Question

graphing substitution for #10-15, use any method to solve each system. graphs are provided on the back of this page. 10. y = -8x + 2 -x + 7y = 14 graphing substitution elimination 11. y = -\frac{2}{3}x + 3 y = \frac{5}{3}x - 4 graphing substitution elimination 12. (-4x + 5y = 26) / 2 8x + y = -30 -4x +13 = 52 8x +5 = -30 19 = 19 22/11 = 2

Explanation:

Let's solve each system one by one. We'll use the appropriate method (substitution, elimination, or graphing) for each.

Problem 10: \(

$$\begin{cases} y = -8x + 2 \\ -x + 7y = 14 \end{cases}$$

\)

Step 1: Substitute \( y \) from the first equation into the second equation.

We know \( y = -8x + 2 \), so substitute this into \( -x + 7y = 14 \):
\[
-x + 7(-8x + 2) = 14
\]

Step 2: Simplify and solve for \( x \).

\[
-x - 56x + 14 = 14 \\
-57x + 14 = 14 \\
-57x = 0 \\
x = 0
\]

Step 3: Substitute \( x = 0 \) back into \( y = -8x + 2 \) to find \( y \).

\[
y = -8(0) + 2 = 2
\]
So the solution for problem 10 is \( (0, 2) \).

Problem 11: \(

$$\begin{cases} y = -\frac{2}{3}x + 3 \\ y = \frac{5}{3}x - 4 \end{cases}$$

\)

Step 1: Set the two expressions for \( y \) equal to each other (since both equal \( y \)).

\[
-\frac{2}{3}x + 3 = \frac{5}{3}x - 4
\]

Step 2: Multiply both sides by 3 to eliminate fractions.

\[
-2x + 9 = 5x - 12
\]

Step 3: Solve for \( x \).

\[
9 + 12 = 5x + 2x \\
21 = 7x \\
x = 3
\]

Step 4: Substitute \( x = 3 \) into one of the original equations (e.g., \( y = -\frac{2}{3}x + 3 \)) to find \( y \).

\[
y = -\frac{2}{3}(3) + 3 = -2 + 3 = 1
\]
So the solution for problem 11 is \( (3, 1) \).

Problem 12: \(

$$\begin{cases} -4x + 5y = 26 \\ 8x + y = -30 \end{cases}$$

\)

Step 1: Let's use elimination. Multiply the first equation by 2 to make the coefficients of \( x \) opposites.

First equation multiplied by 2: \( -8x + 10y = 52 \)
Second equation: \( 8x + y = -30 \)

Step 2: Add the two equations to eliminate \( x \).

\[
(-8x + 10y) + (8x + y) = 52 + (-30) \\
11y = 22 \\
y = 2
\]

Step 3: Substitute \( y = 2 \) into the second equation \( 8x + y = -30 \) to find \( x \).

\[
8x + 2 = -30 \\
8x = -32 \\
x = -4
\]
So the solution for problem 12 is \( (-4, 2) \).

Final Answers:

• Problem 10: \( \boldsymbol{(0, 2)} \)
• Problem 11: \( \boldsymbol{(3, 1)} \)
• Problem 12: \( \boldsymbol{(-4, 2)} \)

Answer:

Let's solve each system one by one. We'll use the appropriate method (substitution, elimination, or graphing) for each.

Problem 10: \(

$$\begin{cases} y = -8x + 2 \\ -x + 7y = 14 \end{cases}$$

\)

Step 1: Substitute \( y \) from the first equation into the second equation.

We know \( y = -8x + 2 \), so substitute this into \( -x + 7y = 14 \):
\[
-x + 7(-8x + 2) = 14
\]

Step 2: Simplify and solve for \( x \).

\[
-x - 56x + 14 = 14 \\
-57x + 14 = 14 \\
-57x = 0 \\
x = 0
\]

Step 3: Substitute \( x = 0 \) back into \( y = -8x + 2 \) to find \( y \).

\[
y = -8(0) + 2 = 2
\]
So the solution for problem 10 is \( (0, 2) \).

Problem 11: \(

$$\begin{cases} y = -\frac{2}{3}x + 3 \\ y = \frac{5}{3}x - 4 \end{cases}$$

\)

Step 1: Set the two expressions for \( y \) equal to each other (since both equal \( y \)).

\[
-\frac{2}{3}x + 3 = \frac{5}{3}x - 4
\]

Step 2: Multiply both sides by 3 to eliminate fractions.

\[
-2x + 9 = 5x - 12
\]

Step 3: Solve for \( x \).

\[
9 + 12 = 5x + 2x \\
21 = 7x \\
x = 3
\]

Step 4: Substitute \( x = 3 \) into one of the original equations (e.g., \( y = -\frac{2}{3}x + 3 \)) to find \( y \).

\[
y = -\frac{2}{3}(3) + 3 = -2 + 3 = 1
\]
So the solution for problem 11 is \( (3, 1) \).

Problem 12: \(

$$\begin{cases} -4x + 5y = 26 \\ 8x + y = -30 \end{cases}$$

\)

Step 1: Let's use elimination. Multiply the first equation by 2 to make the coefficients of \( x \) opposites.

First equation multiplied by 2: \( -8x + 10y = 52 \)
Second equation: \( 8x + y = -30 \)

Step 2: Add the two equations to eliminate \( x \).

\[
(-8x + 10y) + (8x + y) = 52 + (-30) \\
11y = 22 \\
y = 2
\]

Step 3: Substitute \( y = 2 \) into the second equation \( 8x + y = -30 \) to find \( x \).

\[
8x + 2 = -30 \\
8x = -32 \\
x = -4
\]
So the solution for problem 12 is \( (-4, 2) \).

Final Answers:

• Problem 10: \( \boldsymbol{(0, 2)} \)
• Problem 11: \( \boldsymbol{(3, 1)} \)
• Problem 12: \( \boldsymbol{(-4, 2)} \)