Solving Systems of Equations Algebraically
- Solve one equation for one variable.
- Substitute that answer into the second equation and solve for the variable in that equation.
- Using that answer, substitute it back into the first equation and solve for the second variable.
Example:
y + x = 3 ... y - 3x = 5
y + x = 3 = y = 3 - x
Substitute y = 3 - x into the seond equation and solve:
- y - 3x = 5
- (3 - x) - 3x = 5
- -4x = 2
- x = -1/2
Using that answer, substitute it in the first equation and solve for the second variable.
- y + x = 3
- y + (-1/2) = 3
- y=3 1/2
So (-1/2, 3 1/2) is the solution.
- Add or subtract one equation from the other so that a variable with cancel out and you are left with one variable equaling something.
- Once you have the value for one variable, plug it back into either equation to get the answer to the other variable.
Example:
3x + y = 9
4x - y = 5
Add the two equations:
(3x + y) + (4x - y) = 9 + 5
The y's cancel out and you are left with:
7x = 14
So then x=2
Plug x=2 back into either equation to get the answer to y:
- 3x + y = 9
- 3(2)+ y = 9
- 6 + y = 9
- y = 3
- So then (2, 3) is the solution.
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