Question #d9f0b

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Question #d9f0b

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Answer 1 · 2018-02-04T18:49:31

$\frac{2}{2 x + 11} - \frac{1}{x - 4} = \frac{1}{2}$ $2/(2x+11)-1/(x-4)=1/2$

$\frac{2 (x - 4) - 1 (2 x + 11)}{(2 x + 11) (x - 4)} = \frac{1}{2}$ $(2(x-4) - 1(2x+11))/((2x+11)(x-4) )=1/2$

$\frac{2 x - 8 - 2 x - 11}{2 x^{2} + 11 x - 8 x - 44} = \frac{1}{2}$ $(2x-8-2x-11)/(2x^2+11x-8x-44) = 1/2$

$\frac{- 19}{2 x^{2} + 3 x - 44} = \frac{1}{2}$ $(-19)/(2x^2+3x-44) = 1/2$

$- 38 = 2 x^{2} + 3 x - 44$ $-38=2x^2+3x-44$

$= 2 x^{2} + 3 x + 38 - 44$ $=2x^2+3x+38-44$

$= 2 x^{2} + 3 x - 6$ $=2x^2+3x-6$

Answer 2 · 2018-02-04T18:57:21

Get rid of the denominators by multiplying through by the LCM.

Move all the terms to one side and simplify.

Explanation:

You need to get rid of the denominators in the given equation.
You can do this by multiplying each term by the LCM of the denominators, which is $2 (2 x + 11) (x - 4)$ $color(blue)(2(2x+11)(x-4))$ ..

Each denominator will cancel.

The left side will look like this:

$(2xxcolor(blue)(2cancel((2x+11))(x-4)))/cancel((2x+11)) -(1xxcolor(blue)(2(2x+11)cancel((x-4))))/cancel((x-4))$

and the right side like this: $(1xxcolor(blue)(cancel2(2x+11)(x-4)))/cancel2$

This leaves the equation without fractions.

$4(x-4) -2(2x+11) = (2x+11)(x-4)$

Multiply out the brackets:

$4x-16 -4x-22 = 2x^2 +3x-44$

Move all the terms to the right hand side.

$0 = 2x^2 +3x -44+38$

$0 = 2x^2 +3x -6$

This is the required equation.

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