Question

1. \\(\

$$\begin{cases}x + y = 1\\\\-2x + y = 4\\end{cases}$$

\\) 11. \\(\

$$\begin{cases}4x - 9y = -21\\\\4x + 3y = -9\\end{cases}$$

\\) 13. \\(\

$$\begin{cases}x + 2y = 13\\\\-x + y = 5\\end{cases}$$

\\) 14. \\(\

$$\begin{cases}x + 4y = 47\\\\3x - 4y = -19\\end{cases}$$

\\)

Explanation:

Let's solve each system of equations one by one using the elimination or substitution method.

Problem 10

We have the system:
\[

$$\begin{cases} x + y = 1 \quad (1) \\ -2x + y = 4 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( y \)

Subtract equation (1) from equation (2):
\[
(-2x + y) - (x + y) = 4 - 1
\]
\[
-2x + y - x - y = 3
\]
\[
-3x = 3
\]

Step 2: Solve for \( x \)

Divide both sides by \(-3\):
\[
x = \frac{3}{-3} = -1
\]

Step 3: Substitute \( x = -1 \) into equation (1)

\[
-1 + y = 1
\]
Add \( 1 \) to both sides:
\[
y = 1 + 1 = 2
\]

So the solution for problem 10 is \( x = -1, y = 2 \).

Problem 11

We have the system:
\[

$$\begin{cases} 4x - 9y = -21 \quad (1) \\ 4x + 3y = -9 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( x \)

Subtract equation (1) from equation (2):
\[
(4x + 3y) - (4x - 9y) = -9 - (-21)
\]
\[
4x + 3y - 4x + 9y = -9 + 21
\]
\[
12y = 12
\]

Step 2: Solve for \( y \)

Divide both sides by \( 12 \):
\[
y = \frac{12}{12} = 1
\]

Step 3: Substitute \( y = 1 \) into equation (2)

\[
4x + 3(1) = -9
\]
\[
4x + 3 = -9
\]
Subtract \( 3 \) from both sides:
\[
4x = -9 - 3 = -12
\]
Divide both sides by \( 4 \):
\[
x = \frac{-12}{4} = -3
\]

So the solution for problem 11 is \( x = -3, y = 1 \).

Problem 13

We have the system:
\[

$$\begin{cases} x + 2y = 13 \quad (1) \\ -x + y = 5 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( x \)

Add equation (1) and equation (2):
\[
(x + 2y) + (-x + y) = 13 + 5
\]
\[
x + 2y - x + y = 18
\]
\[
3y = 18
\]

Step 2: Solve for \( y \)

Divide both sides by \( 3 \):
\[
y = \frac{18}{3} = 6
\]

Step 3: Substitute \( y = 6 \) into equation (2)

\[
-x + 6 = 5
\]
Subtract \( 6 \) from both sides:
\[
-x = 5 - 6 = -1
\]
Multiply both sides by \(-1\):
\[
x = 1
\]

So the solution for problem 13 is \( x = 1, y = 6 \).

Problem 14

We have the system:
\[

$$\begin{cases} x + 4y = 47 \quad (1) \\ 3x - 4y = -19 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( y \)

Add equation (1) and equation (2):
\[
(x + 4y) + (3x - 4y) = 47 + (-19)
\]
\[
x + 4y + 3x - 4y = 28
\]
\[
4x = 28
\]

Step 2: Solve for \( x \)

Divide both sides by \( 4 \):
\[
x = \frac{28}{4} = 7
\]

Step 3: Substitute \( x = 7 \) into equation (1)

\[
7 + 4y = 47
\]
Subtract \( 7 \) from both sides:
\[
4y = 47 - 7 = 40
\]
Divide both sides by \( 4 \):
\[
y = \frac{40}{4} = 10
\]

So the solution for problem 14 is \( x = 7, y = 10 \).

Final Answers

• Problem 10: \( \boldsymbol{x = -1, y = 2} \)
• Problem 11: \( \boldsymbol{x = -3, y = 1} \)
• Problem 13: \( \boldsymbol{x = 1, y = 6} \)
• Problem 14: \( \boldsymbol{x = 7, y = 10} \)

Answer:

Let's solve each system of equations one by one using the elimination or substitution method.

Problem 10

We have the system:
\[

$$\begin{cases} x + y = 1 \quad (1) \\ -2x + y = 4 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( y \)

Subtract equation (1) from equation (2):
\[
(-2x + y) - (x + y) = 4 - 1
\]
\[
-2x + y - x - y = 3
\]
\[
-3x = 3
\]

Step 2: Solve for \( x \)

Divide both sides by \(-3\):
\[
x = \frac{3}{-3} = -1
\]

Step 3: Substitute \( x = -1 \) into equation (1)

\[
-1 + y = 1
\]
Add \( 1 \) to both sides:
\[
y = 1 + 1 = 2
\]

So the solution for problem 10 is \( x = -1, y = 2 \).

Problem 11

We have the system:
\[

$$\begin{cases} 4x - 9y = -21 \quad (1) \\ 4x + 3y = -9 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( x \)

Subtract equation (1) from equation (2):
\[
(4x + 3y) - (4x - 9y) = -9 - (-21)
\]
\[
4x + 3y - 4x + 9y = -9 + 21
\]
\[
12y = 12
\]

Step 2: Solve for \( y \)

Divide both sides by \( 12 \):
\[
y = \frac{12}{12} = 1
\]

Step 3: Substitute \( y = 1 \) into equation (2)

\[
4x + 3(1) = -9
\]
\[
4x + 3 = -9
\]
Subtract \( 3 \) from both sides:
\[
4x = -9 - 3 = -12
\]
Divide both sides by \( 4 \):
\[
x = \frac{-12}{4} = -3
\]

So the solution for problem 11 is \( x = -3, y = 1 \).

Problem 13

We have the system:
\[

$$\begin{cases} x + 2y = 13 \quad (1) \\ -x + y = 5 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( x \)

Add equation (1) and equation (2):
\[
(x + 2y) + (-x + y) = 13 + 5
\]
\[
x + 2y - x + y = 18
\]
\[
3y = 18
\]

Step 2: Solve for \( y \)

Divide both sides by \( 3 \):
\[
y = \frac{18}{3} = 6
\]

Step 3: Substitute \( y = 6 \) into equation (2)

\[
-x + 6 = 5
\]
Subtract \( 6 \) from both sides:
\[
-x = 5 - 6 = -1
\]
Multiply both sides by \(-1\):
\[
x = 1
\]

So the solution for problem 13 is \( x = 1, y = 6 \).

Problem 14

We have the system:
\[

$$\begin{cases} x + 4y = 47 \quad (1) \\ 3x - 4y = -19 \quad (2) \end{cases}$$

\]

Step 1: Eliminate \( y \)

Add equation (1) and equation (2):
\[
(x + 4y) + (3x - 4y) = 47 + (-19)
\]
\[
x + 4y + 3x - 4y = 28
\]
\[
4x = 28
\]

Step 2: Solve for \( x \)

Divide both sides by \( 4 \):
\[
x = \frac{28}{4} = 7
\]

Step 3: Substitute \( x = 7 \) into equation (1)

\[
7 + 4y = 47
\]
Subtract \( 7 \) from both sides:
\[
4y = 47 - 7 = 40
\]
Divide both sides by \( 4 \):
\[
y = \frac{40}{4} = 10
\]

So the solution for problem 14 is \( x = 7, y = 10 \).

Final Answers

• Problem 10: \( \boldsymbol{x = -1, y = 2} \)
• Problem 11: \( \boldsymbol{x = -3, y = 1} \)
• Problem 13: \( \boldsymbol{x = 1, y = 6} \)
• Problem 14: \( \boldsymbol{x = 7, y = 10} \)