How do you solve this system of equation: #x + 3y = 1 and - 3x - 3y = - 15#?

Question

4780 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T14:39:31

The solutions are #x=7# and #y=-2#

By substitution

#{(x+3y=1),(-3x-3y=-15):}#

#<=>#, #{(x=1-3y),(x+y=5):}#

#<=>#, #{(x=1-3y),(1-3y+y=5):}#

#<=>#, #{(x=1-3y),(1-2y=5):}#

#<=>#, #{(x=1-3y),(2y=-4):}#

#<=>#, #{(x=1-3*-2=7),(y=-2):}#

Or graphically

graph{(x+3y-1)(x+y-5)=0 [-5.704, 12.076, -4.195, 4.694]}

Answer 2 · 2018-07-27T14:39:46

#x=7, \ \ \ y=-2#

Given equations

#x+3y=1\ ......(1)# &

#-3x-3y=-15#

#x+y=5\ .........(2)#

Subtracting (2) from (1), we get

#x+3y-(x+y)=1-5#

#2y=-4#

#y=-2#

setting the value of #y# in (1), we get

#x+3(-2)=1#

#x-6=1#

#x=7#

Hence the solution is #x=7, \ \ \ y=-2#