Question

$$\
LATEXBLOCK0
$$

11.

$$\
LATEXBLOCK1
$$

12.

$$\
LATEXBLOCK2
$$

$$\
LATEXBLOCK3
$$

14.

$$\
LATEXBLOCK4
$$

15.

$$\
LATEXBLOCK5
$$

Explanation:

To solve these systems of linear equations, we can use methods like substitution or elimination. Let's solve each one:

System 1:

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

Step 1: Subtract the first equation from the second to eliminate $y$.

$$(-2x + y) - (x + y) = 4 - 1$$
$$-2x + y - x - y = 3$$
$$-3x = 3$$

Step 2: Solve for $x$.

$$x = \frac{3}{-3} = -1$$

Step 3: Substitute $x = -1$ into the first equation to find $y$.

$$-1 + y = 1$$
$$y = 1 + 1 = 2$$
Solution: $x = -1$, $y = 2$

System 11:

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

Step 1: Subtract the first equation from the second to eliminate $x$.

$$(4x + 3y) - (4x - 9y) = -9 - (-21)$$
$$4x + 3y - 4x + 9y = 12$$
$$12y = 12$$

Step 2: Solve for $y$.

$$y = \frac{12}{12} = 1$$

Step 3: Substitute $y = 1$ into the second equation to find $x$.

$$4x + 3(1) = -9$$
$$4x + 3 = -9$$
$$4x = -12$$
$$x = \frac{-12}{4} = -3$$
Solution: $x = -3$, $y = 1$

System 12:

$$\begin{cases} -2x + 5y = 7 \\ 2x - 5y = -7 \end{cases}$$

Step 1: Add the two equations to eliminate $x$ and $y$.

$$(-2x + 5y) + (2x - 5y) = 7 + (-7)$$
$$0 = 0$$
This is an identity, meaning the two equations are equivalent (they represent the same line).
Solution: Infinitely many solutions (the lines are coincident).

System (next row, first):

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

Step 1: Add the two equations to eliminate $x$.

$$(x + 2y) + (-x + y) = 13 + 5$$
$$3y = 18$$

Step 2: Solve for $y$.

$$y = \frac{18}{3} = 6$$

Step 3: Substitute $y = 6$ into the second equation to find $x$.

$$-x + 6 = 5$$
$$-x = -1$$
$$x = 1$$
Solution: $x = 1$, $y = 6$

System 14:

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

Step 1: Add the two equations to eliminate $y$.

$$(x + 4y) + (3x - 4y) = 47 + (-19)$$
$$4x = 28$$

Step 2: Solve for $x$.

$$x = \frac{28}{4} = 7$$

Step 3: Substitute $x = 7$ into the first equation to find $y$.

$$7 + 4y = 47$$
$$4y = 40$$
$$y = \frac{40}{4} = 10$$
Solution: $x = 7$, $y = 10$

System 15:

$$\begin{cases} -10x + 2y = -7 \\ -10x + 2y = 0 \end{cases}$$

Step 1: Subtract the first equation from the second.

$$(-10x + 2y) - (-10x + 2y) = 0 - (-7)$$
$$0 = 7$$
This is a contradiction, meaning the lines are parallel and never intersect.
Solution: No solution.

Final Answers:

1. $\boldsymbol{x = -1}$, $\boldsymbol{y = 2}$
2. $\boldsymbol{x = -3}$, $\boldsymbol{y = 1}$
3. Infinitely many solutions

Next row (first): $\boldsymbol{x = 1}$, $\boldsymbol{y = 6}$

1. $\boldsymbol{x = 7}$, $\boldsymbol{y = 10}$
2. No solution

Answer:

To solve these systems of linear equations, we can use methods like substitution or elimination. Let's solve each one:

System 1:

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

Step 1: Subtract the first equation from the second to eliminate $y$.

$$(-2x + y) - (x + y) = 4 - 1$$
$$-2x + y - x - y = 3$$
$$-3x = 3$$

Step 2: Solve for $x$.

$$x = \frac{3}{-3} = -1$$

Step 3: Substitute $x = -1$ into the first equation to find $y$.

$$-1 + y = 1$$
$$y = 1 + 1 = 2$$
Solution: $x = -1$, $y = 2$

System 11:

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

Step 1: Subtract the first equation from the second to eliminate $x$.

$$(4x + 3y) - (4x - 9y) = -9 - (-21)$$
$$4x + 3y - 4x + 9y = 12$$
$$12y = 12$$

Step 2: Solve for $y$.

$$y = \frac{12}{12} = 1$$

Step 3: Substitute $y = 1$ into the second equation to find $x$.

$$4x + 3(1) = -9$$
$$4x + 3 = -9$$
$$4x = -12$$
$$x = \frac{-12}{4} = -3$$
Solution: $x = -3$, $y = 1$

System 12:

$$\begin{cases} -2x + 5y = 7 \\ 2x - 5y = -7 \end{cases}$$

Step 1: Add the two equations to eliminate $x$ and $y$.

$$(-2x + 5y) + (2x - 5y) = 7 + (-7)$$
$$0 = 0$$
This is an identity, meaning the two equations are equivalent (they represent the same line).
Solution: Infinitely many solutions (the lines are coincident).

System (next row, first):

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

Step 1: Add the two equations to eliminate $x$.

$$(x + 2y) + (-x + y) = 13 + 5$$
$$3y = 18$$

Step 2: Solve for $y$.

$$y = \frac{18}{3} = 6$$

Step 3: Substitute $y = 6$ into the second equation to find $x$.

$$-x + 6 = 5$$
$$-x = -1$$
$$x = 1$$
Solution: $x = 1$, $y = 6$

System 14:

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

Step 1: Add the two equations to eliminate $y$.

$$(x + 4y) + (3x - 4y) = 47 + (-19)$$
$$4x = 28$$

Step 2: Solve for $x$.

$$x = \frac{28}{4} = 7$$

Step 3: Substitute $x = 7$ into the first equation to find $y$.

$$7 + 4y = 47$$
$$4y = 40$$
$$y = \frac{40}{4} = 10$$
Solution: $x = 7$, $y = 10$

System 15:

$$\begin{cases} -10x + 2y = -7 \\ -10x + 2y = 0 \end{cases}$$

Step 1: Subtract the first equation from the second.

$$(-10x + 2y) - (-10x + 2y) = 0 - (-7)$$
$$0 = 7$$
This is a contradiction, meaning the lines are parallel and never intersect.
Solution: No solution.

Final Answers:

1. $\boldsymbol{x = -1}$, $\boldsymbol{y = 2}$
2. $\boldsymbol{x = -3}$, $\boldsymbol{y = 1}$
3. Infinitely many solutions

Next row (first): $\boldsymbol{x = 1}$, $\boldsymbol{y = 6}$

1. $\boldsymbol{x = 7}$, $\boldsymbol{y = 10}$
2. No solution