How do you solve the following system using substitution?: #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.

#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)