How do you solve #5/x + 3/y = 4# and #25/x - 2/y = 3#?
1 Answer
Explanation:
Right from the start, you know that
Now, start by rewriting your equations to get rid of the fractions. The common denominator will be
#(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
#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
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
#color(red)(cancel(color(black)(-17x))) = color(red)(cancel(color(black)(-17x))) * y implies y = color(green)(1)#
Take this value of
#5 * 1 + 3x = 4 * x * 1#
#4x - 3x = 5 implies x = color(green)(5)#
Since you have
#{(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())#