College Algebra 9th txtbk

430 CHAPTER 7 SYSTEMS OF EQUATIONS AND MATRICES
The equation x 2 paired with either of the two original equations produces an equivalent
system. So, we can substitute x 2 back into either of the two original equations to solve
for y. We choose the second equation.
2(2) 5y 1
5y 5
y 1
SOLUTION
x 2, y 1, or (2, 1).
CHECK
3x 2y 8 2x 5y 1
3(2) 2(1) ? 8 2(2) 5(1) ? 1
8 ✓ 8 1 ✓ 1
MATCHED PROBLEM 4
Solve using elimination by addition:
6x 3y 3
5x 4y 7
When a system has three equations, we will use elimination to reduce to a system with two
equations and two variables, then solve like we did in Example 4. To help you follow a
solution, we will number the equations as E 1 , E 2 , and so on.
EXAMPLE 5 Solution Using Elimination by Addition
x 2y 3z 2
3x 5y 4z 15
2x 3y 2z 2
E 1
E 2
E 3
SOLUTION
Since the coefficient of x in E 1 is 1, our calculations will be simplified if we use E 1 to eliminate
x from the other equations. First we eliminate x from E 2 by multiplying E 1 by 3 and
adding the result to E 2 .
Equivalent System
3x 6y
9z 6
3E 1
x 2y 3z 2
E 1
3x 5y 4z 15
E 2
11y 13z 9
E 4
11y 13z 9
E 4
2x 3y 2z 2
E 3
Now we use E 1 to eliminate x (the same variable eliminated above) from E 3 by multiplying
E 1 by 2 and adding the result to E 3 .
Equivalent System
2x 4y 6z 4
2E 1
x 2y 3z 2
E 1
2x 3y 2z 2
E 3
11y 13z 9
E 4
y 8z 6
E 5
y 8z 6
E 5
Notice that E 4 and E 5 form a system of two equations with two variables. Next we use E 5
to eliminate y from E 4 and replace E 4 with the result.
11y 88z 66
11y 13z 9
75z 75
11E 5
E 4
E 6