1. algebra

$\frac {4}{a-7} = \frac {-2a}{a+3}$

2. $
\frac {4}{a-7} = \frac {-2a}{a+3}
$

3. $\frac{4}
{{a - 7}} = \frac{{ - 2a}}
{{a + 3}}$

$4\left( {a + 3} \right) = - 2a\left( {a - 7} \right)$

$4a + 12 = - 2a^2 + 14a$

$2a^2 - 10a + 12 = 0$

$2\left( {a^2 - 5a + 6} \right) = 0$

$2\left( {a - 3} \right)\left( {a - 2} \right) = 0$

$a - 3 = 0 \Rightarrow a = 3$

or

$a - 2 = 0 \Rightarrow a = 2$

$
\frac {4}{a-7} = \frac {-2a}{a+3}
$
Just like the last problem, we have equal ratios, so we cross multiply:

$4(a + 3) = -2a(a - 7)$

Evaluating, we get:

$4a + 12 = -2a^2 + 14a$

Bring the right side over to the left side, we get:

$2a^2 - 10a + 12 = 0$

That's a quadratic equation. If you need to solve that, go here and enter your coefficients: