Elimination
We have:
4b + 3r |
= |
23 |
2b + 5r |
= |
29 |
Since we have two variables and want to …nd an equation with just one unknown, our strategy is known as elimination (sometimes reduction)
we must get rid of one of the variables (unknowns)
This can be done in essentially 2 di¤erent ways:
solve one equation for one unknown and then substitute for it into the other equation
add or subtract equations
Steve Sugden (Bond University) |
|
10 May 2011 |
6 / 13 |
Elimination method 1
We illustrate this method on the system:
4b + 3r |
= |
23 |
2b + 5r |
= |
29 |
Solving the second equation for b we get
b = 29 5r 2
Now substitute this expression for b into the …rst equation:
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2 |
|
4 |
29 5r |
+ 3r = 23 |
Steve Sugden (Bond University) |
|
10 May 2011 |
7 / 13 |
Elimination method 1
Simplifying:
2 (29 5r ) + 3r |
= 23 |
|
58 10r + 3r |
= |
23 |
58 7r |
= |
23 |
7r |
= |
58 23 |
7r |
= |
35, so r = 5 |
We have done the elimination step (we eliminated the unknown b and thus were able to solve for r.
The next step is called back-substitution: we put the value of r just found back into one of the equations and calculate the value of b.
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b = |
29 5r |
= |
29 25 |
= 2 |
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What is the next step? |
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Steve Sugden (Bond University) |
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10 May 2011 8 / 13 |
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Elimination method 2
We illustrate this method on the system:
4b + 3r |
= |
23 |
2b + 5r |
= |
29 |
Multiply the second equation by 2 to get:
4b + 3r |
= |
23 |
4b + 10r |
= |
58 |
Now subtract the …rst equation from the second to get 7r = 35, whence r = 5.
To get the value for b, we now put r = 5 in either of the original equations and solve for b.
Steve Sugden (Bond University) |
|
10 May 2011 |
9 / 13 |