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

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

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Answer 1 · 2016-02-08T02:46:09

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:

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

Put on equal denominators:

$\frac{1 (4 y)}{x (4 y)} + \frac{5 (4 x)}{y (4 x)} = \frac{3 (x y)}{4 (x y)}$ $(1(4y))/(x(4y)) + (5(4x))/(y(4x)) = (3(xy))/(4(xy))$

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

$4 y + 20 x = 3 x y$ $4y + 20x = 3xy$

Solving for y:

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

Factor out y:

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

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

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

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

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

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

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

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

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

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

$32 x (x - 3) = 0$ $32x(x - 3) = 0$

x = 0 and x = 3

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

Solving the equation to find y:

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

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

$4 y - 9 y = 60$ $4y - 9y = 60$

$- 5 y = 60$ $-5y = 60$

$y = - 12$ $y = -12$

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

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

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How do you solve the following system using substitution?: 2x−2y=12,1x+5y=342/x - 2/y = 1/2, 1/x + 5/y = 3/4

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