Solve Equation

spacewater

Junior Member
Joined
Jul 10, 2009
Messages
67
problem
72x+18x2x1=4\displaystyle \frac{7}{2x+1}-\frac{8x}{2x-1}=-4
steps
14x716x2+8x(2x+1)(2x1)=4\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 for16x2+8x\displaystyle 16x^2 +8x to 16x28x\displaystyle 16x^2 -8x?

16x2+22x7=4(2x+1)(2x1)\displaystyle -16x^2+22x-7=-4(2x+1)(2x-1)

16x2+22x7=16x2+4\displaystyle -16x^2+22x-7=-16x^2+4

22x7=4\displaystyle 22x-7=4 added 16x2\displaystyle 16x^2to both side to cancel out and have 4 by itself

22x=11\displaystyle 22x = 11

Final Answer
x=2\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
 
spacewater said:
14x716x2+8x(2x+1)(2x1)=4\displaystyle \frac{14x-7-16x^2+8x}{(2x+1)(2x-1)}=-4
numerator: your + 8x should be - 8x
 
spacewater said:
problem
72x+18x2x1=4\displaystyle \frac{7}{2x+1}-\frac{8x}{2x-1}=-4
steps
14x716x2+8x(2x+1)(2x1)=4\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 for16x2+8x\displaystyle 16x^2 +8x to 16x28x\displaystyle 16x^2 -8x?
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There is NO cross-multiplication - in fact there is nothing mathematically valid called "cross-multiplication

14x78x(2x+1)(2x+1)(2x1)=4\displaystyle \frac{14x-7-8x\cdot (2x+1)}{(2x+1)(2x-1)} \, = \, -4

14x7(16x2+8x)(2x+1)(2x1)=4\displaystyle \frac{14x-7 \, - \, (16x^2 + 8x)}{(2x+1)(2x-1)} \, = \, -4

14x716x28x(2x+1)(2x1)=4\displaystyle \frac{14x-7 \, - \, 16x^2 - 8x}{(2x+1)(2x-1)} \, = \, -4

716x2+6x(2x+1)(2x1)=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)

7+16x26x(2x+1)(2x1)(2x+1)(2x1)=4(2x+1)(2x1)\displaystyle \frac{7 \, + \, 16x^2 \, - \, 6x}{(2x+1)(2x-1)} \cdot (2x+1)\cdot (2x-1) \, = \, 4 \cdot (2x+1)\cdot (2x-1)

7+16x26x=4(4x21)\displaystyle 7 \, + \, 16x^2 \, - \, 6x \, = \, 4\cdot (4x^2 - 1)

7+16x26x=16x24\displaystyle 7 \, + \, 16x^2 \, - \, 6x \, = \, 16x^2 - 4

6x=11\displaystyle 6x = 11

and continue.....there is really no faster way for this problem
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16x2+22x7=4(2x+1)(2x1)\displaystyle -16x^2+22x-7=-4(2x+1)(2x-1)

16x2+22x7=16x2+4\displaystyle -16x^2+22x-7=-16x^2+4

22x7=4\displaystyle 22x-7=4 added 16x2\displaystyle 16x^2to both side to cancel out and have 4 by itself

22x=11\displaystyle 22x = 11

Final Answer
x=2\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
 
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