1. Systems of Linear Equations and Problem Solving
Chapter 8
Systems of Linear Equations and Problem Solving

2. Solving by Substitution or Elimination
8.2
The Substitution Method
The Elimination Method

3. The Substitution Method
Algebraic (nongraphical) methods for solving systems are often superior to graphing, especially when fractions are involved. One algebraic method, the substitution method, relies on having a variable isolated.

4. Solve the system (1) (2) x + y = 2 x + (x – 1) = 2 Solution
The equations are numbered for reference.
Solution
Equation (2) says that y and x – 1 name the same number. Thus we can substitute x – 1 for y in equation (1):
x + y = 2
x + (x – 1) = 2
Equation (1)
Substituting x – 1 for y
We solve the last equation:

5. x + (x – 1) = 2 2x – 1 = 2 2x = 3 x = . y = 3/2 – 1 y = .
Solution (continued)
x + (x – 1) = 2
2x – 1 = 2
2x = 3
x = .
Now return to the original pair of equations and substitute 3/2 for x in either equation so that we can solve for y
Equation (2)
Substituting 3/2 for x
y = 3/2 – 1
y = .

6. We obtain the ordered pair
Solution (continued)
We obtain the ordered pair
We can substitute the ordered pair into the original pair of equations to check that it is the solution.

7. Next, proceed as in the last example, by substituting:
Solve the system
(1)
(2)
Solution
First, select an equation to solve for one variable. To isolate y, subtract 3x from both sides of equation (1):
(1)
(3)
Next, proceed as in the last example, by substituting:

8. 2x – 3y = 7 2x – 3(5 – 3x) = 7 2x – 15 + 9x = 7 11x = 22 x = 2.
Solution (continued)
2x – 3y = 7
Equation (2)
2x – 3(5 – 3x) = 7
Substituting 5 – 3x for y
2x – x = 7
11x = 22
x = 2.
We can substitute 2 for x in either equation (1), (2), or (3). It is easiest to use (3) because it has already been solved for y:
y = 5 – 3(2)
y = –1.

9. We obtain the ordered pair (2, –1).
Solution (continued)
We obtain the ordered pair (2, –1).
We can substitute the ordered pair into the original pair of equations to check that it is the solution.

10. We can substitute 7y + 2 for x in equation (1) and solve:
Solve the system
(1)
(2)
Solution
We can substitute 7y + 2 for x in equation (1) and solve:
7y + 2 = 7y + 5
Substituting 7y + 2 for x
2 = 5.
Subtracting 7y from both sides
When the y terms drop out, the result is a contradiction. We state that the system has no solution.

11. The Elimination Method
The elimination method for solving systems of equations makes use of the addition principle: If a = b, then a + c = b + c.

12. Solve the system (1) (2) Solution
Note that according to equation (2), –x + y and 9 are the same number. Thus we can work vertically and add –x + y to the left side of equation (1) and 9 to the right side:
x + y = 5 (1)
–x + y = 9 (2)
0x + 2y = 14.
Adding

13. Next, we substitute 7 for y in equation (1) and solve for x:
Solution (continued)
This eliminates the variable x, and leaves an equation with just one variable, y for which we solve:
2y = 14
y = 7
x + 7 = 5
Substituting. We also could have used equation (2).
Next, we substitute 7 for y in equation (1) and solve for x:
x = –2

14. We obtain the ordered pair (–2, 7).
Solution (continued)
We obtain the ordered pair (–2, 7).
We can substitute the ordered pair into the original pair of equations to check that it is the solution.

15. Solve the system
(1)
(2)
Solution
To clear the fractions in equation (2), we multiply both sides of equation (2) by 12 to get equation (3):
(3)

16. Solution (continued) (1) (3) Now we solve the system
Notice that if we add equations (1) and (3), we will not eliminate any variables. If the –2y in equation (1) were changed to –4y, we would. To accomplish this change, we multiply both sides of equation (1) by 2:
2x – 4y = –12
Multiply eqn. (1) by 2
3x + 4y = 12
(3)
5x + 0y = 0
Adding
x = 0.
Solving for x

17. We obtain the ordered pair (0, 3).
Solution (continued)
Substituting 0 for x in equation (3)
Then
3(0) + 4y = 12
4y = 12
y = 3
We obtain the ordered pair (0, 3).
We can plug the ordered pair into the original pair of equations to check that it is the solution.

18. Solve the system (1) (2) Solution 4x – 6y = 4 – 4x + 6y = –4 0 = 0
Multiply equation (1) by 2
– 4x + 6y = –4
0 = 0
Add
Note that what remains is an identity. Any pair that is a solution of equation (1) is also a solution of equation (2). The equations are dependent and the solution set is infinite:

19. Rules for Special Cases
When solving a system of two linear equations in two variables:
1. If an identity is obtained, such as 0 = 0, then the system has an infinite number of solutions. The equations are dependent and, since a solution exists, the system is consistent.
2. If a contradiction is obtained, such as 0 = 7, then the system has no solution. The system is inconsistent.