We are asked to solve several equations. Let's solve problem 71. a). The equation is $(3y - 2c)^2 - (y - 2c)(2y + c) = 7y^2 - 12c^2$.

1. Problem Description

We are asked to solve several equations. Let's solve problem

7

1. a).

The equation is $(3y - 2c)^2 - (y - 2c)(2y + c) = 7y^2 - 12c^2$.

2. Solution Steps

First, we expand the terms in the equation:

$(3y - 2c)^2 = (3y)^2 - 2(3y)(2c) + (2c)^2 = 9y^2 - 12yc + 4c^2$

$(y - 2c)(2y + c) = y(2y + c) - 2c(2y + c) = 2y^2 + yc - 4yc - 2c^2 = 2y^2 - 3yc - 2c^2$

Substituting these into the equation gives:

$9y^2 - 12yc + 4c^2 - (2y^2 - 3yc - 2c^2) = 7y^2 - 12c^2$

$9y^2 - 12yc + 4c^2 - 2y^2 + 3yc + 2c^2 = 7y^2 - 12c^2$

$7y^2 - 9yc + 6c^2 = 7y^2 - 12c^2$

Subtracting $7y^2$ from both sides:

$-9yc + 6c^2 = -12c^2$

$-9yc = -12c^2 - 6c^2$

$-9yc = -18c^2$

If $c = 0$, then the equation holds for any value of $y$.

If $c \ne 0$, then we can divide both sides by $-9c$:

$y = \frac{-18c^2}{-9c}$

$y = 2c$

3. Final Answer

If $c = 0$, then $y$ can be any real number.

If $c \neq 0$, then $y = 2c$.

So, the solution is $y=2c$ if $c \neq 0$, and $y$ can be any real number if $c=0$.

We can also express the solution as $y = 2c$.