Advanced Mathematical Concepts: Precalculus with Applications, Student Edition
Advanced Mathematical Concepts: Precalculus with Applications, Student Edition
1st Edition
ISBN: 9780078682278
Author: McGraw-Hill, Berchie Holliday
Publisher: Glencoe/McGraw-Hill
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Chapter 5.6, Problem 40E
To determine

To solve: The given system of equations algebraically.

Expert Solution & Answer
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Answer to Problem 40E

  x=0, y=4, z=2

Explanation of Solution

Given information: System of equations:

  4x+y+2z=03x+4y2z=202x+5y+3z=14

Calculation:

  4x+y+2z=0(equation 1)3x+4y2z=20(equation 2)2x+5y+3z=14(equation 3)

Start by combining equations 1 and 2 to get,

  7x+5y=20(z's cancel out)

Next we want to combine equations 2 and 3 in way to get rid of the z term there also. But we must multiply each equation in such a way as to make that possible.

  3(3x+4y2z=20)  9x+12y6z=602(2x+5y+3z=14)  4x+10y+6z=28

Combine these

  9x+12y6z=604x+10y+6z=28

Gives us:

  5x+22y=88

Now we use this equation and the equation we got by combining (1) and (2) to solve for x and y ,

  5x+22y=887x+5y=20

To get a common coefficient on the x term the top equation must be multiplied by 7 and the bottom equation by 5.

  7(5x+22y=88)=35x+154y=6165(7x+5y=20)=35x+25y=100

Now subtract the bottom row from the top,

  35x+154y=616(35x+25y=100)129y=516y=4

Plugging back into any of the previous equations that only had x and y in it can get us x .

  7x+5y=207x+5(4)=20, 7x=0, x=0

Now plug this x and y into one of the three original equations to solve for z , I will use equation (1).

  4(0)+4+2z=02x=4z=2

Thus, x=0, y=4, z=2

Chapter 5 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

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