problem
2x+17−2x−18x=−4
steps
(2x+1)(2x−1)14x−7−16x2+8x=−4 <--- question: when you cross multiply and there is a negative sign do you have to switch the sign for
16x2+8x to
16x2−8x?
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There is NO cross-multiplication - in fact there is nothing mathematically valid called "cross-multiplication
(2x+1)(2x−1)14x−7−8x⋅(2x+1)=−4
(2x+1)(2x−1)14x−7−(16x2+8x)=−4
(2x+1)(2x−1)14x−7−16x2−8x=−4
(2x+1)(2x−1)−7−16x2+6x=−4
Multiply both sides by the common LCM ofthe denominator ? (2x+1)(2x-1)
(2x+1)(2x−1)7+16x2−6x⋅(2x+1)⋅(2x−1)=4⋅(2x+1)⋅(2x−1)
7+16x2−6x=4⋅(4x2−1)
7+16x2−6x=16x2−4
6x=11
and continue.....there is really no faster way for this problem
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−16x2+22x−7=−4(2x+1)(2x−1)
−16x2+22x−7=−16x2+4
22x−7=4 added
16x2to both side to cancel out and have 4 by itself
22x=11
Final Answer
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