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Fed [463]
3 years ago
5

How do you solve 23=-7-5x

Mathematics
1 answer:
melomori [17]3 years ago
4 0

23=−7−5x

23=−5x−7

−5x−7=23

−5x−7+7=23+7

−5x=30

(−5x/−5)=(30/−5)

x=−6

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Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –
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Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

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Let

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<em>Alternative Method</em>

The formula to solve a quadratic equation of the form

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in this problem we have

(u)^2+(u)-2=0

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substitute in the formula

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u=\frac{-1-3} {2}=-2

the solutions are

u=-2,u=1

<em>Find the solutions of  the original equation</em>

For u=-2

-2=(x+3) ----> x=-2-3=-5

For u=1

1=(x+3) ----> x=1-3=-2

therefore

The solutions of the original equation are

x=-5 and x=-2

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