How do you solve the system of equations #y= 3x - 13# and #- 11x + 4y = - 46#?

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Answer 1 · 2017-06-15T02:47:23

See a solution process below:

Step 1) Because the first equation is already solved for #y# we can substitute #(3x - 13)# for #y# in the second equation and solve for #x#:

#-11x + 4y = -46# becomes:

#-11x + 4(3x - 13) = -46#

#-11x + (4 * 3)x - (4 * 13) = -46#

#-11x + 12x - 52 = -46#

#(-11 + 12)x - 52 = -46#

#1x - 52 = -46#

#x - 52 = -46#

#x - 52 + color(red)(52) = -46 + color(red)(52)#

#x - 0 = 6#

#x = 6#

Step 2) Substitute #6# for #x# in the first equation and calculate #y#:

#y = 3x - 13# becomes:

#y = (3 * 6) - 13#

#y = 18 - 13#

#y = 5#

The solution is: #x = 6# and #y = 5# or #(6, 5)#