Algebra - Two Variables
Variable Substitution
You can solve a system of equations with a method called substitution. Here’s how to use it.
You have two equations, so “solve†one equation by writing A in terms of B. That is, use the
rules of algebra to rearrange the equation to get “A =†on one side. Then substitute this
expression into the other equation wherever variable A appears. This results in one equation
and one unknown.
Example:
y = 4x
x + y = 90
What are the values of x and y?
Solution:
| The first equation y
= 4x tells you that “4x†is another name for y. |
|
| |
| Substitute 4x for y: |
x + y = 90 |
| |
x + 4x = 90 |
| |
5x = 90 |
| Divide both sides by 5: |
5x/5 = 90/5 |
| |
x = 18 |
| To find y, substitute 18 for x in either
of the original equations: |
y = 4x |
|
y = 4(18) |
| |
y = 72 |
| Substitute the values of x and y into
the other equation: |
x + y = 90 ? |
| |
18 + 72 = 90 ? Yes! |
Example:
Solve the following system of two equations:
Q + N = 33.
25Q + 5N = 505.
| Look at the equations.
What constant should we use?
|
|
| -5(Q + N) = -5(33) |
| Let’s multiply the second equation by
-5: |
–5Q –5N = -165 |
| Add both equations together: |
25Q + 5N = 505
|
| |
-5Q – 5N = -165 |
| |
20Q = 340 |
| Solve for Q: |
20Q/20 = 340/20 |
| |
Q = 17 |
| Substitute Q = 17 to solve for N: |
17 + N = 33 |
| |
N = 33 – 17 = 16 |
| Check the result with both: |
25(17) + 5(16) = 505? Yes! |
| |
17 + 16 = 33? Yes!
|
