How do you solve (3x-1)/3-(x-3)/15=(2x+3)/23x−13−x−315=2x+32?

1 Answer
James M.
Apr 19, 2018

Isolate the first part:

(3x - 1)/3 - (x-3)/153x−13−x−315

Get the denominator (both bottom numbers) equal buy multiplying the tops and bottoms. 15 is a common multiple of 3 and 15 so multiply the first fraction by 5 and the second by 1.

(5(3x - 1))/15 - (x-3)/15 5(3x−1)15−x−315 Now expand and subtract the two

(15x -5 )/15 - (x - 3)/1515x−515−x−315

(15x - 5 - x + 3)/15 = (14x - 2)/1515x−5−x+315=14x−215

Put it back in the whole equation:

(14x - 2)/15 = (2x + 3)/214x−215=2x+32

Both have a common multiple of 30, so, multiply the first part by 2 and the second by 15.

(2(14x - 2))/30 = (15(2x + 3))/30 2(14x−2)30=15(2x+3)30

(28x - 4)/30 = (30x + 45)/3028x−430=30x+4530

28x -4 = 30x + 4528x−4=30x+45
-2x = 49−2x=49

x = -24.5x=−24.5

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