How do solve the following linear system?: # 3x - y = -6, 4x+3y=29 #?

Question

1452 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-11T06:22:29

#color(red)[(x = 11/13) and (y = 111/13)# OR

We can simplify the fractions and write the solutions as

#color(red)[(x ~~ 0.8) and (y ~~ 8.5)#

We are given the linear systems of equations given below:

#3x -y = -6# #color(blue)(Eqn.1)#

#4x +3y = 29# #color(blue)(Eqn.2)#

Multiply #color(blue)(Eqn.1)# by #3#

Hence, #color(blue)(Eqn.1)# yields #color(blue)(Eqn.3)#

#9x -3y = -18# #color(blue)(Eqn.3)#

#4x +3y = 29# #color(blue)(Eqn.2)#

When we add #color(blue)(Eqn.3)# and #color(blue)(Eqn.2)# we get

#9x -cancel(3y) = -18# #color(blue)(Eqn.3)#
#4x +cancel(3y) = 29# #color(blue)(Eqn.2)#

#rArr 13x = 11#

Therefore, #color(red)(x = 11/13)#

Substitute this value of #color(red)(x)# in #color(blue)(Eqn.1)#

#3x -y = -6# #color(blue)(Eqn.1)#

#rArr 3(11/13) - y = -6#

#rArr (33/13) - y = -6#

#rArr - y = -6 - (33/13)#

#rArr - y = (-78 - 33)/13#

#rArr - y = -111/13#

Divide both sides by #-1# to get

#rArr y =111/13#

Hence, our final solutions are : -

#color(red)[(x = 11/13) and (y = 111/13)# OR

We can simplify the fractions and write the solutions as

#color(red)[(x ~~ 0.8) and (y ~~ 8.5)#