Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x2 =t and solve for x1 in terms of t.) 0.02x1 - 0.05x2 = -0.22 0.03x1 + 0.04x2 = 0.59 (x1, x2) =
Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x2 =t and solve for x1 in terms of t.) 0.02x1 - 0.05x2 = -0.22 0.03x1 + 0.04x2 = 0.59 (x1, x2) =
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.7: Solving Systems Of Linear And Quadratic Equations
Problem 29HP
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Question
![Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x2 = t and solve for x1 in
terms of t.)
0.02x1 - 0.05x2 = -0.22
0.03x1 + 0.04x2
0.59
(X1, X2) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5aace38-c620-4ba9-b3fb-a4ecb29bb252%2F3b01362c-e99f-4ff9-b0a8-90255c4b82f0%2F7w8zvld_processed.png&w=3840&q=75)
Transcribed Image Text:Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x2 = t and solve for x1 in
terms of t.)
0.02x1 - 0.05x2 = -0.22
0.03x1 + 0.04x2
0.59
(X1, X2) =
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