Solve Equation

[spacewater]

problem
\(\displaystyle \frac{7}{2x+1}-\frac{8x}{2x-1}=-4\)
steps
\(\displaystyle \frac{14x-7-16x^2+8x}{(2x+1)(2x-1)}=-4\) <--- question: when you cross multiply and there is a negative sign do you have to switch the sign for\(\displaystyle 16x^2 +8x\) to \(\displaystyle 16x^2 -8x\)?

\(\displaystyle -16x^2+22x-7=-4(2x+1)(2x-1)\)

\(\displaystyle -16x^2+22x-7=-16x^2+4\)

\(\displaystyle 22x-7=4\) added \(\displaystyle 16x^2\)to both side to cancel out and have 4 by itself

\(\displaystyle 22x = 11\)

Final Answer
\(\displaystyle x = 2\)

My final answer does not correspond to the answersheet.. can anybody tell me what i did wrong and if there is a faster/easier way to approach these kind of problems please show it to me thanks

 

[Denis]

\(\displaystyle \frac{14x-7-16x^2+8x}{(2x+1)(2x-1)}=-4\)

numerator: your + 8x should be - 8x

 

[Deleted member 4993]

problem
\(\displaystyle \frac{7}{2x+1}-\frac{8x}{2x-1}=-4\)
steps
\(\displaystyle \frac{14x-7-16x^2+8x}{(2x+1)(2x-1)}=-4\) <--- question: when you cross multiply and there is a negative sign do you have to switch the sign for\(\displaystyle 16x^2 +8x\) to \(\displaystyle 16x^2 -8x\)?
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There is NO cross-multiplication - in fact there is nothing mathematically valid called "cross-multiplication

\(\displaystyle \frac{14x-7-8x\cdot (2x+1)}{(2x+1)(2x-1)} \, = \, -4\)

\(\displaystyle \frac{14x-7 \, - \, (16x^2 + 8x)}{(2x+1)(2x-1)} \, = \, -4\)

\(\displaystyle \frac{14x-7 \, - \, 16x^2 - 8x}{(2x+1)(2x-1)} \, = \, -4\)

\(\displaystyle \frac{-7 \, - \, 16x^2 \, + \, 6x}{(2x+1)(2x-1)} \, = \, -4\)

Multiply both sides by the common LCM ofthe denominator ? (2x+1)(2x-1)

\(\displaystyle \frac{7 \, + \, 16x^2 \, - \, 6x}{(2x+1)(2x-1)} \cdot (2x+1)\cdot (2x-1) \, = \, 4 \cdot (2x+1)\cdot (2x-1)\)

\(\displaystyle 7 \, + \, 16x^2 \, - \, 6x \, = \, 4\cdot (4x^2 - 1)\)

\(\displaystyle 7 \, + \, 16x^2 \, - \, 6x \, = \, 16x^2 - 4\)

\(\displaystyle 6x = 11\)

and continue.....there is really no faster way for this problem
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\(\displaystyle -16x^2+22x-7=-4(2x+1)(2x-1)\)

\(\displaystyle -16x^2+22x-7=-16x^2+4\)

\(\displaystyle 22x-7=4\) added \(\displaystyle 16x^2\)to both side to cancel out and have 4 by itself

\(\displaystyle 22x = 11\)

Final Answer
\(\displaystyle x = 2\)

My final answer does not correspond to the answersheet.. can anybody tell me what i did wrong and if there is a faster/easier way to approach these kind of problems please show it to me thanks