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

Here are the step-by-step solutions for each part of your worksheet.

Part A: Solve Graphically

*Note: Since I cannot draw on graph paper for you, I will describe how to plot these lines and calculate the exact intersection points so you can check your work.*

System 1:
$$ \begin{cases} x - y = 5 \\ x + 5y = -7 \end{cases} $$

1. First Equation ($x - y = 5$):
* Find two points to draw the line.
* If $x = 0$, then $-y = 5 \rightarrow y = -5$. Point: $(0, -5)$.
* If $y = 0$, then $x = 5$. Point: $(5, 0)$.
* Draw a line through $(0, -5)$ and $(5, 0)$.

2. Second Equation ($x + 5y = -7$):
* Find two points.
* If $x = 3$, then $3 + 5y = -7 \rightarrow 5y = -10 \rightarrow y = -2$. Point: $(3, -2)$.
* If $x = -2$, then $-2 + 5y = -7 \rightarrow 5y = -5 \rightarrow y = -1$. Point: $(-2, -1)$.
* Draw a line through these points.

3. Intersection: The lines cross at $(3, -2)$.

System 2:
$$ \begin{cases} x + y = 4 \\ 2x - 3y = 3 \end{cases} $$

1. First Equation ($x + y = 4$):
* Points: $(0, 4)$ and $(4, 0)$. Draw a line connecting them.

2. Second Equation ($2x - 3y = 3$):
* Points: If $x=0, y=-1$ $\rightarrow (0, -1)$. If $x=3, 6-3y=3 \rightarrow y=1$ $\rightarrow (3, 1)$. Draw a line connecting them.

3. Intersection: The lines cross at $(3, 1)$.

---

Part B: Substitution Method

*Instructions: Solve one variable in terms of the other, substitute it into the second equation, and solve.*

a.
$$ \begin{cases} x = -y + 7 \\ x - y = -9 \end{cases} $$

1. The first equation is already solved for $x$: $x = -y + 7$.
2. Substitute $(-y + 7)$ for $x$ in the second equation:
$$ (-y + 7) - y = -9 $$
3. Combine like terms:
$$ -2y + 7 = -9 $$
4. Subtract 7 from both sides:
$$ -2y = -16 $$
5. Divide by -2:
$$ y = 8 $$
6. Plug $y = 8$ back into the first equation to find $x$:
$$ x = -(8) + 7 $$
$$ x = -1 $$
Solution: $(-1, 8)$

b.
$$ \begin{cases} y = 2x \\ 4x + 3y = 20 \end{cases} $$

1. The first equation is already solved for $y$: $y = 2x$.
2. Substitute $(2x)$ for $y$ in the second equation:
$$ 4x + 3(2x) = 20 $$
3. Multiply:
$$ 4x + 6x = 20 $$
4. Combine terms:
$$ 10x = 20 $$
5. Divide by 10:
$$ x = 2 $$
6. Plug $x = 2$ back into the first equation to find $y$:
$$ y = 2(2) $$
$$ y = 4 $$
Solution: $(2, 4)$

c.
$$ \begin{cases} y = 2x + 5 \\ 3x - 2y = -5 \end{cases} $$

1. The first equation is already solved for $y$: $y = 2x + 5$.
2. Substitute $(2x + 5)$ for $y$ in the second equation:
$$ 3x - 2(2x + 5) = -5 $$
3. Distribute the -2:
$$ 3x - 4x - 10 = -5 $$
4. Combine terms:
$$ -x - 10 = -5 $$
5. Add 10 to both sides:
$$ -x = 5 $$
6. Multiply by -1:
$$ x = -5 $$
7. Plug $x = -5$ back into the first equation to find $y$:
$$ y = 2(-5) + 5 $$
$$ y = -10 + 5 $$
$$ y = -5 $$
Solution: $(-5, -5)$

---

Part C: Elimination Method

*Instructions: Add or subtract equations to eliminate one variable.*

a.
$$ \begin{cases} 7x - 2y = 4 \\ 5x + y = 15 \end{cases} $$

1. Look at the $y$ terms: $-2y$ and $+y$. If we multiply the bottom equation by 2, they will cancel out when added.
2. Multiply the second equation by 2:
$$ 2(5x + y = 15) \rightarrow 10x + 2y = 30 $$
3. Now add the top equation and the new bottom equation:
$$ (7x - 2y) + (10x + 2y) = 4 + 30 $$
$$ 17x = 34 $$
4. Divide by 17:
$$ x = 2 $$
5. Plug $x = 2$ into the original second equation ($5x + y = 15$):
$$ 5(2) + y = 15 $$
$$ 10 + y = 15 $$
$$ y = 5 $$
Solution: $(2, 5)$

b.
$$ \begin{cases} 5x + 2y = 6 \\ -2x + y = -6 \end{cases} $$

1. Look at the $y$ terms: $+2y$ and $+y$. If we multiply the bottom equation by -2, they will cancel out when added.
2. Multiply the second equation by -2:
$$ -2(-2x + y = -6) \rightarrow 4x - 2y = 12 $$
3. Add the top equation and the new bottom equation:
$$ (5x + 2y) + (4x - 2y) = 6 + 12 $$
$$ 9x = 18 $$
4. Divide by 9:
$$ x = 2 $$
5. Plug $x = 2$ into the original second equation ($-2x + y = -6$):
$$ -2(2) + y = -6 $$
$$ -4 + y = -6 $$
6. Add 4 to both sides:
$$ y = -2 $$
Solution: $(2, -2)$

Final Answer:
Part A:
1. Intersection at $(3, -2)$
2. Intersection at $(3, 1)$

Part B:
a. $x = -1, y = 8$
b. $x = 2, y = 4$
c. $x = -5, y = -5$

Part C:
a. $x = 2, y = 5$
b. $x = 2, y = -2$