MHB Stuck on this simultaneous equation

[ai93]

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

let $$\frac{1}{x}=a$$ let $$\frac{1}{y}=b$$

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

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

(5)-(6)
15a cancels out. Leaving 19b=76
$$b=\frac{76}{19}$$ $$x=4$$
But I have the answer wrong. I got this question from my textbook, there is no explantion the answer is $$x=\frac{1}{2} and y=\frac{1}{4}$$ 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]

$$\frac{3}{x}+\frac{2}{y}=14\tag 1$$

$$\frac{5}{x}-\frac{3}{y}=-2\tag 2$$

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

$$3y+2x=14xy$$

$$5y-3x=-2xy\implies 14xy=-35y+21x$$

Thus, we have:

$$3y+2x=-35y+21x$$

$$38y=19x$$

$$x=2y$$

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

$$\frac{3}{2y}+\frac{2}{y}\cdot\frac{2}{2}=14$$

$$\frac{7}{2y}=14$$

$$\frac{1}{2y}=2$$

$$4y=1$$

$$y=\frac{1}{4}\implies x=2\cdot\frac{1}{4}=\frac{1}{2}$$

Hence:

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