Question

7)
x + 2y = 5
-4x + 2y = -30
8)
8x - 2y = 58
6x - 2y = 40

Explanation:

Problem 7: Solve the system of equations $\boldsymbol{x + 2y = 5}$ and $\boldsymbol{-4x + 2y = -30}$

Step 1: Subtract the two equations to eliminate \( y \)

We have the equations:

1. \( x + 2y = 5 \)
2. \( -4x + 2y = -30 \)

Subtract equation 2 from equation 1:
\[

$$\begin{align*} (x + 2y) - (-4x + 2y) &= 5 - (-30)\\ x + 2y + 4x - 2y &= 5 + 30\\ 5x &= 35 \end{align*}$$

\]

Step 2: Solve for \( x \)

Divide both sides of \( 5x = 35 \) by 5:
\[
x = \frac{35}{5} = 7
\]

Step 3: Substitute \( x = 7 \) into the first equation to find \( y \)

Substitute \( x = 7 \) into \( x + 2y = 5 \):
\[
7 + 2y = 5
\]
Subtract 7 from both sides:
\[
2y = 5 - 7 = -2
\]
Divide both sides by 2:
\[
y = \frac{-2}{2} = -1
\]

Problem 8: Solve the system of equations $\boldsymbol{8x - 2y = 58}$ and $\boldsymbol{6x - 2y = 40}$

Step 1: Subtract the two equations to eliminate \( y \)

We have the equations:

1. \( 8x - 2y = 58 \)
2. \( 6x - 2y = 40 \)

Subtract equation 2 from equation 1:
\[

$$\begin{align*} (8x - 2y) - (6x - 2y) &= 58 - 40\\ 8x - 2y - 6x + 2y &= 18\\ 2x &= 18 \end{align*}$$

\]

Step 2: Solve for \( x \)

Divide both sides of \( 2x = 18 \) by 2:
\[
x = \frac{18}{2} = 9
\]

Step 3: Substitute \( x = 9 \) into the second equation to find \( y \)

Substitute \( x = 9 \) into \( 6x - 2y = 40 \):
\[
6(9) - 2y = 40
\]
\[
54 - 2y = 40
\]
Subtract 54 from both sides:
\[
-2y = 40 - 54 = -14
\]
Divide both sides by -2:
\[
y = \frac{-14}{-2} = 7
\]

Answer:

s:

• For problem 7: \( x = 7 \), \( y = -1 \)
• For problem 8: \( x = 9 \), \( y = 7 \)