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3 years ago

12

Select all of the correct solutions for the following equation

1 answer:

Luba_88 [7]3 years ago

4 0

Answer:

(1,3)

Step-by-step explanation:

2x-2=4x-y-1

subtract 2x on both sides

-2=2x-y-1

add y on both sides

y-2=2x-1

add 2 on both sides

y=2x+1

now we can just kinda plug and check.

if you plug in 1, you get 3, so A

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Soloha48 [4]

Answer:

The solutions are:

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Step-by-step explanation:

To find the solutions to the equation $\left(x^2+5x+6\right)\left(x^2-3x-4\right)=0$ you need to:

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Break the expression into groups

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Factor out $x$ from $(x^2+2x)$

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Factor out 3 from $3x+6$

$3x+6=3\left(x+2\right)$

$x^2+5x+6=x\left(x+2\right)+3\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\\\x^2+5x+6=\left(x+2\right)\left(x+3\right)$

Factor $\left(x^2-3x-4\right)$

$x^2-3x-4=\left(x^2+x\right)+\left(-4x-4\right)\\\\x^2-3x-4=x\left(x+1\right)-4\left(x+1\right)\\\\x^2-3x-4=\left(x+1\right)\left(x-4\right)$

Therefore

$\left(x^2+5x+6\right)\left(x^2-3x-4\right)=\left(x+2\right)\left(x+3\right)\left(x+1\right)\left(x-4\right)=0$

Using the Zero Factor Theorem:

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