In this lesson, we will discuss how to solve systems of equations using the substitution method. This method is achieved by solving one of the equations for one of the variables (usually x or y). Once one variable has been found it can be “plugged” back into the other equation to solve for the remaining variable. Below is how the substitution method works.
Example 1:
y = -5
5x + 4y = -20
In this case the y variable has been isolated for us. We can “plug” the information given for y = to find x.
5x + 4(-5) = -20
Now let’s distribute, combine like-terms, and solve for x
5x – 20 = -20
+20 +20
5x = 0
Remember in order to isolate a variable you must divide both sides of the equation by the coefficient of that variable
The final answer for this problem is (0, -5). Answers are reported in coordinate form. Remember for this problem we were already given what y is equal to. So you only needed to find the answer to x. In some special cases the answer can be infinite solutions or no solution (we will discuss problems like these later on).
Example 2:
x + 7y = 0
2x – 8y = 22
First decide which variable you would like to isolate. For this example we will isolate x first (you can choose y to be first if you would like). Deciding which variable to isolate first is completely up to you.
x + 7y = 0
In order to isolate I must move everything to the other side of the equation sign except x (we are isolating this variable first).
x + 7y = 0
-7y -7y
Remember to change the sign of the term that has been moved.
x = -7y + 0
Now that we have found what x is equal to we can “substitute” x into the other equation to find y.
Distribute and combine like-terms .
-14y + 0 – 8y = 22
After combing like-terms the sentence should read
-22y + 0 = 22
Now isolate and solve for y
Lastly plug the answer for y back into the other equation to find x.
x + 7y = 0
x + 7(-1) = 0
x -7 = 0
+7 +7
x = 7
The final answer is (7, -1).
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