8(2y-5)-4(3y-7)-6a=-2y-12

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вопрос

To solve this equation, we first need to distribute the constants outside the parentheses to the terms inside the parentheses:

8(2y-5) = 16y - 40
4(3y-7) = 12y - 28

Substitute the results back into the original equation:

16y - 40 - 12y + 28 - 6a = -2y - 12

Now, combine like terms:

16y - 12y - 6a - 40 + 28 = -2y - 12
4y - 6a - 12 = -2y - 12

Add 2y to both sides:

4y + 2y - 6a - 12 = -12
6y - 6a - 12 = -12

Add 12 to both sides:

6y - 6a - 12 + 12 = 0
6y - 6a = 0

Divide by 6:

y - a = 0

Therefore, the solution to the equation is y - a = 0.