What is the final step of completing a solve by substitution problem?

1 Answer
Aug 20, 2016

I'm not sure where exactly you mean "final" in the solving process. So, I've prepared a couple of problems that I will work through slowly and carefully, showing all the steps to the final answer.

Example 1: Solve the following system of equations-#2x + y = 5, 3x + 2y = 9#

Since we want to solve with substitution, we must solve for one variable in one of the equations. I think it would be easiest to solve for #y# in the first equation.

#y = 5 - 2x#

We can now substitute into the other equation:

#3x + 2(5 - 2x) = 9#

#3x + 10 - 4x = 9#

#-x = -1#

#x = 1#

We must now find the value of #y#. This is found by inserting #x = 1# into one of the equations and solving for #y#.

#y = 5 - 2x#

#y = 5 - 2(1)#

#y = 3#

Hence, our solution set is #{1, 3}#.

Example 2: Find all real values of #x# and #y# that satisfy the following system of equations: #3y = -2x^2 + 2, 2x^2 - 3y^2 = -4#

Once again, as with the last example, we need to solve for one of the variables in one of the equations. It looks easiest to isolate the #y# in the first equation, however solving this equation won't be as neat as solving the previous one.

#y = -2/3x^2 + 2/3#

We can now substitute into equation #2.

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

#2x^2 - 3(4/9x^4 - 8/9x^2 + 4/9) = -4#

#2x^2 - 4/3x^4 + 8/3x^2 -4/3 = -4#

#-4/3x^4 +14/3x^2 + 8/3 = 0#

Solve using a graphing calculator. If it's a standard one, like a TI84, use #y_1 = -4/3x^4 + 14/3x^2 + 8/3 = 0# and #y_2 = 0#, and press "calc" followed by "intersect".

This will give you real roots of #2# and #-2#. All that is left to do is solve for #y#.

#y = -2/3(2)^2 + 2/3" AND "-2/3(-2)^2 + 2/3#

#y = -8/3 + 2/3" AND "-8/3 + 2/3#

#y = -2" AND " -2#

Hence, our solution set is #{2, -2}# and #{-2, -2}#.

Use the following practice exercises to develop your comfort with the skills dealt with in this answer.

Practice exercises:

  1. Find the real values of #x# and #y# that satisfy the following systems of equations.

a) #2x - 3y = 4, x + 2y = 9#

b) #3x + y = -2, x^2 = y#

c) #2x^2 - 3y^2 = -10, x^2 - 2x + 3y = 5#

Hopefully this helps, and good luck!