Subtract $ -x + 1 $ from both sides:
$ x^2 + 3x + 1 - (-x + 1) = 0 $
$ x^2 + 3x + 1 + x - 1 = 0 $
$ x^2 + 4x = 0 $

## Step 3: Factor the equation
Factor out $ x $:
$ x(x + 4) = 0 $

## Step 4: Solve for $ x $
Set each factor equal to zero:
$ x = 0 $ or $ x + 4 = 0 $
So $ x = 0 $ or $ x = -4 $

## Step 5: Find the corresponding $ y $-values
For $ x = 0 $:
Substitute into $ y = -x + 1 $: $ y = -0 + 1 = 1 $

For $ x = -4 $:
Substitute into $ y = -x + 1 $: $ y = -(-4) + 1 = 4 + 1 = 5 $

Now let's solve problem 10. The system of equations is:

$$
\begin{cases}
y = x^2 + 1 \
y = -2x
\end{cases}
$$

## Step 1: Set the two equations equal to each other
$ x^2 + 1 = -2x $

## Step 2: Simplify the equation
Add $ 2x $ to both sides:
$ x^2 + 2x + 1 = 0 $

## Step 3: Factor the equation
This is a perfect square trinomial:
$ (x + 1)^2 = 0 $

## Step 4: Solve for $ x $
$ x + 1 = 0 $ so $ x = -1 $

## Step 5: Find the corresponding $ y $-value
Substitute $ x = -1 $ into $ y = -2x $:
$ y = -2(-1) = 2 $

For problem 11, we need to find the intersection points of the parabola (blue) and the line (red) from the graph.

Looking at the graph:
- The line intersects the parabola at two points. Let's find the coordinates.
- From the graph, when $ x = -5 $, let's check the $ y $-value. Wait, maybe better to see the grid. Wait, the red line has a slope. Let's assume the red line is $ y = x - 1 $ or something? Wait, no, looking at the graph, the red line passes through (0, -1) and (1, 0)? Wait, no, the grid: O is the origin (0,0). The red line goes through (0, -1) and (1, 0)? Wait, maybe the red line is $ y = x - 1 $? Wait, no, let's check the intersection points.

Wait, the blue parabola: let's find its equation. The vertex is at (-2, 3) maybe? Wait, the graph shows the parabola opening downward, vertex at (-2, 3), passing through (0, 0) and (-5, 0)? Wait, no, the x-intercepts: one at (-5, 0) and (0, 0)? Wait, the blue parabola crosses the x-axis at (-5, 0) and (0, 0), vertex at (-2.5, 3.125)? Wait, maybe the equation is $ y = -x^2 - 5x $? Let's check: when $ x = 0 $, $ y = 0 $; when $ x = -5 $, $ y = -25 + 25 = 0 $. Vertex at $ x = -b/(2a) = -(-5)/(2*(-1)) = 5/(-2) = -2.5 $, $ y = -(-2.5)^2 -5*(-2.5) = -6.25 + 12.5 = 6.25 $. But the graph shows the vertex at (-2, 3)? Maybe my estimation is off.

But the red line: let's see, it passes through (0, -1) and (1, 0), so slope 1, equation $ y = x - 1 $.

Now find intersection of $ y = -x^2 -5x $ and $ y = x - 1 $:

Set equal: $ -x^2 -5x = x - 1 $

$ -x^2 -6x + 1 = 0 $

$ x^2 +6x -1 = 0 $

Using quadratic formula: $ x = [-6 ± √(36 + 4)]/2 = [-6 ± √40]/2 = [-6 ± 2√10]/2 = -3 ± √10 $

But from the graph, the intersection points seem to be at (-5, -6) and (0, -1)? Wait, maybe the red line is $ y = x - 1 $, and the blue parabola is $ y = -x^2 -5x $. Let's check $ x = 0 $: $ y = 0 $ (parabola) and $ y = -1 $ (line) – no, that's not. Wait, maybe the blue parabola is $ y = -x^2 - 3x $? Let's check: when $ x = 0 $, $ y = 0 $; when $ x = -3 $, $ y = -9 + 9 = 0 $. Vertex at $ x = -(-3)/(2*(-1)) = 3/(-2) = -1.5 $, $ y = -(-1.5)^2 -3*(-1.5) = -2.25 + 4.5 = 2.25 $. Still not matching.

Alternatively, maybe the red line is $ y = x - 1 $, and the blue parabola is $ y = -x^2 - 4x $. Let's check: $ x = 0 $, $ y = 0 $; $ x = -4 $, $ y = -16 + 16 = 0 $. Vertex at $ x = -(-4)/(2*(-1)) = 4/(-2) = -2 $, $ y = -(-2)^2 -4*(-2) = -4 + 8 = 4 $. Ah, that matches the graph: vertex at (-2, 4), x-intercepts at (-4, 0) and (0, 0). Yes! So the blue parabola is $ y = -x^2 -4x $, and the red line: let's find its equation. The red line passes through (0, -1) and (1, 0)? Wait, no, in the graph, the red line intersects the blue parabola at two points. Let's find the intersection points of $ y = -x^2 -4x $ and the red line.

Looking at the graph, the red line goes through (0, -1) and (1, 0), so slope 1, equation $ y = x - 1 $.

Set $ -x^2 -4x = x - 1 $

$ -x^2 -5x + 1 = 0 $

$ x^2 +5x -1 = 0 $

Solutions: $ x = [-5 ± √(25 + 4)]/2 = [-5 ± √29]/2 $

But from the graph, the intersection points seem to be at (-5, -6) and (0, -1)? Wait, when $ x = -5 $, $ y = -(-5)^2 -4*(-5) = -25 + 20 = -5 $, and $ y = -5 -1 = -6 $ – not equal. Wait, maybe the red line is $ y = x - 1 $, and the blue parabola is $ y = -x^2 -5x $. Then at $ x = -5 $, $ y = -25 +25 = 0 $, and $ y = -5 -1 = -6 $ – no.

Alternatively, maybe the red line is $ y = x - 1 $, and the blue parabola is $ y = -x^2 - 5x $. Then intersection at $ x = -5 $: $ y = 0 $ and $ y = -6 $ – no.

Wait, maybe the red line is $ y = x - 1 $, and the blue parabola is $ y = -x^2 - 4x $. Then intersection:

$ -x^2 -4x = x - 1 $

$ -x^2 -5x + 1 = 0 $

$ x^2 +5x -1 = 0 $

$ x = [-5 ± √29]/2 ≈ [-5 ± 5.385]/2 $

So $ x ≈ (0.385)/2 ≈ 0.192 $ or $ x ≈ (-10.385)/2 ≈ -5.192 $

Then $ y ≈ 0.192 - 1 ≈ -0.808 $ or $ y ≈ -5.192 - 1 ≈ -6.192 $

Which matches the graph: one intersection near (0, -1) and one near (-5, -6). So the intersection points are approximately (0.19, -0.81) and (-5.19, -6.19), but from the grid, maybe the exact points are (-5, -6) and (0, -1)? Wait, when $ x = 0 $, $ y = -0^2 -4*0 = 0 $ (parabola) and $ y = 0 -1 = -1 $ (line) – no. Maybe the red line is $ y = x - 0 $? No, it crosses the y-axis at -1.

Alternatively, maybe the red line is $ y = x - 1 $, and the blue parabola is $ y = -x^2 - 5x $. Then at $ x = -5 $, $ y = 0 $ (parabola) and $ y = -6 $ (line) – not equal. I think I made a mistake in the parabola equation. Let's re-examine the graph:

The blue parabola: vertex at (-2, 3), x-intercepts at (-5, 0) and (0, 0). So the equation is $ y = a(x + 5)x $, vertex at $ x = -b/(2a) = -5/2 = -2.5 $, but the vertex is at (-2, 3), so that's not. Wait, the graph shows the vertex at (-2, 3), so $ y = a(x + 2)^2 + 3 $. It passes through (0, 0), so:

$ 0 = a(0 + 2)^2 + 3 $

$ 0 = 4a + 3 $

$ a = -3/4 $

So the equation is $ y = -3/4(x + 2)^2 + 3 $

Simplify: $ y = -3/4(x^2 +4x +4) + 3 = -3/4x^2 -3x -3 + 3 = -3/4x^2 -3x $

Now the red line: let's find its equation. It passes through (0, -1) and (1, 0), so slope 1, equation $ y = x - 1 $

Find intersection:

$ -3/4x^2 -3x = x - 1 $

Multiply both sides by 4:

$ -3x^2 -12x = 4x - 4 $

$ -3x^2 -16x + 4 = 0 $

$ 3x^2 +16x -4 = 0 $

Solutions: $ x = [-16 ± √(256 + 48)]/6 = [-16 ± √304]/6 = [-16 ± 4√19]/6 = [-8 ± 2√19]/3 $

$ √19 ≈ 4.358 $, so $ -8 + 2*4.358 ≈ -8 + 8.716 ≈ 0.716 $, divided by 3 ≈ 0.239

$ -8 - 8.716 ≈ -16.716 $, divided by 3 ≈ -5.572

Then $ y = 0.239 - 1 ≈ -0.761 $ and $ y = -5.572 - 1 ≈ -6.572 $

So the intersection points are approximately (0.24, -0.76) and (-5.57, -6.57). From the grid, maybe the exact points are (-5, -6) and (0, -1), but that's an approximation.

# Answer:
Problem 9: The solutions are $ (0, 1) $ and $ (-4, 5) $

Problem 10: The solution is $ (-1, 2) $

Problem 11: The intersection points are approximately $ (-5.57, -6.57) $ and $ (0.24, -0.76) $ (or the exact points from the graph's grid, possibly $ (-5, -6) $ and $ (0, -1) $ if estimated from the grid)

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QUESTION IMAGE

Question

Explanation:

Step 1: Set the two equations equal to each other

Step 2: Simplify the equation

Step 3: Factor the equation

Step 4: Solve for \( x \)

Step 5: Find the corresponding \( y \)-values

Step 1: Set the two equations equal to each other

Step 2: Simplify the equation

Step 3: Factor the equation

Step 4: Solve for \( x \)

Step 5: Find the corresponding \( y \)-value

Answer: