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7.5
Solving Systems of Linear Equations by Elimination
7.5 Solving Systems of Linear Equations by Elimination
• 1. Solve linear systems by elimination.
• 2. Multiply when using the elimination method.
• 3. Use an alternative method to find the second
value in a solution.
• 4. Solve special systems by elimination.
Objective 1
Solve linear systems by elimination.
•An
algebraic method
that depends on the addition
Elimination
Method
property of equality can also be used to solve systems.
Adding the same quantity to each side of an equation
results in equal sums:
•
If A = B, then A + C = B + C.
• We can take this addition a step further. Adding equal
quantities, rather than the same quantity, to each
side of an equation also results in equal sums:
• If A = B and C = D, then A + C = B + D.
• The elimination method uses the addition property
of equality to solve systems of equations.
Classroom
Example 1
Using the Elimination Method
• Solve the system by the elimination method.
3x  y  7
To find the corresponding
2x  y  3
y-value, substitute 2 for x
in either of the two
equations of the system.
3x  y  7
2x  y  3
5x
 10
x2
2x  y  3
2(2)  y  3
4 y 3
y  1
The solution set is {(2, –1)}.
Solving a Linear System by Elimination
•
Step 1 Write both equations in standard form
Ax + By = C.
•
Step 2 Transform the equations as needed so
that the coefficients of one pair of variable
terms are opposites. Multiply one or both
equations by appropriate numbers so that the
sum of the coefficients of either the x- or the yterms is 0.
•
Step 3 Add the new equations to eliminate a
variable. The sum should be an equation with
just one variable.
Solving a Linear System by Elimination
•
Step 4 Solve the equation from Step 3 for the
remaining variable.
•
Step 5 Find the other value. Substitute the
result from Step 4 into either of the original
equations, and solve for the other variable.
•
Step 6 Check the values in both of the original
equations. Then write the solution set as a set
containing an ordered pair.
Classroom
Example 2
Using the Elimination Method
• Solve the system.
x  2  y
2 x  y  10
• Write both equations in standard form.
• Eliminate y.
x y 2
2 x  y  10
x y 2
2 x  y  10
•
3x
 12
x4
The solution set is {(4, –2)}.
Find the value of y.
2 x  y  10
2(4)  y  10
8  y  10
2  y
Objective 2
Multiply when using the elimination
method.
Multiply when using elimination.
• Sometimes we need to multiply each side of one or
both equations in a system by a number before
adding will eliminate a variable.
Classroom
Example 3
Using the Elimination Method
• Solve the system.
4 x  5 y  18 (1)
3 x  2 y  2 (2)
• Multiply each equation by a suitable number so that
the coefficients of one of the two variables are
opposites.
Eliminate y.
• Multiply equation (1) by 2
• and equation (2) by 5.
8 x  10 y  36
2  4 x  5 y   2(18 )
15 x  10 y  10
5  3 x  2 y   5(2)
23 x
 46
x  2
Classroom
Example 3
Using the Elimination Method (cont.)
• Solve the system.
4 x  5 y  18 (1)
3x  2 y  2 (2)
• Find the value of y by substituting 2 for x in either
equation (1) or equation (2).
4 x  5 y  18 (1)
4(2)  5 y  18
8  5 y  18
5 y  10
y2
Check that the solution set of the system is {(–2, 2)}.
Multiply when using elimination.
• In the previous example, we eliminated the variable y.
Alternatively, we could multiply each equation of the
system by a suitable number so that the variable x is
eliminated.
• CAUTION!
• When using the elimination method, remember to
multiply both sides of an equation by the same
nonzero number.
Objective 3
Use an alternative method to find
the second value in a solution.
Finding the Second Value
• Sometimes it is easier to find the value of the second
variable in a solution using the elimination method
twice.
Classroom
Example 4
Finding the Second Value Using an Alternative
Method
• Solve the system.
3y  8  4x
6x  9  2 y
• Write each equation in standard form.
4 x  3 y  8
6x  2 y  9
• Eliminate y by multiplying equation (1) by 2 and
equation (2) by 3.
8 x  6 y  16 To solve for y, start
18 x  6 y  27
over with the
original equations
26 x
 11
11
and eliminate x.
x
26
Classroom
Example 4
Finding the Second Value Using an Alternative
Method (cont.)
• Solve the system.
3y  8  4x
6x  9  2 y
4 x  3 y  8
• Each equation in standard form:
6x  2 y  9
• Eliminate x by multiplying equation (1) by 3 and
equation (2) by –2 .
12 x  9 y  24
12 x  4 y  18
13 y  42
42
y
13
 11 42 
The solution set is  , .
 26 13 
Objective 4
Solve special systems by elimination.
Solving Special Systems Using the Elimination
Method
Classroom
Example 5
• Solve each system by the elimination method.
• a. 3 x  y  7
6x  2 y  5
• Multiply equation (1) by –2.
2  3 x  y   2(7)
6x  2 y  5
6 x  2 y  14
6 x  2 y  5
0  19
• The false statement 0 = 19 indicates that the given
system has solution set .
Classroom
Example 5
Solving Special Systems Using the Elimination
Method (cont.)
• Solve each system by the elimination method.
• b. 2 x  5 y  1
4 x  10 y  2
• Multiply equation (1) by 2.
4 x  10 y  2
4 x  10 y  2
00
• A true statement occurs when the equations are
equivalent. The solution set is {(x, y) | 2x + 5y = 1}.