College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 5 - Section 5.1 - Systems of Linear Equations - 5.1 Exercises - Page 492: 110

Answer

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Work Step by Step

From the previous task, solution is $(40,15,30)$. Let's check the solution by substituting in each equation: $$25x+40y+20z=2200$$ $$25(40)+40(15)+20(30)=1000+600+600=2200$$ First is correct. $$4x+2y+3z=280$$ $$4(40)+2(15)+3(30)=160+30+90=280$$ Second is correct. $$3x+2y+1z=180$$ $$3(40)+2(15)+1(30)=120+30+30=180$$ Third is correct.
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