To avoid distributing the 1/7 which would only make more fractions, place the (3x - 2) over the 7:
(3x - 2) / 7 = (x + 10) / 5
cross multiply the 7 and 5:
5(3x - 2) = 7(x + 10)
distribute to get:
15x - 10 = 7x + 70
add 10 to each side:
15x = 7x + 80
subtract 7x from each side:
8x = 80
divide by 8:
x = 10
2007-06-08 16:08:12
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answer #1
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answered by MathGuy 6
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1.) To eliminate any fractions multiply both sides by the Least common denominator ( which is 35).
35[(1/7)(3x-2)] = [(x+10)/5]35
5(3x-2) = 7(x+10)
2.) The rest is basic algebra.
15x-10 = 7x+ 70
8x = 80
x = 10
2007-06-08 23:08:16
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answer #2
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answered by whatcanmaxdo4u?everythingupscant 3
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Multiply both sides by 35:-
5.(3x - 2) = 7.(x + 10)
15x - 10 = 7x + 70
8x = 80
x = 10
2007-06-09 07:33:24
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answer #3
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answered by Como 7
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Distribute the 1/7 on the left side:
(3/7)x - (2/7) = (x + 10)/5
Multiply both sides by 5:
(15/7)x - (10/7) = x + 10
Get all x's on one side and all others on the other:
(8/7)x = 80/7
Multiply both sides by (7/8):
x = 10
2007-06-08 23:08:17
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answer #4
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answered by yeeeehaw 5
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[(1/7)(3x-2)]=(x+10)/5
(3x-2)/7 = (x+10)/5
Multiply all by 35
5(3x-2) = 7(x+10)
15x - 10 = 7x + 70
8x = 80
x = 10
2007-06-08 23:10:05
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answer #5
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answered by Steve A 7
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multiply both sides by 35 to get rid of the fractions
5(3x-2) = 7(x+10)
15x-10 = 7x + 70
8x = 80
x= 10
2007-06-08 23:09:09
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answer #6
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answered by steffers27 5
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(3x-2)/7 = (x+10)/5
cross multiply to give you:
5*(3x-2) = 7*(x+10)
distribute:
15x-10 = 7x-70
solve for x:
15x = 7x + 80
8x = 80
x = 10
2007-06-08 23:11:57
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answer #7
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answered by slay_09 2
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3x/7 - 2/7 = x/5 +2
3x/7 -x/5 = 10/5 + 2/7
(15x -7x)/35 = (70 + 10)/35
15x-7x = 70+10
8x = 80
x=10
2007-06-08 23:09:25
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answer #8
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answered by skipper 7
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written differently (3x-2)/7=(x+10)/5
Multiply both sides by 35 (5*7)
15x-10=7x+70
8X=80
x=10
2007-06-08 23:12:21
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answer #9
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answered by blizebliz 5
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Treat it like a proportion. Use the distributive property for both sides first. Combine all the like terms until you have x on one side and a number on the other.
2007-06-08 23:06:55
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answer #10
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answered by Anonymous
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