How do you solve the system of linear equations #2y+6=x# and #3y+10=x#?

1 Answer
Feb 20, 2016

Answer:

Since x is already isolated in both equations we can easily solve for y by substitution.

Explanation:

#2y + 6 = 3y + 10#

#2y - 3y = 10 - 6#

#-y = 4#

#y = -4#

Now solve for x by substituting -4 for y in either one of the equations.

#2(-4) + 6 = x#

#-8 + 6 = x#

#-2 = x#

The solution set is #{-2, -4}#.

Here are a few of the most fundamental rules for solving systems of equations by substitution.

  1. Always solve for the easiest variable in the equation. For example, in #6x + 3y = -9#, it would be easiest to solve for y because you don't end up with fractions, which can become long and tricky to work with.

  2. Always only replace the value of x or y in the equation. Don't get rid of any coefficients x or y may have. Example:

If you want to substitute 2x + 3 = y into #4x - 2y = -3#, you must only replace y:

#4x - 2(2x + 3) = -3

You would then distribute and then solve.

Practice exercises:

  1. Solve the following linear systems of equations by substitution.

a). #x + 3y = -3, 2x - 2y = 10#

b). #2x + 3y = 10, -3x + 4y = 36#

Good luck!