Simultaneous Equations: 2x + y = 1, 1/x - 1/y = 2

[val1]

Hi
Can you help me to solve this simultaneous equation please?

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

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


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


Subst. into (2)

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


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

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


\(\displaystyle \L 1 - 2x - x = 2x - 4x^2\)


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


Using the quadratic formula:
\(\displaystyle \L x = \frac{{ - 5 \pm \sqrt {41} }}{8}\)

I do not think this answer is right. Please can you check it for me?

Thank you

 

[galactus]

Re: Simultaneous Equation help please

Hi
Can you help me to solve this simultaneous equation please?

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

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


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


Subst. into (2)

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


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

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


\(\displaystyle \L 1 - 2x - x = 2x - 4x^2\)


4x^2 - 5x + 1 = 0
I believe you merely have your signs backwards



Using the quadratic formula:
\(\displaystyle \L x = \frac{{ - 5 \pm \sqrt {41} }}{8}\)

I do not think this answer is right. Please can you check it for me?

Thank you

 

[val1]

oops!
Thanks