Question

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

Answer 1

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:

`1/x + 5/y = 3/4`

Put on equal denominators:

`(1(4y))/(x(4y)) + (5(4x))/(y(4x)) = (3(xy))/(4(xy))`

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

`4y + 20x = 3xy`

Solving for y:

`4y - 3xy = -20x`

Factor out y:

`y(4 - 3x) = -20x`

`y = (-20x) / (4 - 3x)`

Substitute `(-20x) / (4 - 3x)` for y in the other equation

`2/x - 2/((-20x) / (4 - 3x)) = 1/2`

Once placed on a common denominator of `(-40x^2)/(4 - 3x)`:

`(-80x)/(4 - 3x) - 4x = (-20x^2)/(4 - 3x)`

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

`(-80x)/(4 - 3x) - (4x(4 - 3x))/(4 - 3x) = (-20x^2)/(4 - 3x)`

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

`32x^2 - 96x` = 0

`32x(x - 3) = 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:

`1/3 + 5/y = 3/4`

`(4y)/(12y) + 60/(12y) = (9y)/(12y)`

`4y - 9y = 60`

`-5y = 60`

`y = -12`

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