Question

1. $\begin{cases}2x - 5y = -22\\x + 7y = 27end{cases}$2. $\begin{cases}y = -3x - 13\\9x - 2y = -4end{cases}$3. $\begin{cases}x + y = 15\\4x + 4y = 10end{cases}$4. $\begin{cases}x - 3y = -15\\2x + y = 5end{cases}$5. $\begin{cases}x - 2y = -8\\6x + 7y = 9end{cases}$6. $\begin{cases}x + 2y = -9\\5x + 6y = -13end{cases}$ $\boldsymbol{x=7}$7. $\begin{cases}3x + y = 19\\4x - 8y = 16end{cases}$8. $\begin{cases}x - 4y = 31\\x - y = 4end{cases}$9. $\begin{cases}3x - 5y = -12\\y = -x - 4end{cases}$10. $\begin{cases}2x + 6y = -28\\x + 3y = -14end{cases}$

Explanation:

Problem 1

Step1: Isolate $x$ from Eq2

From $x + 7y = 27$, get $x = 27 - 7y$

Step2: Substitute $x$ into Eq1

$2(27 - 7y) - 5y = -22$
$54 - 14y - 5y = -22$
$-19y = -76$
$y = 4$

Step3: Solve for $x$

$x = 27 - 7(4) = 27 - 28 = -1$

Problem 2

Step1: Substitute $y$ into Eq2

$9x - 2(-3x -13) = -4$
$9x + 6x + 26 = -4$

Step2: Solve for $x$

$15x = -30$
$x = -2$

Step3: Solve for $y$

$y = -3(-2) -13 = 6 -13 = -7$

Problem 3

Step1: Simplify Eq2

Divide $4x + 4y = 10$ by 4: $x + y = \frac{10}{4} = 2.5$

Step2: Compare with Eq1

Eq1: $x + y =15$, Eq2: $x + y=2.5$. Contradiction, no solution.

Problem 4

Step1: Isolate $y$ from Eq2

$y = 5 - 2x$

Step2: Substitute $y$ into Eq1

$x - 3(5 - 2x) = -15$
$x -15 +6x = -15$
$7x = 0$
$x=0$

Step3: Solve for $y$

$y =5 -2(0)=5$

Problem 5

Step1: Isolate $x$ from Eq1

$x = 2y -8$

Step2: Substitute $x$ into Eq2

$6(2y -8) +7y =9$
$12y -48 +7y =9$
$19y =57$
$y=3$

Step3: Solve for $x$

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

Problem 6

Step1: Eliminate $x$: Multiply Eq1 by 5

$5x +10y = -45$

Step2: Subtract Eq2 from new Eq1

$(5x+10y)-(5x+6y)=-45-(-13)$
$4y=-32$
$y=-8$

Step3: Solve for $x$

$x +2(-8)=-9$
$x -16=-9$
$x=7$

Problem 7

Step1: Isolate $y$ from Eq1

$y=19-3x$

Step2: Substitute $y$ into Eq2

$4x -8(19-3x)=16$
$4x -152 +24x=16$
$28x=168$
$x=6$

Step3: Solve for $y$

$y=19-3(6)=19-18=1$

Problem 8

Step1: Subtract Eq2 from Eq1

$(x-4y)-(x-y)=31-4$
$-3y=27$
$y=-9$

Step2: Solve for $x$

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

Problem 9

Step1: Substitute $y$ into Eq1

$3x -5(-x-4)=-12$
$3x +5x +20=-12$

Step2: Solve for $x$

$8x=-32$
$x=-4$

Step3: Solve for $y$

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

Problem 10

Step1: Simplify Eq1

Divide $2x+6y=-28$ by 2: $x+3y=-14$

Step2: Compare with Eq2

Eq1 and Eq2 are identical, infinitely many solutions: $x=-14-3y$, $y$ is any real number.

Answer:

1. $x=-1,\ y=4$
2. $x=-2,\ y=-7$
3. No solution
4. $x=0,\ y=5$
5. $x=-2,\ y=3$
6. $x=7,\ y=-8$
7. $x=6,\ y=1$
8. $x=-5,\ y=-9$
9. $x=-4,\ y=0$
10. Infinitely many solutions: $x=-14-3y$ (where $y$ is any real number)