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3 years ago

Solve the equation. 8b – 7 = 7b – 2

Mathematics

1 answer:

lidiya [134]3 years ago

8 0

<u> 8b - 7 = 7b - 2</u>

Subtract 7b from each side: b - 7 = -2

Add 7 to each side: <em> b = 5</em>

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What is the difference? StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction Sta

katovenus [111]

Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

$\frac{x}{x^2-16}-\frac{3}{x-4}$

To solve the given expression

$\frac{x}{x^2-16}-\frac{3}{x-4}$

$=\frac{x}{x^2-4^2}-\frac{3}{x-4}$

$=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}$ ( using the property $a^2-b^2=(a+b)(a-b)$ )

$=\frac{x-3(x+4)}{(x+4)(x-4)}$

$=\frac{x-3x-12}{(x+4)(x-4)}$ ( by using distributive property )

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

$=\frac{-2(x+6)}{(x+4)(x-4)}$

$\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is $\frac{-2(x+6)}{(x+4)(x-4)}$

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-(5a+6)=2(3a+8) what is the answer

ValentinkaMS [17]

Answer:

a = -2

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-(5a+6)=2(3a+8)

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