8.2
SOLVING BY SUBSTITUTION OR ELIMINATION
Homework:
p. 485; Do Every Other Odd Problems 3-35.
| Nancy J. McCormick's Home Page | DSPM 0850 Class Notes |
A. Solving Systems by the
Substitution Method
1. Take one of the equations and get either of the
variables by itself. (Look for coefficients of 1 or –1 to make the work
easier).
2. Take the expression found in step one that is
equal to one of the variables and substitute the expression into the other
equation in place of that variable.
Example:
Solve the system 
using
substitution.
In the first equation of the system, we see that the variable x
has a coefficient of one. This means that we could solve the first equation
for x (get x by itself) without having to divide.
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We substitute into the other equation, putting |
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Remove the parentheses. Combine the like terms on left. Add 45 to both sides Divide both sides by 26. |
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Put the value for y into the equation that is solved for x and perform the calculations to get the value of x. We have the solution of the system. |
was
subtracted from both sides of the first equation.
in
place of the variable 
.
Example:
Solve the system 
using
substitution.
In the second equation of the system, we see that y
has a coefficient of negative one. Dividing by –1 will not result in
fractions.
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Subtract 3x from both sides. Divide each term of the equation by –1. |
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We substitute into the other equation, putting |
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Remove parentheses and combine the like terms on the left side of the equation. Add 14 to both sides and divide both sides by 8. |
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Put the value for x into the equation that is solved for y and perform the calculations to get the value of y. We have the solution of the system. |
in
place of the variable 
.
B. Solving Systems By Elimination
1. Get the variable expressions on the left side
and the constants on the right side in both equations.
2. If it is not already the case, multiply one or
both equations by some constant so that the coefficients of one of the
variables are opposites (the coefficients will add to zero).
3. Add the two equations to eliminate one of the
variables so that the value of the variable that remains can be determined.
Example:
Solve the system 
using
elimination.
We see that the coefficients of x have opposite signs and if we multiply the
first equation by the constant 2, the coefficients of x will be –4 and 4
which indeed are opposites and will add to zero.
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Multiply the first equation by 2. Be careful to multiply each term of the equation by 2, not forgetting the constant on the right. No change is made to the second equation. |
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Add the two equations, which results in one equation having only the variable y. Solve to find the value of y. |
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Plug the value of 2 in place of the y in either of the original equations to find the value of x. |
Remember that you can check your results by substituting the x and y-coordinates into the equations to verify that these values make both equations true. You can also check your results by using the grapher. We will check the results in the example above by graphing and determining that the point of intersection is the point (-2,2).
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Get the variable y by itself in each of the equations. Enter these equations into the Y= window of the grapher. |
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GRAPH. |
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Use 2nd CALC, intersect to verify that (-2,2) is the solution of the system. |
Solving graphically, you can tell when the lines are parallel or when they are
the same line.
When solving algebraically, you can also determine when a system is
inconsistent (parallel lines with no solution). In this case, whether solving
by substitution or by elimination, you will get a false statement such as 3=0.
However, when solving algebraically you get the true statement 0=0, the system
is dependent (the same line) and has infinite solutions.
Example:
Solve the system 
using
elimination.
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Since the coefficients of x or of y are not multiples of each other, we must multiply both equations by some constant. |
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I chose to eliminate the y, multiplying the first equation by 5 and multiplying the second equation by –2. |
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Adding the two equations and solving the resulting equation, we find the value of x. |
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Since the value of x is a fraction, it might be easier to find the value of y by repeating the elimination process and eliminate the variable x this time. |
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Adding the two equations
and solving the resulting equation, we find the value of y. |
