Algebra 1 Section 7.3

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

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

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

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.

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

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

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

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

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

Example 4 2a + 3c = 82 3a + 5c = 130 3( ) + 5c = 130 + 41 -3c 2 2( ) (2) -9c + 246 + 10c = 260

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

Since we know that c = 14 and Example 4 2a + 3c = 82 3a + 5c = 130 Since we know that c = 14 and a = + 41 -3c 2 a = + 41 -3(14) 2 = -21 + 41 = 20

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

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

Example 5 The total principal is $20,000. x + y = 20,000 The total interest is $1175. 0.04x + 0.09y = 1175 y = -x + 20,000

Example 5 0.04x + 0.09y = 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

Mr. Peterson invested $12,500 at 2% and $7500 at 4.5%. Example 5 Since we know that x = 12,500... y = -12,500 + 20,000 y = 7500 Mr. Peterson invested $12,500 at 2% and $7500 at 4.5%.

Homework: pp. 291-293