SOLUTION: Solving systems of equations by graphing - Studypool - Free Printable

We are asked to solve each system of equations by graphing. Since we can't draw on the image directly, I'll guide you through solving each system algebraically (to find the point of intersection), and then explain how to graph it step-by-step. The solution to a system by graphing is the point where the two lines intersect.

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🔧 General Strategy:

1. Convert each equation to slope-intercept form: $ y = mx + b $
2. Plot both lines on the same coordinate plane.
3. Find the point of intersection — that’s the solution.
4. If lines are parallel → no solution; if they're the same line → infinitely many solutions.

Let’s solve each one:

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1)

$$
\begin{align*}
(1)\quad & 2x + 3y = -6 \\
(2)\quad & -2x + 3y = 6
\end{align*}
$$

#### Step 1: Solve for $ y $

From (1):
$ 3y = -2x - 6 \Rightarrow y = -\frac{2}{3}x - 2 $

From (2):
$ 3y = 2x + 6 \Rightarrow y = \frac{2}{3}x + 2 $

#### Graph:
- Line 1: slope $-\frac{2}{3}$, y-intercept $-2$
- Line 2: slope $\frac{2}{3}$, y-intercept $2$

They have different slopes, so they will intersect.

#### Find intersection:
Set equations equal:
$$
-\frac{2}{3}x - 2 = \frac{2}{3}x + 2
$$
Add $ \frac{2}{3}x $ to both sides:
$$
-2 = \frac{4}{3}x + 2
$$
Subtract 2:
$$
-4 = \frac{4}{3}x \Rightarrow x = -3
$$
Plug into one equation:
$ y = -\frac{2}{3}(-3) - 2 = 2 - 2 = 0 $

✔ Solution: $ (-3, 0) $

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2)

$$
\begin{align*}
(1)\quad & -2x + 3y = -9 \\
(2)\quad & x + y = 2
\end{align*}
$$

#### Solve for $ y $:

(1): $ 3y = 2x - 9 \Rightarrow y = \frac{2}{3}x - 3 $

(2): $ y = -x + 2 $

#### Set equal:
$$
\frac{2}{3}x - 3 = -x + 2
$$
Add $ x $:
$ \frac{5}{3}x - 3 = 2 $

Add 3:
$ \frac{5}{3}x = 5 \Rightarrow x = 3 $

Then $ y = -3 + 2 = -1 $

✔ Solution: $ (3, -1) $

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3)

$$
\begin{align*}
(1)\quad & -x + 3y = 6 \\
(2)\quad & 4x + 3y = -9
\end{align*}
$$

Solve for $ y $:

(1): $ 3y = x + 6 \Rightarrow y = \frac{1}{3}x + 2 $

(2): $ 3y = -4x - 9 \Rightarrow y = -\frac{4}{3}x - 3 $

Set equal:
$$
\frac{1}{3}x + 2 = -\frac{4}{3}x - 3
$$
Add $ \frac{4}{3}x $:
$ \frac{5}{3}x + 2 = -3 $

Subtract 2:
$ \frac{5}{3}x = -5 \Rightarrow x = -3 $

Then $ y = \frac{1}{3}(-3) + 2 = -1 + 2 = 1 $

✔ Solution: $ (-3, 1) $

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4)

$$
\begin{align*}
(1)\quad & -x + 3y = 6 \\
(2)\quad & -5x + 3y = 6
\end{align*}
$$

Solve for $ y $:

(1): $ 3y = x + 6 \Rightarrow y = \frac{1}{3}x + 2 $

(2): $ 3y = 5x + 6 \Rightarrow y = \frac{5}{3}x + 2 $

Same y-intercept, different slopes → intersect.

Set equal:
$$
\frac{1}{3}x + 2 = \frac{5}{3}x + 2
\Rightarrow \frac{1}{3}x = \frac{5}{3}x
\Rightarrow 0 = \frac{4}{3}x \Rightarrow x = 0
$$

Then $ y = \frac{1}{3}(0) + 2 = 2 $

✔ Solution: $ (0, 2) $

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5)

$$
\begin{align*}
(1)\quad & -5x + 4y = -16 \\
(2)\quad & x + 4y = 8
\end{align*}
$$

Solve for $ y $:

(1): $ 4y = 5x - 16 \Rightarrow y = \frac{5}{4}x - 4 $

(2): $ 4y = -x + 8 \Rightarrow y = -\frac{1}{4}x + 2 $

Set equal:
$$
\frac{5}{4}x - 4 = -\frac{1}{4}x + 2
\Rightarrow \frac{6}{4}x = 6 \Rightarrow \frac{3}{2}x = 6 \Rightarrow x = 4
$$

Then $ y = -\frac{1}{4}(4) + 2 = -1 + 2 = 1 $

✔ Solution: $ (4, 1) $

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6)

$$
\begin{align*}
(1)\quad & 2x + y = 2 \\
(2)\quad & -x + y = -4
\end{align*}
$$

Solve for $ y $:

(1): $ y = -2x + 2 $

(2): $ y = x - 4 $

Set equal:
$$
-2x + 2 = x - 4
\Rightarrow -3x = -6 \Rightarrow x = 2
$$

Then $ y = -2(2) + 2 = -4 + 2 = -2 $

✔ Solution: $ (2, -2) $

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7)

$$
\begin{align*}
(1)\quad & 5x + 3y = 12 \\
(2)\quad & -x + 3y = -6
\end{align*}
$$

Solve for $ y $:

(1): $ 3y = -5x + 12 \Rightarrow y = -\frac{5}{3}x + 4 $

(2): $ 3y = x - 6 \Rightarrow y = \frac{1}{3}x - 2 $

Set equal:
$$
-\frac{5}{3}x + 4 = \frac{1}{3}x - 2
\Rightarrow -\frac{6}{3}x = -6 \Rightarrow -2x = -6 \Rightarrow x = 3
$$

Then $ y = \frac{1}{3}(3) - 2 = 1 - 2 = -1 $

✔ Solution: $ (3, -1) $

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8)

$$
\begin{align*}
(1)\quad & -x + 2y = -6 \\
(2)\quad & 3x + 2y = 2
\end{align*}
$$

Solve for $ y $:

(1): $ 2y = x - 6 \Rightarrow y = \frac{1}{2}x - 3 $

(2): $ 2y = -3x + 2 \Rightarrow y = -\frac{3}{2}x + 1 $

Set equal:
$$
\frac{1}{2}x - 3 = -\frac{3}{2}x + 1
\Rightarrow \frac{4}{2}x = 4 \Rightarrow 2x = 4 \Rightarrow x = 2
$$

Then $ y = \frac{1}{2}(2) - 3 = 1 - 3 = -2 $

✔ Solution: $ (2, -2) $

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✔ Final Answers (Solutions):

| Problem | Solution |
|--------|----------|
| 1) | $ (-3, 0) $ |
| 2) | $ (3, -1) $ |
| 3) | $ (-3, 1) $ |
| 4) | $ (0, 2) $ |
| 5) | $ (4, 1) $ |
| 6) | $ (2, -2) $ |
| 7) | $ (3, -1) $ |
| 8) | $ (2, -2) $ |

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📌 How to Graph Each System:

For each problem:
1. Use the slope-intercept form ($ y = mx + b $) to plot the y-intercept.
2. Use the slope to plot another point.
3. Draw the line.
4. Repeat for the second equation.
5. The intersection point is your answer.

Example for #1:
- Line 1: $ y = -\frac{2}{3}x - 2 $: Start at (0, -2), go down 2, right 3.
- Line 2: $ y = \frac{2}{3}x + 2 $: Start at (0, 2), go up 2, right 3.
- They cross at $ (-3, 0) $

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