For what value of c do the following equations have infinitely many solutions: cx + 3y − (c−3) = 0 and 12x + cy − c = 0?

The general form of a linear equation is $Ax + By + C = 0$.
The given equations are:

1. $cx + 3y - (c-3) = 0$
2. $12x + cy - c = 0$

Comparing these to $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
From equation 1: $a_1 = c$, $b_1 = 3$, $c_1 = -(c-3) = 3-c$
From equation 2: $a_2 = 12$, $b_2 = c$, $c_2 = -c$

For a system of linear equations to have infinitely many solutions, the ratios of their coefficients must be equal:

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Substitute the coefficients:

$$\frac{c}{12} = \frac{3}{c} = \frac{3-c}{-c}$$

$$\frac{c}{12} = \frac{3}{c}$$

Cross-multiply:

$$c \times c = 12 \times 3$$ $$c^2 = 36$$

Take the square root of both sides:

$$c = \pm \sqrt{36}$$ $$c = \pm 6$$

So, $c$ can be 6 or $-6$.

$$\frac{3}{c} = \frac{3-c}{-c}$$

Cross-multiply:

$$3(-c) = c(3-c)$$ $$-3c = 3c - c^2$$

Rearrange the terms to form a quadratic equation:

$$c^2 - 3c - 3c = 0$$ $$c^2 - 6c = 0$$

Factor out $c$:

$$c(c - 6) = 0$$

This gives two possible values for $c$:

$$c = 0 \quad \text{or} \quad c - 6 = 0 \implies c = 6$$

From Step 3, $c = 6$ or $c = -6$.
From Step 4, $c = 0$ or $c = 6$.
The value of $c$ that is common to both sets of solutions is $c=6$.

Let's verify with the first and third ratios as well:

$$\frac{c}{12} = \frac{3-c}{-c}$$

Cross-multiply:

$$c(-c) = 12(3-c)$$ $$-c^2 = 36 - 12c$$

Rearrange the terms:

$$c^2 - 12c + 36 = 0$$

This is a perfect square trinomial, $(c-6)^2$:

$$(c-6)^2 = 0$$ $$c-6 = 0$$ $$c = 6$$

All three conditions consistently yield $c=6$.