# 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

 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

 Nov 15, 2021 Solve this system of equations NEW by: Anonymous 3.6x+5.2y-z= -2.2 3.2x- 4.8y + 3.9z= 8.1 6.4z + 4.1y +2.3x= 5.1 x= y= z=

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