Stuck on this simultaneous equation

[ai93]

A question I am working on
\(\displaystyle \frac{3}{x}+\frac{2}{y}=14\) (1)
\(\displaystyle \frac{5}{x}-\frac{3}{y}=-2\) (2)

let \(\displaystyle \frac{1}{x}=a\) let \(\displaystyle \frac{1}{y}=b\)

\(\displaystyle \therefore3a+2b=14\) (3) (x5)
\(\displaystyle 5a-3b=-2\) (4) (x3)

\(\displaystyle =15a+10b=70\) (5)
\(\displaystyle 15a-9b=-6\) (6)

(5)-(6)
15a cancels out. Leaving 19b=76
\(\displaystyle b=\frac{76}{19}\) \(\displaystyle x=4\)
But I have the answer wrong. I got this question from my textbook, there is no explantion the answer is \(\displaystyle x=\frac{1}{2} and y=\frac{1}{4}\) But I can't seem to get there I don't know what I am doing wrong. Corrections please?

 

[Siron]

Well, your solution is correct, substituting $b=4$ into one of the equations gives $a=2$. Hence back-substitution gives
$$a=2=\frac{1}{x} \Rightarrow x=\frac{1}{2}$$
$$b=4=\frac{1}{y} \Rightarrow y = \frac{1}{4}$$
which are the desired results.

 

[ai93]

Well, your solution is correct, substituting $b=4$ into one of the equations gives $a=2$. Hence back-substitution gives
$$a=2=\frac{1}{x} \Rightarrow x=\frac{1}{2}$$
$$b=4=\frac{1}{y} \Rightarrow y = \frac{1}{4}$$
which are the desired results.

Ohh yes. That explains it. Thank you!

 

[Siron]

Ohh yes. That explains it. Thank you!

Nice! You're welcome.

 

[MarkFL]

\(\displaystyle \frac{3}{x}+\frac{2}{y}=14\tag 1\)

\(\displaystyle \frac{5}{x}-\frac{3}{y}=-2\tag 2\)

You could multiply both equations by $xy$ to get:

\(\displaystyle 3y+2x=14xy\)

\(\displaystyle 5y-3x=-2xy\implies 14xy=-35y+21x\)

Thus, we have:

\(\displaystyle 3y+2x=-35y+21x\)

\(\displaystyle 38y=19x\)

\(\displaystyle x=2y\)

Go back to (1), and substitute for $x$:

\(\displaystyle \frac{3}{2y}+\frac{2}{y}\cdot\frac{2}{2}=14\)

\(\displaystyle \frac{7}{2y}=14\)

\(\displaystyle \frac{1}{2y}=2\)

\(\displaystyle 4y=1\)

\(\displaystyle y=\frac{1}{4}\implies x=2\cdot\frac{1}{4}=\frac{1}{2}\)

Hence:

\(\displaystyle (x,y)=\left(\frac{1}{2},\frac{1}{4}\right)\)