System of Equations: 4/(3x)+3/(2y)=5/12 and 3/(2x)-2/(3y)=4

[catalyst0435]

Having some serious trouble with:

4/(3x)+3/(2y)=5/12 and 3/(2x)-2/(3y)=4

What's my first step?

 

[soroban]

Re: System of Equations

Hello, catalyst0435!

Solve: \(\displaystyle \L\:\begin{array}{cc}[1] \\ \\ \\ \\ \\ \\ \\[2]\end{array}\;\begin{array}{ccc}\frac{4}{3x}\,+\,\frac{3}{2y}\:=\:\frac{5}{12} \\ \\ \\ \frac{3}{2x}\,-\,\frac{2}{3y}\:=\:4\end{array}\)

Multiply [1] by \(\displaystyle \frac{2}{3}:\L\;\;\frac{8}{9x}\,+\,\frac{1}{y}\:=\:\frac{5}{18}\)

Multiply [2] by \(\displaystyle \frac{3}{2}:\L\;\;\frac{9}{4x}\,-\,\frac{1}{y}\:=\:6\)

Add: \(\displaystyle \L\:\frac{8}{9x}\,+\,\frac{9}{4x}\:=\:\frac{5}{18}\,+\,6\;\;\Rightarrow\;\;\frac{113}{36x}\:=\:\frac{113}{18}\)

Hence: \(\displaystyle \L\:36x\:=\:18\;\;\Rightarrow\;\;\fbox{x\,=\,\frac{1}{2}}\)


Substitute into [2]: \(\displaystyle \L\:\frac{3}{2(\frac{1}{2})}\,-\,\frac{2}{3y}\:=\:4\;\;\Rightarrow\;\;3\,-\,\frac{2}{3y}\:=\:4\)

Therefore: \(\displaystyle \L\,-\frac{2}{3y}\:=\:1\;\;\Rightarrow\;\;\fbox{y\:=\:-\frac{2}{3}}\)