1. Algebra 1
Section 7.3

2. Substitution Method
Sometimes a common factor can be found, making this method easier to use.

3. Example 1 3x + 7y = 34 2x + 4y = 20 2x = -4y + 20 x = -2y + 10
3(-2y + 10) + 7y = 34
y + 30 = 34
y = 4

4. Example 1 Since we know that y = 4... x = -2y + 10 x = -2(4) + 10
Solution: (2, 4)

5. Substitution Method
You cannot always avoid fractions when solving by substitution.
If the coefficient of a term is a factor of the like term in the other equation, solve for that variable.

6. Example 2 2x + 3y = 16 4x + 5y = 28 2x = -3y + 16 x = -3y + 16 2

7. Example 2 -3y + 16 4( ) + 5y = 28 2 2(-3y + 16) + 5y = 28

8. Example 2 Since we know that y = 4... -3y + 16 x = 2 -3(4) + 16 x = 2
Solution: (2, 4)

9. Example 4 Let a = adult ticket price c = child ticket price
Moores: 2a + 3c = 82
Smythes: 3a + 5c = 130

10. Example 4 2a + 3c = 82 3a + 5c = 130 2a = -3c + 82 -3c + 82 a = 2 -3c

11. Example 4 2a + 3c = 82 3a + 5c = 130 3( ) + 5c = 130 + 41 -3c 2
2( )
(2)
-9c c = 260

12. Example 4 2a + 3c = 82 3a + 5c = 130 -9c + 246 + 10c = 260

13. Since we know that c = 14 and
Example 4
2a + 3c = a + 5c = 130
Since we know that c = 14 and
a =
-3c 2
a =
-3(14) 2 =
= 20

14. Example 4 Let a = adult ticket price c = child ticket price
The discounted adult tickets are $20.
The discounted child tickets are $14.

15. Example 5 Let x = principal in 2% account
y = principal in 4.5% account
P r t I 2% x
0.02 2
0.04x
4.5% y
0.045 2
0.09y

16. Example 5 The total principal is $20,000. x + y = 20,000
The total interest is $1175.
0.04x y = 1175
y = -x + 20,000

17. Example 5
0.04x y = 1175
4x + 9y = 117,500
4x + 9(-x + 20,000) = 117,500
4x – 9x + 180,000 = 117,500
-5x + 180,000 = 117,500
-5x = -62,500
x = 12,500

18. Mr. Peterson invested $12,500 at 2% and $7500 at 4.5%.
Example 5
Since we know that x = 12,500...
y = -12, ,000
y = 7500
Mr. Peterson invested $12,500 at 2% and $7500 at 4.5%.

19. Homework: pp