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Consider the linear system: 5 x + 3 y + 3 z = 1 4у — 3 z %3D 2 2 z = 3 Using backwards substitution, the solution is y = z = You must enter your answers as whole numbers or fractions, not decimals. Fractions must be entered as ratios eg. 5/2, not 2 1/2. You can verify that your solution is correct by substituting the values back into the original equations.

Consider the linear system: 5 x + 3 y + 3 z = 1 4у — 3 z %3D 2 2 z = 3 Using backwards substitution, the solution is y = z = You must enter your answers as whole numbers or fractions, not decimals. Fractions must be entered as ratios eg. 5/2, not 2 1/2. You can verify that your solution is correct by substituting the values back into the original equations.

Intermediate Algebra
Intermediate Algebra
19th Edition
ISBN:9780998625720
Author:Lynn Marecek
Publisher:Lynn Marecek
Chapter4: Systems Of Linear Equations
Section4.3: Solve Mixture Applications With Systems Of Equations
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Consider the linear system: 5 x + 3 y + 3 z = 1 4у — 3 z %3D 2 2 z = 3 Using backwards substitution, the solution is y = z = You must enter your answers as whole numbers or fractions, not decimals. Fractions must be entered as ratios eg. 5/2, not 2 1/2. You can verify that your solution is correct by substituting the values back into the original equations.
Transcribed Image Text:Consider the linear system: 5 x + 3 y + 3 z = 1 4у — 3 z %3D 2 2 z = 3 Using backwards substitution, the solution is y = z = You must enter your answers as whole numbers or fractions, not decimals. Fractions must be entered as ratios eg. 5/2, not 2 1/2. You can verify that your solution is correct by substituting the values back into the original equations.
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