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2 Using the substitution method, Ken solves the following system of equations algebraically. 2x - y = 5 3x + 2y =-3 Which equivalent equation could Ken use? 1) 3x + 2(2x – 5) = -3 2) 3x + 2(5 – 2x) = -3 3) 3y++2y --3 5 -y+2y = -3 4) 3.

2 Using the substitution method, Ken solves the following system of equations algebraically. 2x - y = 5 3x + 2y =-3 Which equivalent equation could Ken use? 1) 3x + 2(2x – 5) = -3 2) 3x + 2(5 – 2x) = -3 3) 3y++2y --3 5 -y+2y = -3 4) 3.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter6: Systems Of Linear Equations And Inequalities
Section6.5: Applying Systems Of Linear Equations
Problem 1AGP
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2 Using the substitution method, Ken solves the following system of equations algebraically. 2x – y = 5 3x + 2y = -3 Which equivalent equation could Ken use? 1) 3x + 2(2x – 5) = -3 3x + 2(5 – 2x) =-3 2) 3) 3{y+) (y++2y =-3 4) 3 + 2y = -3
Transcribed Image Text:2 Using the substitution method, Ken solves the following system of equations algebraically. 2x – y = 5 3x + 2y = -3 Which equivalent equation could Ken use? 1) 3x + 2(2x – 5) = -3 3x + 2(5 – 2x) =-3 2) 3) 3{y+) (y++2y =-3 4) 3 + 2y = -3
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