General Contents
Detailed Contents
Index
Programmed tutorial: Substitution Method for Linear Systems: Example 1
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If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Solve
x
+ 7
y
= 4
Equation 1
3
x
2
y
= 11
Equation 2
General Contents
Detailed Contents
Index
How do we proceed?
When we have
one of the coefficients equal to 1, as in Equation 1, the
Substitution Method
is an easy way to solve.
How does that apply here?
Well solve one equation for x and substitute the result in the other equation.
Solve equation 1 for
x
.
We
have
x
+ 7
y
= 4.
Subtract 7
y
from each side.
x
= 4 7
y
Substitute this expression for
x
in equation 2.
(3
x
) 2
y
= 11 becomes
3(47
y
) 2
y
= 11
Simplify this equation.
12 21
y
2
y
= 12 23
y
= 11
Add 12 to both sides.
23
y
= 11 + 12 = 23.
Divide both sides by 23.
y
= 23/(23) = 1
Now we have a value for
y
. Are we done?
No, we need to determine
x
.
How do we do that?
W
e substitute this value of
y
in either of the original equations.
Do it.
Let's use the first equation.
x
+ 7
y
=
4 becomes
x
+ 7(1) =
4, or
x
7 = 4.
Add 7 to both sides.
Summarize these results.
The equations
x
+ 7
y
=
4
3
x
2
y
= 11
have (3, 1) as a solution.
Do we need to check our results?
Yes.
In
both
original equations?
Yes.
Do it.
Using
x
= 3,
y
= 1, check Equation 1.
x
+ 7
y
= 3 + 7(1) = 3 7 = 4 OK.
Now check Equation 2.
3
x
2
y
= 3*3 2*(1) = 9 + 2 =
11 OK
What is the graphical interpretation of these results?
Each equation graphs as a line. The solution is the intersection of the lines.
Plot the lines on paper and then check your graph by clicking the Next button.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
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