How do you solve #x / 2 - y / 3 = 5/6# and #x / 5 - y / 4 = 29 / 10# using substitution?

1 Answer
Jul 16, 2016

Answer:

#x= -13, y = -22#

Explanation:

Before we can even think of solving, let's change the equations into a more user-friendly form!.

Multiply each by the LCM of the denominators to get rid of the fractions.

#(color(magenta)6 xx x) / 2 - (color(magenta)6 xx y) / 3 = (color(magenta)6 xx 5)/6 " and " "x / 5 - y / 4 = 29 / 10#

#3x - 2y = 5 " and "(color(blue)20 xx x) / 5 - (color(blue)20 xx y) / 4 = (color(blue)20 xx 29)/10#

#3x - 2y = 5 " and " 4x -5y = 58#

Make one of the variables in one equation the subject.

#color(magenta)(x = (5+2y)/3 )" or " y = (3x-5)/2#

(one is not really easier than the other to substitute.)

#4 (color(magenta)((5+2y)/3 )) -5y = 58#

#((20+8y))/3 - 5y = 58" " xx3#

#20+8y -15y = 174#

#20-174 = 7y#

#-154 = 7y#

#y =-22 " now use this value to find x"#

#x = (5+2y)/3 rArr " " x = (5+2(-22))/3#

#x = -13#

Check these values in the second equation to verify.