How do you solve the following linear system: # 3y-15=x , y-3=-2x #?

1 Answer
May 30, 2018

#x=-6/7#, #y=4 5/7#

Explanation:

We have
1) #3y-15=x#
2) #y-3=-2x#

We can use substitution, i.e. insert #x=3y-15# from 1) into 2) and solve for y and afterwards x.

Here we have another way which is basically as simple, since we have
3) from 1): #2x=6y-30# (we multiply each term with 2)
From 2): #-2x=y-3#

Add 3) and 2 to get rid of the x terms:
4) #2x-2x=6y-30+y-3#
This gives #7y-33=0# or #7y=33#
#y=33/7#
#x=3y-15=3*33/7-15 =(99-15*7)/7=-6/7#

Check:
2)Left side: #y-3=33/7-3=4 5/7 - 3= 1 5/7#
Right side: #-2x=-2*(-6/7)=12/7=1 5/7#