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:

$$
LATEXBLOCK0
$$

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:

$$
LATEXBLOCK1
$$

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:

$$
LATEXBLOCK2
$$

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:

$$
LATEXBLOCK3
$$

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:

$$
LATEXBLOCK0
$$

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:

$$
LATEXBLOCK1
$$

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:

$$
LATEXBLOCK2
$$

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:

$$
LATEXBLOCK3
$$

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: \( \boldsymbol{x = 7, y = 10} \)