How do you solve the system of equations #y= 2x + 8# and #3x + 5y = 1#?

Question

2105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-26T10:51:42

By arranging the equations

Arrange the equations (multiply the 1st by 3 and the 2nd by 2):

#3y - 6x = 24#
#6x + 10y = 2#

Combine these:
#13 y = 26#

#y= 26/13 = 2#

Put this in the first original equation:
#2x = y-8#

#2x = 2-8#

#2x = -6#

#x=-3#

Answer 2 · 2018-06-26T10:54:24

#x=-3# and #y=2#

After putting first equation into second,

#3x+5*(2x+8)=1#

#13x+40=1#

#13x=-39#, so #x=-3#

Thus, #y=2*(-3)+8=2#