How do you solve the following system?: #4x + 5y = 2 , y + 5 = 3x #

2 Answers
Mar 5, 2018

Answer:

#x = 27/19# and #y = -14/19#

Explanation:

Let's isolate #y# in the second equation

#y = 3x - 5#

Now we can solve for #x# in the other equation

#4x + 5 (3x - 5) = 2#

#4x + 15x -25 = 2#

#19x = 27#

#x = 27/19#

Now we solve for #y#

#y = 3x - 5#

#y = 3 (27/19) - 5#

#y = 81/19 - 5/1 xx 19/19#

#y = 81/19 - 95/19#

#y = -14/19#

To check our work, let's substitute #27/19# for #x# and #-14/19# for #y# in the first equation and solve.

#4(27/19) + 5(-14/19) = 2#

#108/19 - 70/19 = 38/19#

#38/19 = 2 = 2#!

We were right!

Mar 5, 2018

Answer:

#color(blue)(x = 27/19), color(purple)(y = -14 / 19)#

Explanation:

#4x + 5y = 2# Eqn (1)

#y + 5 = 3x#

#y = 3x - 5# Eqn (2)

Substituting value of y term in Eqn (1) in terms of x,

#4x + 5(3x - 5) = 2#

#4x + 15x - 25 = 2#

Rearranging variables on L H S, constants on R H S,

#4x + 15x = 27#

#19x = 27# or #x = 27/19#

Substituting value of x in Eqn (2),

# y = 3x - 5 = (3 * (27 / 19)) - 5 = (81 - 95) / 19 = -14/19#