Look at an easy example first.
- Solve a system in which
coefficients of one variable are opposites.
-
- 1st: Rearrange the equations so terms line up as: Ax + By = C
Solve:
2y + x = 6 4x - 2y = 12 |
- becomes
Solve: |
x | + 2y | = 6 |
4x | - 2y | = 12 |
-
- 2nd: Multiply none, or one, or both equations by constant(s)
- so that the coefficients of one of the variables are opposites.
-
- In this example, the above condition is already achieved.
-
- The y coefficients are opposites, so, nothing more need be done.
Solve: |
x | + 2y | = 6 |
4x | - 2y | = 12 |
-
- 3rd: Add the two equations together to eliminate one of the variables.
Solve: |
x | + 2y | = 9 |
4x | - 2y | = -3 |
- becomes
Solve: |
x | + 2y | = 9 |
4x | - 2y | = -3 |
|
5x | | =6 |
-
- 4th: Solve.
Solve: |
5x | = | 6 |
5x/5 | = | 6/5 |
x | = | 6/5 |
-
- 5th: Use one of the equations and this value to solve for the other variable.
- It doesn't matter which equation is used to do this.
Solve: |
x | + 2y | = 9 |
6/5 | + 2y | = 9 |
| 2y | = 9 - 6/5 |
| 2y | = 39/5 |
| 2y/2 | = (39/5)/2 |
| y | = 39/10 |
-
- or
Solve: |
4x | - 2y | = -3 |
4(6/5) | - 2y | = -3 |
24/5 | - 2y | = -3 |
| - 2y | = -3 -24/5 |
| - 2y | = -39/5 |
| - 2y/(-2) | = (-39/5)/(-2) |
| y | = 39/10 |
-
- 5th ALTERNATE METHOD: Repeat steps 1 through 4 eliminating the other variable.
- This alternate step avoids the messy substitution and computation with fractions as show in step 5 above.
Solve: |
x | + 2y | = 9 |
4x | - 2y | = -3 |
- becomes
Solve: |
-4(x | + 2y | = 9) |
4x | - 2y | = -3 |
- becomes
Solve: |
-4x | -8y | = -36 |
4x | - 2y | = -3 |
|
| -10y | =-39 |
| -10y/(-10) | =-39/(-10) |
| y | =39/10 |
-
- 6th: State the solution and include values for both variables.
- If you use an ordered pair to do this, usually write in in alphabetical order.
-
- or
-
Solve A System in which Coefficients of One Variable Are Opposites.
The 2nd step: "Multiply none, or one, or both equations by constant(s) so that the
coefficients of one of the variables are opposites." is the most fun!
Through it's simple arithmetic, a sophisticated combination of equations is reduced to a single
easy equation.
It's a lot like finding a common demonimator in which the denominators are not erxactly the
same but opposites.
-
Solve: | 6x | +3y | =2 |
5x | +4y | =8 | | when multiplied |
Solve: |
4(6x | +3y | =2) | -3(5x | +4y | =8) | |
becomes | Solve: |
24x | +12 | =8 | -15x | -12y | =-24 | |
then | |
-
- or
-
-
Solve: | 6x | +3y | =2 |
5x | +4y | =8 | | when multiplied |
Solve: |
5(6x | +3y | =2) | -6(5x | +4y | =8) | |
becomes | Solve: |
30x | +15 | =10 | -30x | -24y | =-48 | |
then | |
-
- or
-
-
Solve: | 4x | +5y | =3 |
x | +3y | =6 | |
when multiplied |
Solve: | 4x | +5y | =3 |
-4(x | +3y | =6) | | becomes |
Solve: |
4x | +5y | =3 | -4x | -12y | =-24 | |
then | |
-
- or
-
-
Solve: | 4x | +5y | =3 |
x | +3y | =6 | |
when multiplied |
Solve: | -3(4x | +5y | =3) |
5(x | +3y | =6) | | becomes |
Solve: |
-12x | -15y | =-9 | 5x | +15y | =30 | |
then | |
Or, these last examples which show how
an inconsistent system and a dependent system look when using
linear combination to solve the system.
-
Solve: | 4x | +2y | =1 |
4x | +2y | =6 | |
when multiplied |
Solve: | 4x | +2y | =1 |
-(4x | +2y | =6) | |
becomes |
Solve: | 4x | +2y | =1 |
-4x | -2y | =-24 | | then |
| 0 | =-23 |
never true | inconsistent system no solution |
|
-
- or
-
-
Solve: | 4x | +2y | =1 |
-8x | -4y | =-2 | | when multiplied |
Solve: | 2(4x | +2y | =1) |
-8x | -4y | =-2 | | becomes |
Solve: |
8x | +4y | =2 | -8x | -4y | =-2 | |
then | | 0 | =0 |
always true | dependent system infinite number of solutions |
|
Solve the system. Mouseover the arrow to check answer. |
x - 2y = -2 -3x - 6y = -6
|
y = -2x + 4 y = -2x + 5
|
5x + 5y = 10 10x + 10y = 12
|
y = 4x - 1 8y = 8x + 8
|
-2x - 2y = -4 3x - 8y = -12
|
-3x+3y=6 6x+9y=12
|
4x + 4y = 10 x + y = 12
|
5x+15y=10 3x+9y=6
|
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- ©
2008, A. Azzolino
- www.mathnstuff.com/math/algebra/asystem.htm
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