How do you solve the system of linear equations #5x - 4y = 6# and #2x + 3y = 2#?

1 Answer
Dec 6, 2017

Answer:

#x= 26/23 and y = -2/23#

Explanation:

#5x - 4y = 6# --------------- Let this be equation (1).

#2x + 3y = 2# --------------Let this be equation (2).

To eliminate one variable, we must make the coefficient of that variable same in both the equations. Say, if we want to eliminate #y# in order to find #x#,

Multiply equation(1) by 3 and multiply equation(2) by 4, we get the following set:

(1) x 3 # => 15x - 12y = 18# -----------(1')and

(2) x 4 #=> 8x +12y =8#--------------(2'),

We see that the coefficients of #y# are equal and opposite in sign, so we add these two equations(1') and(2'), in order to eliminate #y#:

#=> 15x+8x -cancel(12y)+cancel(12y) =18+8#

#=> 23x =26#

#=> x= 26/23= 1 3/23#

Substituting value of #x# in any one equation, say in (1),

#=> 5x - 4y = 6#

# => 5 xx 26/23 -4y = 6#

# => 130/23 -4y =6#

#=> -4y = 6-130/23#

# => -4y = 6(23/23) - 130/23= (138-130)/23#

#=> y = 1/-4 xx 8/23#

#=> y = 1/-cancel4^1 xx cancel8^2/23#

#=> y =-2/23#

So we have, #x= 26/23 and y = -2/23#

Let us cross check by substituting values of #x# and #y# in equation (2),

#2x + 3y = 2#

#=># Left hand side =# 2(26/23) + 3(-2/23)#

#= 52/23 - 6/23 = 46/23 = 2= #Right hand side =# 2#
Hence values obtained are correct.