Solved Worksheet #7: Systems of Linear Equations in Two | Chegg.com - Free Printable

Let's solve each part of Worksheet #7: Systems of Linear Equations in Two Variables step by step.

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A. Solve graphically and check

We are given two systems to solve graphically. Since we can't draw graphs here, I'll explain how to graph them and find the solution, then verify algebraically.

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#### System 1:
$$
\begin{cases}
x - y = 5 \quad \text{(1)}\\
x + 5y = -7 \quad \text{(2)}
\end{cases}
$$

Step 1: Rewrite in slope-intercept form (y = mx + b)

From (1):
$$
x - y = 5 \Rightarrow -y = -x + 5 \Rightarrow y = x - 5
$$

From (2):
$$
x + 5y = -7 \Rightarrow 5y = -x - 7 \Rightarrow y = -\frac{1}{5}x - \frac{7}{5}
$$

Graphing:
- Line 1: $ y = x - 5 $ → slope = 1, y-intercept = -5
- Line 2: $ y = -\frac{1}{5}x - \frac{7}{5} $ → slope = -0.2, y-intercept = -1.4

Find intersection point (solution):

Set equations equal:
$$
x - 5 = -\frac{1}{5}x - \frac{7}{5}
$$

Multiply both sides by 5:
$$
5x - 25 = -x - 7 \\
5x + x = -7 + 25 \\
6x = 18 \Rightarrow x = 3
$$

Plug into $ y = x - 5 $:
$ y = 3 - 5 = -2 $

✔ Solution: $ (3, -2) $

Check in both equations:

1. $ x - y = 3 - (-2) = 5 $ ✔
2. $ x + 5y = 3 + 5(-2) = 3 - 10 = -7 $ ✔

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#### System 2:
$$
\begin{cases}
x + y = 4 \quad \text{(1)}\\
2x - 3y = 3 \quad \text{(2)}
\end{cases}
$$

Rewrite in slope-intercept:

(1): $ y = -x + 4 $

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

Graphing:
- Line 1: slope = -1, y-int = 4
- Line 2: slope = 2/3, y-int = -1

Solve algebraically:

Use substitution or elimination.

Substitute $ y = -x + 4 $ into (2):

$$
2x - 3(-x + 4) = 3 \\
2x + 3x - 12 = 3 \\
5x = 15 \Rightarrow x = 3
$$

Then $ y = -3 + 4 = 1 $

✔ Solution: $ (3, 1) $

Check:

1. $ x + y = 3 + 1 = 4 $ ✔
2. $ 2x - 3y = 6 - 3 = 3 $ ✔

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B. Substitution Method

#### a.
$$
\begin{cases}
x = -y + 7 \quad \text{(1)}\\
x - y = -9 \quad \text{(2)}
\end{cases}
$$

Substitute (1) into (2):
$$
(-y + 7) - y = -9 \\
-2y + 7 = -9 \\
-2y = -16 \Rightarrow y = 8
$$

Now plug back into (1):
$ x = -8 + 7 = -1 $

✔ Solution: $ (-1, 8) $

Check:
1. $ x = -y + 7 \Rightarrow -1 = -8 + 7 = -1 $ ✔
2. $ x - y = -1 - 8 = -9 $ ✔

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#### b.
$$
\begin{cases}
y = 2x \quad \text{(1)}\\
4x + 3y = 20 \quad \text{(2)}
\end{cases}
$$

Substitute (1) into (2):
$$
4x + 3(2x) = 20 \\
4x + 6x = 20 \Rightarrow 10x = 20 \Rightarrow x = 2
$$

Then $ y = 2(2) = 4 $

✔ Solution: $ (2, 4) $

Check:
1. $ y = 2x \Rightarrow 4 = 4 $ ✔
2. $ 4(2) + 3(4) = 8 + 12 = 20 $ ✔

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#### c.
$$
\begin{cases}
y = 2x + 5 \quad \text{(1)}\\
3x - 2y = -5 \quad \text{(2)}
\end{cases}
$$

Substitute (1) into (2):
$$
3x - 2(2x + 5) = -5 \\
3x - 4x - 10 = -5 \\
-x - 10 = -5 \Rightarrow -x = 5 \Rightarrow x = -5
$$

Then $ y = 2(-5) + 5 = -10 + 5 = -5 $

✔ Solution: $ (-5, -5) $

Check:
1. $ y = 2x + 5 \Rightarrow -5 = -10 + 5 = -5 $ ✔
2. $ 3(-5) - 2(-5) = -15 + 10 = -5 $ ✔

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C. Elimination Method

#### a.
$$
\begin{cases}
7x - 2y = 4 \quad \text{(1)}\\
5x + y = 15 \quad \text{(2)}
\end{cases}
$$

We want to eliminate one variable. Let's eliminate $ y $. Multiply (2) by 2:

$$
2(5x + y) = 2(15) \Rightarrow 10x + 2y = 30 \quad \text{(2')}
$$

Now add (1) and (2'):

$$
(7x - 2y) + (10x + 2y) = 4 + 30 \\
17x = 34 \Rightarrow x = 2
$$

Plug into (2): $ 5(2) + y = 15 \Rightarrow 10 + y = 15 \Rightarrow y = 5 $

✔ Solution: $ (2, 5) $

Check:
1. $ 7(2) - 2(5) = 14 - 10 = 4 $ ✔
2. $ 5(2) + 5 = 10 + 5 = 15 $ ✔

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#### b.
$$
\begin{cases}
5x + 2y = 6 \quad \text{(1)}\\
-2x + y = -6 \quad \text{(2)}
\end{cases}
$$

Eliminate one variable. Let's eliminate $ y $. Multiply (2) by 2:

$$
2(-2x + y) = 2(-6) \Rightarrow -4x + 2y = -12 \quad \text{(2')}
$$

Now subtract (2') from (1):

$$
(5x + 2y) - (-4x + 2y) = 6 - (-12) \\
5x + 2y + 4x - 2y = 6 + 12 \\
9x = 18 \Rightarrow x = 2
$$

Now plug into (2): $ -2(2) + y = -6 \Rightarrow -4 + y = -6 \Rightarrow y = -2 $

✔ Solution: $ (2, -2) $

Check:
1. $ 5(2) + 2(-2) = 10 - 4 = 6 $ ✔
2. $ -2(2) + (-2) = -4 - 2 = -6 $ ✔

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✔ Final Answers Summary

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A. Graphical Solutions

1. $ x - y = 5 $, $ x + 5y = -7 $ → $ \boxed{(3, -2)} $
2. $ x + y = 4 $, $ 2x - 3y = 3 $ → $ \boxed{(3, 1)} $

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B. Substitution Method

a. $ x = -y + 7 $, $ x - y = -9 $ → $ \boxed{(-1, 8)} $
b. $ y = 2x $, $ 4x + 3y = 20 $ → $ \boxed{(2, 4)} $
c. $ y = 2x + 5 $, $ 3x - 2y = -5 $ → $ \boxed{(-5, -5)} $

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C. Elimination Method

a. $ 7x - 2y = 4 $, $ 5x + y = 15 $ → $ \boxed{(2, 5)} $
b. $ 5x + 2y = 6 $, $ -2x + y = -6 $ → $ \boxed{(2, -2)} $

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All solutions have been checked and verified.

Let me know if you'd like a visual explanation for the graphs!