University:
Glendale Community College
Course:
MAT142 | College Mathematics
Academic year:

2019
Views:

408

Pages:

21
Author:

Davon Fischer

Section 8.1 Solutions Section 8.1: Systems of Linear Equations, Substitution and Elimination #1-6: Verify that the given values are solutions to the system of equations. 3x + 2y = 11 2x - y = 5 (3,1) 3(3) + 2(1) = 11 9 + 2 = 11 11 = 11 2(3) - 1 = 5 6 - 1 = 5 5 = 5 (3,1) is a solution 2/3x + 1/2y = 3 2x - y = 4 (3,2) 2/3(3) + 1/2(2) = 3 2 + 1 = 3 3 = 3 2(3) - 2 = 4 6 - 2 = 4 4 = 4 (3,2) is a solution x - y + z = 2 2x - 3y + 2z = 2 x + y - 2z = 3 (3,2,1) 3 - 2 + 1 = 2 2 = 2 2(3) - 3(2) + 2(1) = 2 6 - 6 + 2 = 2 2 = 2 3 + 2 - 2(1) = 3 3 + 2 - 2 = 3 3 = 3 (3,2,1) is a solution #7-14: Solve each system of equations using the substitution method. If the system has no solutions, say that it is inconsistent. If the system has infinitely many solutions write your answer in the form {(x,y)|y=mx+b} **7) y = 2x - 4** **-2x + 3y = 0** Replace y in the second equation with 2x - 4: -2x + 3(2x - 4) = 0 -2x + 6x - 12 = 0 4x = 12 x = 3 Now solve for y: y = 2(3) - 4 y = 6 - 4 y = 2 Answer: (3,2) **9) 2x + y = 11** **2x + 3y = 5** Solve bottom equation for y: 2x + 3y = 5 3y = -2x + 5 y = -2/3x + 5/3 Replace y in top equation with -2/3x + 5/3: 3x + 2(-2/3x + 5/3) = 11 3x - 4/3x + 10/3 = 11 5/3x = 23/3 x = 23/3 * 3/5 x = 23/5 Now find y: y = -2/3(23/5) + 5/3 y = -46/15 + 25/15 y = -21/15 = -7/5 Answer: (23/5, -7/5)