How do you solve the system of equations #2x - 3y = - 1# and #x + 4y = 5#?

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Answer 1 · 2017-06-04T15:06:32

See a solution process below:

Step 1) Solve the second equation for #x#:

#x + 4y = 5#

#x + 4y - color(red)(4y) = 5 - color(red)(4y)#

#x + 0 = 5 - 4y#

#x = 5 - 4y#

Step 2) Substitute #(5 - 4y)# for #x# in the first equation and solve for #x#:

#2x - 3y = -1# becomes:

#2(5 - 4y) - 3y = -1#

#(2 * 5) - (2 * 4y) - 3y = -1#

#10 - 8y - 3y = -1#

#10 + (-8 - 3)y = -1#

#10 - 11y = -1#

#-color(red)(10) + 10 - 11y = -color(red)(10) - 1#

#0 - 11y = -11#

#-11y = -11#

#(-11y)/color(red)(-11) = -11/color(red)(-11)#

#(color(red)(cancel(color(black)(-11)))y)/cancel(color(red)(-11)) = 1#

#y = 1#

Step 3) Substitute #1# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 5 - 4y# becomes:

#x = 5 - (4 xx 1)#

#x = 5 - 4#

#x = 1#

The solution is: #x = 1# and #y = 1# or #(1, 1)#