General Contents
Detailed Contents
Index
Programmed tutorial: Substitution Method for Linear Systems: Example 2
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Use the
substitution method
to solve
5
x
4
y
= 2
Equation 1
2
x
3
y
= 5
Equation 2
[Here is an
elimination method version
.]
General Contents
Detailed Contents
Index
This problem illustrates how this method is less convenient than the
elimination method
when all of the coefficients are different from 1.
We will have to deal with a fraction.
First, let's solve equation 2 for y.
Subtract 2
x
from both sides.
3
y
= 5 2
x
.
Divide by -3 on both sides.
Substitute this expression for
y
in equation 1.
5
x
4(
y
) = 2 becomes
Multiply both sides by 3 to clear the fraction.
Distribute the 3.
15
x
4*(5 + 2
x
) = 6
Now what do we do?
Multiply and collect like terms as we usually do.
Do it.
We get
15
x
20 8
x
= 6, or
7
x
20 = 6.
Add 20 to both sides.
7
x
= 6 + 20 = 14.
Divide by 7 on both sides to get
x
= 14 / 7 = 2
What do we do next?
We find
y
by using this value of
x
in either original equation.
Let's use it in equation 1.
With
x
= 2,
5
x
4
y
= 2 becomes
5(2) 4
y
= 2, or
10 4
y
= 2.
Subtract 10 from both sides.
4
y
= 2 10 = 12.
Divide by 4
y
= 12 / (4) = 3, as before.
Check this result.
We check
x
= 2,
y
= 3 in
both
of the original equations.
In Equation 1:
5(2) 4(3) = 10 12 = 2 OK
In Equation 2:
2(2) 3(3) = 4 9 = 5 OK.
Summarize these results.
The point (2, 3) solves the original equations.
If we plot the two lines represented by the original equations, which line will (2, 3) be on?
It will be on both lines, at their intersection.
Plot these lines on paper to see this. Check your graph by clicking the Next button.
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