Question

1. $\begin{cases} 4x + 9y = 5 \\ -4x + 7y = 11 end{cases}$
2. $\begin{cases} 10x - 3y = 18 \\ -2x + 3y = 6 end{cases}$
3. $\begin{cases} x - y = 10 \\ 3x + y = 18 end{cases}$
4. $\begin{cases} x = 3y + 11 \\ 2x - 3y = 16 end{cases}$
5. $\begin{cases} 4y = 2x - 8 \\ 5x - 4y = 20 end{cases}$
6. $\begin{cases} 3x - 4y = -10 \\ 3x - 4y = -13 end{cases}$
7. $\begin{cases} 2x + y = -10 \\ -y = 2x + 10 end{cases}$

Explanation:

Problem 5

Step1: Add the two equations

$(4x + 9y) + (-4x + 7y) = 5 + 11$
$16y = 16$

Step2: Solve for $y$

$y = \frac{16}{16} = 1$

Step3: Substitute $y=1$ into first equation

$4x + 9(1) = 5$
$4x = 5 - 9 = -4$

Step4: Solve for $x$

$x = \frac{-4}{4} = -1$

Problem 6

Step1: Add the two equations

$(10x - 3y) + (-2x + 3y) = 18 + 6$
$8x = 24$

Step2: Solve for $x$

$x = \frac{24}{8} = 3$

Step3: Substitute $x=3$ into second equation

$-2(3) + 3y = 6$
$-6 + 3y = 6$
$3y = 12$

Step4: Solve for $y$

$y = \frac{12}{3} = 4$

Problem 7

Step1: Add the two equations

$(x - y) + (3x + y) = 10 + 18$
$4x = 28$

Step2: Solve for $x$

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

Step3: Substitute $x=7$ into first equation

$7 - y = 10$
$-y = 10 - 7 = 3$

Step4: Solve for $y$

$y = -3$

Problem 8

Step1: Substitute $x=3y+11$ into second equation

$2(3y + 11) - 3y = 16$
$6y + 22 - 3y = 16$
$3y + 22 = 16$

Step2: Solve for $y$

$3y = 16 - 22 = -6$
$y = \frac{-6}{3} = -2$

Step3: Substitute $y=-2$ into first equation

$x = 3(-2) + 11 = -6 + 11 = 5$

Problem 9

Step1: Rearrange first equation

$2x - 4y = 8$

Step2: Subtract rearranged first equation from second equation

$(5x - 4y) - (2x - 4y) = 20 - 8$
$3x = 12$

Step3: Solve for $x$

$x = \frac{12}{3} = 4$

Step4: Substitute $x=4$ into first equation

$4y = 2(4) - 8 = 8 - 8 = 0$

Step5: Solve for $y$

$y = \frac{0}{4} = 0$

Problem 10

Step1: Analyze the two equations

The left sides of both equations are identical ($3x-4y$), but equal to different constants ($-10$ and $-13$). This is a contradiction.

Problem 11

Step1: Rearrange the second equation

$-y = 2x + 10 \implies 2x + y = -10$

Step2: Compare with first equation

The rearranged second equation is identical to the first equation, meaning there are infinitely many solutions.

Answer:

1. $x=-1$, $y=1$
2. $x=3$, $y=4$
3. $x=7$, $y=-3$
4. $x=5$, $y=-2$
5. $x=4$, $y=0$
6. No solution (contradictory system)
7. Infinitely many solutions (dependent system, $2x + y = -10$)