Question

1. $\

$$\begin{cases} 5x + 2y = 7 \\\\ 4x + y = 8 \\end{cases}$$

$

1. $\

$$\begin{cases} x - 2y = 3 \\\\ 4x - 8y = 12 \\end{cases}$$

$

1. $\

$$\begin{cases} 2x + y = -2 \\\\ 5x + 3y = -8 \\end{cases}$$

$

1. $\

$$\begin{cases} 2x - 3y = -24 \\\\ x + 6y = 18 \\end{cases}$$

$

1. $\

$$\begin{cases} 8x - y = -6 \\\\ 2x - 3y = 4 \\end{cases}$$

$

1. $\

$$\begin{cases} x + 2y = -2 \\\\ 3x + 4y = 6 \\end{cases}$$

$

Explanation:

Problem 9:

Step1: Isolate $y$ from Eq2

$y = 8 - 4x$

Step2: Substitute $y$ into Eq1

$5x + 2(8 - 4x) = 7$

Step3: Simplify to solve for $x$

$5x + 16 - 8x = 7 \implies -3x = -9 \implies x = 3$

Step4: Find $y$ with $x=3$

$y = 8 - 4(3) = 8 - 12 = -4$

Problem 10:

Step1: Multiply Eq1 by 4

$4x - 8y = 12$

Step2: Compare to Eq2

$4x - 8y = 12$ and $4x - 8y = 12$ are identical, so infinitely many solutions.

Problem 11:

Step1: Isolate $y$ from Eq1

$y = -2 - 2x$

Step2: Substitute $y$ into Eq2

$5x + 3(-2 - 2x) = -8$

Step3: Simplify to solve for $x$

$5x - 6 - 6x = -8 \implies -x = -2 \implies x = 2$

Step4: Find $y$ with $x=2$

$y = -2 - 2(2) = -2 - 4 = -6$

Problem 12:

Step1: Multiply Eq1 by 2

$4x - 6y = -48$

Step2: Add to Eq2 to eliminate $y$

$(4x - 6y) + (x + 6y) = -48 + 18 \implies 5x = -30 \implies x = -6$

Step3: Find $y$ with $x=-6$

$-6 + 6y = 18 \implies 6y = 24 \implies y = 4$

Problem 13:

Step1: Isolate $y$ from Eq1

$y = 8x + 6$

Step2: Substitute $y$ into Eq2

$2x - 3(8x + 6) = 4$

Step3: Simplify to solve for $x$

$2x - 24x - 18 = 4 \implies -22x = 22 \implies x = -1$

Step4: Find $y$ with $x=-1$

$y = 8(-1) + 6 = -8 + 6 = -2$

Problem 14:

Step1: Multiply Eq1 by 2

$2x + 4y = -4$

Step2: Subtract from Eq2 to eliminate $y$

$(3x + 4y) - (2x + 4y) = 6 - (-4) \implies x = 10$

Step3: Find $y$ with $x=10$

$10 + 2y = -2 \implies 2y = -12 \implies y = -6$

Answer:

1. $x=3$, $y=-4$
2. Infinitely many solutions (all points on $x-2y=3$)
3. $x=2$, $y=-6$
4. $x=-6$, $y=4$
5. $x=-1$, $y=-2$
6. $x=10$, $y=-6$