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Solve Linear System for x, y, and z











































Solve Linear System of Equations

   • Find the value for x, y and z:

           2x+2y+4z=48

           2x+9y+7z=105

           x+4y+z=37

Comments for Solve Linear System for x, y, and z

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Oct 10, 2012
Solve Linear System
by: Staff


Answer:

Part I


Solve for x, y, and z


1st Equation: 2x + 2y + 4z = 48


2nd Equation: 2x + 9y + 7z = 105


3rd Equation: x + 4y + z = 37


Subtract the 2nd Equation from the 1st Equation. This will eliminate the variable “x”.

  2x + 2y + 4z =  48

-(2x + 9y + 7z = 105)
----------------------------
0 - 7y - 3z = -57

-7y - 3z = -57


Solve for “z” in terms of y


Add 3z to each side of the equation

-7y - 3z + 3z = -57 + 3z

-7y + 0 = -57 + 3z

-7y = -57 + 3z

Add 57 to each side of the equation

-57 + 3z = -7y

-57 + 3z + 57 = -7y + 57

3z + 57 - 57 = -7y + 57

3z + 0 = -7y + 57

3z = -7y + 57

Divide each side of the equation by 3

3z = 57 - 7y

3z / 3 = (57 - 7y) / 3

z * (3 / 3) = (57 - 7y) / 3

z * (1) = (57 - 7y) / 3

z = (57 - 7y) / 3

z = 19 - (7/3)y




Multiply the 3rd Equation by 2.
2 * [(x + 4y +  z)  =   37]


2x + 8y + 2z = 74


Subtract the new 3rd Equation from the 2nd Equation. This will eliminate the variable “x”.

  

2x + 9y + 7z = 105
-(2x + 8y + 2z = 74)
------------------------------
0 + y + 5z = 31

y + 5z = 31


---------------------------------------------------------

Oct 10, 2012
Solve Linear System
by: Staff


---------------------------------------------------------
Part II

Solve for “y” in terms of z


Subtract 5z from each side of the equation

y + 5z - 5z = 31 - 5z

y + 0 = 31 - 5z

y = 31 - 5z


Solve for the final value of “z”

Substitute 31 - 5z for y in the equation z = 19 - (7/3)y


z = 19 - (7/3)y

z = 19 - (7/3)*( 31 - 5z)

Use the distributive property to eliminate the parentheses

z = 19 - (7/3)*( 31 - 5z)

z = 19 + (-7/3)*31 +(-7/3)*(- 5z)

z = 19 + (-217/3) +(35z/3)

Convert all fractions on the right side
of the equation to the same common
denominator

z = 57/3 + (-217/3) + (35z/3)

Combine fractions where possible

z = 57/3 + (-217/3) + (35z/3)

z = -160/3 + (35z/3)

Multiply each side of the equation by 3

3*z = 3*[-160/3 + (35z/3)]

3z = -160 + 35z

Add 160 to each side of the equation

3z + 160 = -160 + 35z + 160

3z + 160 = 0 + 35z

3z + 160 = 35z

Subtract 3z from each side of the equation

3z + 160 - 3z = 35z - 3z

3z - 3z + 160 = 35z - 3z

0 + 160 = 35z - 3z

160 = 35z - 3z

160 = 32z

32z = 160

Divide each side of the equation by 32

32z / 32 = 160 / 32

z * (32/ 32) = 160 / 32

z * (1) = 160 / 32

z = 160 / 32

z = 5



Solve for the final value of “y”

Substitute 5 for z in the following equation:


y = 31 - 5z

y = 31 – 5*5

y = 31 – 25

y = 6


---------------------------------------------------------

Oct 10, 2012
Solve Linear System
by: Staff


---------------------------------------------------------
Part III


Solve for the final value of “x”

 Substitute 5 for z and 6 for y in the following equation:


x + 4y + z = 37

x + 4*6 + 5 = 37

x + 24 + 5 = 37

x + 29 = 37

Subtract 29 from each side of the equation

x + 29 - 29 = 37 - 29

x + 0 = 37 - 29

x = 37 - 29

x = 8




Final Answer:

                 x = 8

                 y = 6

                 z = 5

------------------------------------------------

Check the answers by substituting the values of x, y, and z into the original equations


1st Equation: 2x + 2y + 4z = 48

2*8 + 2*6 + 4*5 = 48

16 + 12 + 20 = 48

48 = 48, OK


2nd Equation: 2x + 9y + 7z = 105

2*8 + 9*6 + 7*5 = 105

16 + 54 + 35 = 105

105 = 105, OK


3rd Equation: x + 4y + z = 37

8 + 4*6 + 5 = 37

8 + 24 + 5 = 37

37 = 37, OK




The solutions for x, y, and z are valid because they meet the conditions of the three original equations.




Thanks for writing.

Staff
www.solving-math-problems.com



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