How do you solve $4 y^{2} - 3 (y - 4) = 4 y (y - 1) + 3 y$ $4y^2 - 3(y-4)=4y(y-1)+3y$ ?

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How do you solve $4 y^{2} - 3 (y - 4) = 4 y (y - 1) + 3 y$ $4y^2 - 3(y-4)=4y(y-1)+3y$ ?

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Answer 1 · 2018-04-23T05:19:05

When only one variable is there, an equation can be solved by using simplification and $y = 6$ $y=6$ in this equation. You can verify it also by putting $y = 6$ $y=6$ on both sides and the results will be same.

Explanation:

$4 y^{2} - 3 (y - 4) = 4 y (y - 1) + 3 y$ $4y^2-3(y-4)=4y(y-1)+3y$

$4 y^{2} - 3 y + 12 = 4 y^{2} - 4 y + 3 y [- 3 \cdot (- 4) = 12, 4 y \cdot y = 4 y^{2}]$ $4y^2 -3y+12=4y^2-4y+3y " " " " [-3 * (-4)=12, 4y * y=4y^2]$

or

$4 y^{2} - 4 y^{2} - 3 y + 12 = - y$ $4y^2-4y^2 -3y +12=-y$

$- 3 y + 12 = - y$ $-3y+12=-y$

So

$12 = 3 y - y$ $12=3y-y$

$2 y = 12$ $2y=12$

$y = \frac{12}{2}$ $y=12/2$

$y = 6$ $y=6$

Hope it helps...

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How do you solve $4 y^{2} - 3 (y - 4) = 4 y (y - 1) + 3 y$ $4y^2 - 3(y-4)=4y(y-1)+3y$ ?

1 Answer

Explanation:

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How do you solve 4y2−3(y−4)=4y(y−1)+3y4y^2 - 3(y-4)=4y(y-1)+3y?

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Explanation:

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How do you solve $4 y^{2} - 3 (y - 4) = 4 y (y - 1) + 3 y$ $4y^2 - 3(y-4)=4y(y-1)+3y$ ?