How do you solve by substitution #x-3y=-5# and #2x+y=11#?

Question

Ethan Adkins · Accepted Answer

First subtract #2x# from both sides of the second equation to get: #y = 11 - 2x# Substitute this express for #y# into the first equation to get: #-5 = x - 3y = x - 3(11-2x) = x-33+6x# #= 7x - 33# Add #33# to both ends to get: #7x=28# Divide both sides by #7# to get: #x=4# Then #y = 11 - 2x = 11 - (2xx4) = 11 - 8 = 3#

Nathan Axel · Answer

undefined To solve the system of equations by substitution, follow these steps:

1. Solve one of the equations for one variable.
2. Substitute the expression found in step 1 into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found in step 3 back into one of the original equations to solve for the other variable.
5. Verify the solution by checking if it satisfies both equations.

Given the system:

1. Solve the second equation for $ y $: $ y = 11 - 2x $.
2. Substitute $ y = 11 - 2x $ into the first equation: $ x - 3(11 - 2x) = -5 $.
3. Solve for $ x $: $ x - 33 + 6x = -5 $, $ 7x - 33 = -5 $, $ 7x = 28 $, $ x = 4 $.
4. Substitute $ x = 4 $ into the second equation: $ 2(4) + y = 11 $.
5. Solve for $ y $: $ 8 + y = 11 $, $ y = 3 $.

Therefore, the solution to the system of equations is $ x = 4 $ and $ y = 3 $.