3/(x+1) + 2 = 5/(x+1)
If you multiply both sides by x+1, you get
3 + 2(x+1) = 5
Apply the distributive property of multiplication over addition
a(b+c)=ab+ac
and you get
3 + 2x + 2 = 5... oh my... see where this is going? Think you can solve it from here?
2007-12-03 16:42:45
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answer #1
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answered by gugliamo00 7
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(1) Times (x+1) both sides
(2) 3 + 2 (x+1) = 5
(3) 2(x+1)=2
(4) x+1=1
(5) x = 0
2007-12-03 16:39:33
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answer #2
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answered by GetDownWithThe$ickness 4
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multiply both sides of the equation by x + 1
you then come up with:
3 + 2(x +1) = 5
3 + 2x + 2 =5
5 + 2x = 5
2x = 5 - 5
2x = 0
x = 0
2007-12-03 19:14:13
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answer #3
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answered by rayne 1
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3 + 2x + 2 = 5
2x = 0
x = 0
2007-12-04 00:23:24
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answer #4
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answered by Como 7
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x=0
Ask for help on the tougher questions. These you can do, just mechanical. LCD (x+1), 5=5+2x, too easy.
2007-12-03 16:42:44
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answer #5
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answered by Wylie Coyote 6
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multiply both sides by x+1
3+2x+2=5
2x=0
x=0
2007-12-03 16:39:31
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answer #6
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answered by someone else 7
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locate the Least ordinary Denominator: 2(x-a million) + 3(x+a million)..............5 ---------------------- = ------------------- (x+a million)*(x-a million) ..........(x+a million)*(x-a million) The (x+a million)*(x-a million) cancels out, while the two sides of the equation are accelerated with (x+a million) * (x-a million) . you at the instant are left with 2*(x-a million) + 3(x+a million) = 5 boost: 2x - 2 + 3x + 3 = 5 5x + a million = 5 resolve for x: 5x = 4 x = 4/5 or 0.8
2016-12-17 06:29:05
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answer #7
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answered by ? 4
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treat them like ordinary fractions. rearrange:
3/(x+1) -5/(x+1)=-2
-2/(x+1)=-2
-2=-2(x+1)
solve for x
2007-12-03 16:39:02
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answer #8
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answered by Anonymous
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