Solve By Substitution
Return to Roadmap 2
Start by solving by substitution Example 2 from the Chapter 2 Roadmap page. Start with the initial system
of two linear equations in two variables.
| 1x - 4y |
= |
20 |
(1) |
| 3x + 4y |
= |
12 |
(2) |
Solve one of the equations for one of the variables. In this case it
is easiest to solve equation (1) for x.
Then substitute the right side of (3) for x in the other equation (2).
| 3( 4y + 20) + 4y |
= |
12 |
(4) |
Simplify and solve that equation which is now a linear equation in 1 variable, namely, y.
| 12y + 60 + 4y |
= |
12 |
(5) |
| 16y + 60 |
= |
12 |
(6) |
| 16y |
= |
– 48 |
(7) |
| y |
= |
– 3 |
(8) |
Now that we know the value of y, namely, – 3,
we substitute that value into either one of the original equations, or into equation (3), which would be
the easiest one to do.
| x |
= |
4(–3) + 20 |
(9) |
| x |
= |
–12 + 20 |
(10) |
| x |
= |
8 |
(11) |
So our solution is the point where x=8 and y=–3,
that is, the point (8,–3).
The problem shown above was facilitated by the fact that it was quite easy to solve one of the eaquations,
namely (1), for one of the variables without getting any fractions.
Here is another problem, but one where the substitution
method generates some fractions during the solution.
Solve:
| 3x + 5y |
= |
47 |
(12) |
| 4x + 11y |
= |
93 |
(13) |
Solve one of the equations for one of the variables. In this case we
will solve equation (12) for y.
| 5y |
= |
–3x + 47 |
(14) |
| y |
= |
(–3/5)x + 47/5 |
(15) |
Then substitute the right side of (15) for x in the other equation (13).
| 4x +11( (–3/5)x + 47/5 ) |
= |
93 |
(16) |
Simplify and solve that equation which is now a linear equation in 1 variable, namely, y.
| 4x + (–33/5)x + 517/5 |
= |
93 |
(17) |
| 20x + –33x + 517 |
= |
465 |
(18) |
| –13x +517 |
= |
465 |
(19) |
| –13x |
= |
–52 |
(20) |
| x |
= |
4 |
(21) |
Now that we know the value of x, namely, 4,
we substitute that value into either one of the original equations, or into equation (15).
In this case we will use equation (12).
| 3(4) + 5y |
= |
47 |
(22) |
| 12 + 5y |
= |
47 |
(23) |
| 5y |
= |
35 |
(24) |
| y |
= |
7 |
(24) |
So our solution is the point where x=4 and y=7,
that is, the point (4,7).
Return to Roadmap 2
©Roger M. Palay Saline, MI 48176 January, 2017