Question

solve the linear systems of equations. write the solution as an ordered pair. check the solution.

1. $y = 7x - 10$ and $y = - 3$
2. $y = - 8$ and $y = - 2x - 12$
3. $y = 6x$ and $y = 5x + 7$
4. $y = 9x - 9$ and $y = - 9$

Explanation:

1) System: $y=7x-10$ and $y=-3$

Step1: Substitute $y=-3$ into first equation

$-3 = 7x - 10$

Step2: Solve for $x$

$7x = 10 - 3 = 7$
$x = \frac{7}{7} = 1$

Step3: Check solution $(1, -3)$

Left side of first equation: $y=-3$
Right side: $7(1)-10=7-10=-3$ (matches)
Left side of second equation: $y=-3$ (matches given)

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2) System: $y=-8$ and $y=-2x-12$

Step1: Substitute $y=-8$ into second equation

$-8 = -2x - 12$

Step2: Solve for $x$

$2x = -12 + 8 = -4$
$x = \frac{-4}{2} = -2$

Step3: Check solution $(-2, -8)$

Left side of first equation: $y=-8$ (matches given)
Right side of second equation: $-2(-2)-12=4-12=-8$ (matches)

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3) System: $y=6x$ and $y=5x+7$

Step1: Set equations equal to each other

$6x = 5x + 7$

Step2: Solve for $x$

$6x - 5x = 7$
$x = 7$

Step3: Find $y$ using $y=6x$

$y = 6(7) = 42$

Step4: Check solution $(7, 42)$

Right side of first equation: $6(7)=42$ (matches $y$)
Right side of second equation: $5(7)+7=35+7=42$ (matches $y$)

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4) System: $y=9x-9$ and $y=-9$

Step1: Substitute $y=-9$ into first equation

$-9 = 9x - 9$

Step2: Solve for $x$

$9x = -9 + 9 = 0$
$x = \frac{0}{9} = 0$

Step3: Check solution $(0, -9)$

Left side of first equation: $y=-9$
Right side: $9(0)-9=0-9=-9$ (matches)
Left side of second equation: $y=-9$ (matches given)

Answer:

1. Ordered pair: $(1, -3)$
2. Ordered pair: $(-2, -8)$
3. Ordered pair: $(7, 42)$
4. Ordered pair: $(0, -9)$