# How do you solve \frac { 6y } { 2y + 6} + \frac { y + 18} { 3y + 9} = \frac { 7y + 17} { y + 3}?

Apr 19, 2017

No real solution. Read the explanation carefully.

#### Explanation:

$\frac{6 y}{2 y + 6} + \frac{y + 18}{3 y + 9} = \frac{7 y + 17}{y + 3}$

We can rewrite it as:-

$\frac{6 y}{2 \left(y + 3\right)} + \frac{y + 18}{3 \left(y + 3\right)} = \frac{7 y + 17}{y + 3}$

$\implies \frac{9 y + y + 18}{3 \left(y + 3\right)} = \frac{7 y + 17}{y + 3}$

Cross multiply;

$\implies \left(10 y + 18\right) \left(y + 3\right) = 3 \left(7 y + 17\right) \left(y + 3\right)$

$\implies \left(10 y + 18\right) \left(y + 3\right) - 3 \left(7 y + 17\right) \left(y + 3\right) = 0$

Taking $\left(y + 3\right)$ common;

$\implies \left(y + 3\right) \left(10 y + 18 - 21 y - 51\right) = 0$

$\implies \left(y + 3\right) \left(- 11 y - 33\right) = 0$

$\implies - 11 \cdot \left(y + 3\right) \cdot \left(y + 3\right) = 0$

$\implies {\left(y + 3\right)}^{2} = 0$

$\implies y + 3 = 0$ $\implies \textcolor{red}{y = - 3}$

BUT there is a catch.

We might think that $y = - 3$ is the solution but try to substitute this value back into the original equation i.e.

$\frac{6 y}{2 y + 6} + \frac{y + 18}{3 y + 9} = \frac{7 y + 17}{y + 3}$

We find that terms in the denominator become zero.
$2 \cdot \left(- 3\right) + 6 = - 6 + 6 = 0$,
$3 \cdot \left(- 3\right) + 9 = - 9 + 9 = 0$,
$- 3 + 3 = 0$.

That means the expressions $\frac{6 y}{2 y + 6} , \frac{y + 18}{3 y + 9} , \frac{7 y + 17}{y + 3}$ become undefined at $y = - 3$ since division by zero is not defined. Hence we cannot say for sure whether the left-hand-side equals the right-hand-side or not. So $y = - 3$ is not a solution to this equation.

Where did we go wrong?

Remember the step where we cross multiplied $y + 3$. Cross multiplication, or in fact any kind of multiplication on both sides of the equation is allowed only when the number(s) being multiplied is/are non-zero. Therefore cross multiplication in that step is only valid if $y + 3 \ne 0$ or $y \ne - 3$. I.e. we are cross multiplying on the assumption that $y = - 3$ is not a solution and even if it turns out to be a solution after solving the subsequent equations, we are going to neglect it.

$\therefore$ we neglect $y = - 3$ and hence, the given equation has no real solution.

Just to be sure, I checked on wolfram. 