Simulataneous equation help please

[Guest]

Hi
Please can you help me to solve this simultaneous equation?

\(\displaystyle \L \begin{array}{l}
2x + y = 1 \\
\frac{1}{x} - \frac{1}{y} = 2 \\
\end{array}\)
..................................................

\(\displaystyle \L 2x + y = 1\)
\(\displaystyle \L y = 1 - 2x\)

\(\displaystyle \L
\frac{1}{x} - \frac{1}{{1 - 2x}} = 2\)

\(\displaystyle \L
\frac{1}{x} - \frac{1}{{1 - 2x}} - 2 = 0\)

\(\displaystyle \L
\frac{{1 - 2x - x - 2x(1 - 2x)}}{{x(1 - 2x)}} = 0\)

\(\displaystyle \L
\frac{{4x^2 - 5x + 1}}{{x(1 - 2x)}} = 0\)

Can you tell me if this is correct so far?

If it is, when I try to apply the quadratic formula I get a messy answer which doesnt seem right?

Thanks

 

[stapel]

Everything looks fine, though you might want to save yourself a little messiness and multiply through by x(1 - 2x²) (after noting the restriction that x cannot be either of 0 or 1/2). Then you're left with just your quadratic.

Check your work on the Quadratic Formula. The answers shouldn't be "messy". Your quadratic should give you two fairly "neat" solutions.

And don't forget that you can always check by graphing.

Eliz.

_____________
Edit: Ne'mind. Complete solution is below.

 

[soroban]

Hello, bbk!

\(\displaystyle \L \begin{array}{cc} 2x\,+\,y\:=\:1 \\ \frac{1}{x}\,-\,\frac{1}{y}\:=\:2\end{array}\)

I would get rid of those fractions first . . .

Multiply the second equation by \(\displaystyle xy:\L\;\;xy\left(\frac{1}{x}\,-\,\frac{1}{y}\:=\:2\right)\)\(\displaystyle \;\;\Rightarrow\;\;y\,-\,x\:=\:2xy\;\) [1]

The first equation gives us: \(\displaystyle \,y\:=\:1\,-\,2x\;\) [2]

Substitute into [1]: \(\displaystyle \,(1\,-\,2x)\,-\,x\:=\:2x(1\,-\,2x)\;\;\Rightarrow\;\;1\,-\,2x\,-\,x\:=\:2x\,-\,4x^2\)

We have a quadratic equation: \(\displaystyle \,4x^2\,-\,5x\,+\,1\:=\:0\)

\(\displaystyle \;\;\)which factors: \(\displaystyle \,(x\,-\,1)(4x\,-\,1)\:=\:0\)

\(\displaystyle \;\;\)and has roots: \(\displaystyle \,x\,=\,1,\;\frac{1}{4}\)

From [2], we get: \(\displaystyle y\,=\,-1,\;\frac{1}{2}\)


Therefore: \(\displaystyle \L\,(x,y)\:=\1,\,-1),\;\left(\frac{1}{4},\,\frac{1}{2}\right)\)