How do you solve #5/x + 3/y = 4# and #25/x - 2/y = 3#?

1 Answer
Aug 23, 2015

Answer:

#{(x=5), (y = 1) :}#

Explanation:

Right from the start, you know that #x# and #y# cannot be equal to zero, since they represent denominators.

Now, start by rewriting your equations to get rid of the fractions. The common denominator will be #x * y#, so multiply each term accordingly. For the first equation you will get

#(5y)/(xy) + (3x)/(xy) = (4xy)/(xy)#

#5y + 3x = 4xy#

For the second equation you will get

#(25y)/(xy) - (2x)/(xy) = (3xy)/(xy)#

#25y - 2x = 3xy#

Multiply the first equation by #-5# to get

#5y + 3x = 4xy | * (-5)#

#-25y -15y = -20xy#

Your two equations are now

#{(-25y - 15x = -20xy), (25y - 2x = 3xy) :}#

Notice that if you were to add these two equations together, the #y#-terms on the left sides of the equations would cancel out,

You will be left with

#-color(red)(cancel(color(black)(25y))) - 15x + color(red)(cancel(color(black)(25y))) - 2x = -20xy + 3xy#

#-17x = -17xy#

You can further simplify this to get the value of #y#

#color(red)(cancel(color(black)(-17x))) = color(red)(cancel(color(black)(-17x))) * y implies y = color(green)(1)#

Take this value of #y# into one of the two equations and find the value of #x#

#5 * 1 + 3x = 4 * x * 1#

#4x - 3x = 5 implies x = color(green)(5)#

Since you have #x!=0# and #y!=0#, the two solutions to your system of equations will be

#{(x=5), (y=1):}#

You can do a quick check to make sure that the calculations are correct

#5/5 + 3/1 = 4#

#1 + 3 = 4" "color(green)(sqrt())#

and

#25/5 - 2/1 = 3#

#5 - 2 = 3" "color(green)(sqrt())#