# How do you solve 4(y-2/4)=9(y+1/3) ?

Dec 20, 2015

$y = - 1$

#### Explanation:

$4 \left(y - \frac{2}{4}\right) = 9 \left(y + \frac{1}{3}\right)$

According to BEDMAS, work on the brackets first. Reduce $\frac{2}{4}$ to $\frac{1}{2}$.

$4 \left(y - \frac{1}{2}\right) = 9 \left(y + \frac{1}{3}\right)$

Make the bracketed terms have the same denominator.

$4 \left(\frac{2 y}{2} - \frac{1}{2}\right) = 9 \left(\frac{3 y}{3} + \frac{1}{3}\right)$

Subtract the bracketed terms.

$4 \left(\frac{2 y - 1}{2}\right) = 9 \left(\frac{3 y + 1}{3}\right)$

Reduce the fractions by cancelling.

${\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}^{2} \left(\frac{2 y - 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) = {\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{9}}}}^{3} \left(\frac{3 y + 1}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{3}}}}\right)$

Rewrite the equation.

$2 \left(2 y - 1\right) = 3 \left(3 y + 1\right)$

Multiply.

$4 y - 2 = 9 y + 3$

Isolate for $y$ by bringing all terms with $y$ to the left and all without to the right.

$4 y - 9 y = 3 + 2$

Solve.

$- 5 y = 5$

$y = - 1$