Solving Equations by Plugging in Values
A solution is any value of a variable that makes the specified equation true. A solution set is the set of all variables that makes the equation true. The solution set of 2y + 6 = 14 is {4}, because 2(4) + 6 = 14. The solution set of y^{2} + 6 = 5y is {2, 3} because 2^{2} + 6 = 5(2) and 3^{2} + 6 = 5(3). If an equation has no solutions, its solution set is the empty set or null seta set with no members, denoted Ø. For example, the solution set to x^{2} =  9 is Ø, because no number, when squared, is equal to a negative number.
Sometimes we will be given a set of values from which to find a solutiona replacement set. The replacement set is the set of values that may be substituted for the variable. To find the solution set from the replacement set, plug in each value from the replacement set and evaluate both sides of the equation. If the two sides are equal, the equation is true and thus the value is a solution.
Example 1: Find the solution set of 11  5w = 1 from the replacement set {0, 2, 4}.


11  5(0) = 1 ?

11  0 = 1 ?

11 = 1 ? No. Not a solution.


11  5(2) = 1 ?

11  10 = 1 ?

1 = 1 ? Yes. Solution.


11  5(4) = 1 ?

11  20 = 1 ?

9 = 1 ? No. Not a solution.
Thus, the solution set is
{2}.
Example 2: Find the solution set of 15  x = 5 + x from the replacement set { 5, 2, 5}.


15  ( 5) = 5 + ( 5) ?

15 + 5 = 5  5 ?

20 = 0 ? No. Not a solution.


15  2 = 5 + 2 ?

13 = 7 ? No. Not a solution.


15  5 = 5 + 5 ?

10 = 10 ? Yes. Solution
Thus, the solution set is
{5}.
Example 3: Find the solution set of 15 + 3x = 4x + 15  x from the replacement set {1, 2, 3, 4}.


15 + 3(1) = 4(1) + 15  1 ?

15 + 3 = 4 + 15  1 ?

18 = 18 ? Yes. Solution.


15 + 3(2) = 4(2) + 15  2 ?

15 + 6 = 8 + 15  2 ?

21 = 21 ? Yes. Solution.


15 + 3(3) = 4(3) + 15  3 ?

15 + 9 = 12 + 15  3 ?

24 = 24 ? Yes. Solution.


15 + 3(4) = 4(4) + 15  4 ?

15 + 12 = 16 + 15  4 ?

27 = 27 ? Yes. Solution.
Thus, the solution set is
{1, 2, 3, 4}.
Example 4: Find the solution set of x^{2} + 4x = 5 from the replacement set { 5, 0, 1}.


( 5)^{2} + 4( 5) = 5 ?

25 + ( 20) = 5 ?

5 = 5 ? Yes. Solution.


0^{2} + 4(0) = 5 ?

0 + 0 = 5 ?

0 = 5 ? No. Not a solution.


1^{2} + 4(1) = 5 ?

1 + 4 = 5 ?

5 = 5 ? Yes. Solution.
Thus, the solution set is
{ 5, 1}.