How do you solve the system of equations #\frac { y } { 4} - \frac { x } { 3} = 1# and #\frac { x } { 2} + \frac { y } { 5} = 10#?

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Answer 1 · 2017-05-04T21:58:10

# x = 12 #
# y=20 #

We want to find the simultaneous solution of:

# y/4-x/3=1 \ \ \ # ..... [A]

# x/2+y/5=10 # ..... [B]

We can multiply out the fractions, as follows:.

# Eq [A] xx 12 => 3y-4x=12 \ \ \ # ..... [C]
# Eq [B] xx 10 => 5x+2y=100 # ..... [D]

We can eliminate #y# by noting that #6# is a common multiple of #3# and #2#, the coefficient of #y# in the above equations: Thus

# Eq [C] xx 2 => 6y-8x \ \ =24 \ \ \ # ..... [E]
# Eq [D] xx 3 => 15x+6y=300 # ..... [F]

So now we have the same coefficient of #y# in both equation; thus:

# Eq[F]-Eq[E] => 15x-(-8x)=300-24 #
# :. 23x =276 #
# :. x =12 #

And if we substitute this value of #x# into Eq[E] we get:

# 6y-8*12=24 #
# :. 6y=120 #
# :. y=20 #

Check with Eq [F]:

# 15x+6y = 15*12+6*20 = 180+120 = 300 #

Hence the solution is:

# x = 12 #
# y=20 #

Which we can validate graphically:
graph{(y/4-x/3-1)(y/5+x/2-10)=0 [-10, 20, -10, 45]}