7(х-6)/4=5(х+1)/3-3(х+2)

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вопрос

To solve this equation, we will first simplify both sides of the equation:

7(х-6)/4 = 5(х+1)/3 - 3(х+2)

To simplify the left side, distribute the 7:
(7х - 42)/4 = 5(х+1)/3 - 3(х+2)

Next, simplify the right side by distributing the 5 and 3:
(7х - 42)/4 = (5х + 5)/3 - (3х + 6)

Now, find a common denominator for the fractions on the right side, which is 12:
(7х - 42)/4 = (4(5х + 5))/12 - (3(3х + 6))/12

Simplify the fractions on the right side:
(7х - 42)/4 = (20х + 20)/12 - (9х + 18)/12

Combine the fractions on the right side:
(7х - 42)/4 = (20х + 9х + 20 - 18)/12
(7х - 42)/4 = (29х + 2)/12

Now, clear the fractions by multiplying by 12:
12(7х - 42)/4 = 12(29х + 2)/12
3(7х - 42) = 29х + 2

Distribute on the left side:
21x - 126 = 29x + 2

Subtract 21x from both sides:
-126 = 8x + 2

Subtract 2 from both sides:
-128 = 8x

Divide by 8 on both sides:
x = -16

Therefore, the solution to the equation 7(х-6)/4 = 5(х+1)/3 - 3(х+2) is x = -16.