Solving Systems of Equations by Substitution
Example
A dependent system solved by substitution
Solve by substitution:
| 2x + 3y |
= 5 + x + 4y |
| y |
= x - 5 |
Solution
Substitute y = x - 5 into the first equation:
| 2x + 3(x - 5) |
= 5 + x + 4(x - 5) |
| 2x + 3x - 15 |
= 5 + x + 4x - 20 |
| 5x - 15 |
= 5x - 15 |
Because the last equation is an identity, any ordered pair that satisfies y = x
- 5 will also satisfy 2x + 3y = 5 + x + 4y. The equations of this system are dependent.
The solution set to the system is the set of all points that satisfy y = x - 5. We
write the solution set in set notation as
{(x, y) | y = x - 5}.
We can verify this result by writing 2x + 3y = 5 + x + 4y in slope-intercept form:
| 2x + 3y |
= 5 + x + 4y |
| 3y |
= -x + 5 + 4y |
| -y |
= -x + 5 |
| y |
= x - 5 |
Because this slope-intercept form is identical to the slope-intercept form of the
other equation, they are two equations that look different for the same straight
line.
Heplful Hint
The purpose of this Example is to
show what happens when a
dependent system is solved
by substitution. If we had
first written the first equation
in slope-intercept form, we
would have known that
the equations are dependent
and would not have done
substitution.
If a system is dependent, then an identity will result after the substitution. If the
system is inconsistent, then an inconsistent equation will result after the substitution.
The strategy for solving an independent system by substitution can be summarized
as follows.
The Substitution Method
1. Solve one of the equations for one variable in terms of the other.
2. Substitute into the other equation to get an equation in one variable.
3. Solve for the remaining variable (if possible).
4. Insert the value just found into one of the original equations to find the value
of the other variable.
5. Check the two values in both equations.
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