Worksheet for graphing linear systems and identifying solutions.

Let's solve each of the four linear systems by graphing and identifying their solutions. Since I can't see the actual image, I'll work through each system algebraically and describe how to graph them. Then, I’ll determine the solution (the point of intersection) for each pair of equations.

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Problem 1:

$$
\begin{align*}
y &= \frac{12}{5}x + 9 \\
y &= \frac{5}{2}x + 5
\end{align*}
$$

#### Step 1: Find the point of intersection (solution)
Set the two equations equal:
$$
\frac{12}{5}x + 9 = \frac{5}{2}x + 5
$$

Multiply both sides by 10 (LCM of 5 and 2) to eliminate denominators:
$$
10\left(\frac{12}{5}x + 9\right) = 10\left(\frac{5}{2}x + 5\right)
\Rightarrow 24x + 90 = 25x + 50
$$

Solve:
$$
90 - 50 = 25x - 24x \Rightarrow 40 = x
$$

Now plug $ x = 40 $ into one equation to find $ y $. Use the first:
$$
y = \frac{12}{5}(40) + 9 = 96 + 9 = 105
$$

✔ Solution: $ (40, 105) $

> Graphing Tip:
> - First line: slope = $ \frac{12}{5} $, y-intercept = 9 → go up 12, right 5 from (0,9)
> - Second line: slope = $ \frac{5}{2} $, y-intercept = 5 → go up 5, right 2 from (0,5)
> - They intersect at (40, 105)

---

Problem 2:

$$
\begin{align*}
y &= \frac{1}{3}x + 4 \\
y &= \frac{2}{3}x + 6
\end{align*}
$$

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

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

Now find $ y $:
$$
y = \frac{1}{3}(-6) + 4 = -2 + 4 = 2
$$

✔ Solution: $ (-6, 2) $

> Graphing Tip:
> - First line: slope $ \frac{1}{3} $, y-int 4 → up 1, right 3 from (0,4)
> - Second line: slope $ \frac{2}{3} $, y-int 6 → up 2, right 3 from (0,6)
> - Intersect at (-6, 2)

---

Problem 3:

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

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

Add $ \frac{2}{3}x $ to both sides:
$$
\frac{1}{3}x + 6 = -5
\Rightarrow \frac{1}{3}x = -11
\Rightarrow x = -33
$$

Now find $ y $:
$$
y = -\frac{1}{3}(-33) + 6 = 11 + 6 = 17
$$

✔ Solution: $ (-33, 17) $

> Graphing Tip:
> - First line: slope $ -\frac{1}{3} $, y-int 6 → down 1, right 3 from (0,6)
> - Second line: slope $ -\frac{2}{3} $, y-int -5 → down 2, right 3 from (0,-5)
> - Intersect at (-33, 17)

---

Problem 4:

$$
\begin{align*}
y &= -\frac{1}{2}x + 2 \\
y &= -\frac{1}{2}x + 3
\end{align*}
$$

These lines have the same slope ($ -\frac{1}{2} $) but different y-intercepts (2 and 3), so they are parallel and never intersect.

✔ Solution: No solution (lines are parallel)

> Graphing Tip:
> - Both lines have slope $ -\frac{1}{2} $, so same steepness
> - One starts at (0,2), other at (0,3)
> - They run parallel and never cross

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✔ Final Answers:

1. Solution: $ (40, 105) $
2. Solution: $ (-6, 2) $
3. Solution: $ (-33, 17) $
4. Solution: No solution (parallel lines)

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🔍 How to Graph Each System:

For each problem:
1. Plot the y-intercept.
2. Use the slope to find a second point.
3. Draw the line.
4. Find where the two lines cross — that’s the solution.

If the lines are parallel (same slope, different intercepts), no solution exists.

Let me know if you'd like a visual sketch or further explanation!