How do you solve the following system using substitution?: 2/x - 2/y = 1/2, 1/x + 5/y = 3/4

Feb 8, 2016

You will have to solve for one of the variables. I prefer to do y, so that's how I'll show it to you.

Explanation:

$\frac{1}{x} + \frac{5}{y} = \frac{3}{4}$

Put on equal denominators:

$\frac{1 \left(4 y\right)}{x \left(4 y\right)} + \frac{5 \left(4 x\right)}{y \left(4 x\right)} = \frac{3 \left(x y\right)}{4 \left(x y\right)}$

We can now eliminate the denominators because we're on equal denominators.

$4 y + 20 x = 3 x y$

Solving for y:

$4 y - 3 x y = - 20 x$

Factor out y:

$y \left(4 - 3 x\right) = - 20 x$

$y = \frac{- 20 x}{4 - 3 x}$

Substitute $\frac{- 20 x}{4 - 3 x}$ for y in the other equation

$\frac{2}{x} - \frac{2}{\frac{- 20 x}{4 - 3 x}} = \frac{1}{2}$

Once placed on a common denominator of $\frac{- 40 {x}^{2}}{4 - 3 x}$:

$\frac{- 80 x}{4 - 3 x} - 4 x = \frac{- 20 {x}^{2}}{4 - 3 x}$

Now, we must place the -4x on the new common denominator of 4 - 3x.

$\frac{- 80 x}{4 - 3 x} - \frac{4 x \left(4 - 3 x\right)}{4 - 3 x} = \frac{- 20 {x}^{2}}{4 - 3 x}$

Eliminating the denominators, we get the quadratic equation $- 80 x - 16 x + 12 {x}^{2} = - 20 {x}^{2}$

$32 {x}^{2} - 96 x$ = 0

$32 x \left(x - 3\right) = 0$

x = 0 and x = 3

x = 0 cannot be a solution since division by 0 is non defined.

Solving the equation to find y:

$\frac{1}{3} + \frac{5}{y} = \frac{3}{4}$

$\frac{4 y}{12 y} + \frac{60}{12 y} = \frac{9 y}{12 y}$

$4 y - 9 y = 60$

$- 5 y = 60$

$y = - 12$

So, the solution set is (3, -12)