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Solve the system of equations by elimination.\newline3x+y−3z=23x + y - 3z = 2\newline3x−3y−z=−143x - 3y - z = -14\newlinex−2y+2z=6x - 2y + 2z = 6

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Solve a system of equations in three variables using eliminationright chevron icon

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Q. Solve the system of equations by elimination.\newline3x+y−3z=23x + y - 3z = 2\newline3x−3y−z=−143x - 3y - z = -14\newlinex−2y+2z=6x - 2y + 2z = 6
  1. Eliminate z from Equations: First, let's eliminate z from the first two equations.\newline3x+y−3z=23x + y - 3z = 2 (Equation 11)\newline3x−3y−z=−143x - 3y - z = -14 (Equation 22)\newlineMultiply Equation 22 by 33 to make the coefficients of z the same.\newline3∗(3x−3y−z)=3∗(−14)3*(3x - 3y - z) = 3*(-14)\newline9x−9y−3z=−429x - 9y - 3z = -42 (Equation 22 modified)\newlineNow, add Equation 11 and Equation 22 modified.\newline(3x+y−3z)+(9x−9y−3z)=2+(−42)(3x + y - 3z) + (9x - 9y - 3z) = 2 + (-42)\newline12x−8y=−4012x - 8y = -40
  2. Solve for xx and yy: Next, let's eliminate zz from the second and third equations.3x−3y−z=−143x - 3y - z = -14 (Equation 22)x−2y+2z=6x - 2y + 2z = 6 (Equation 33) Multiply Equation 33 by −12-\frac{1}{2} to make the coefficients of zz the same.−12×(x−2y+2z)=−12×6-\frac{1}{2}\times(x - 2y + 2z) = -\frac{1}{2}\times6−12x+y−z=−3-\frac{1}{2}x + y - z = -3 (Equation 33 modified) Now, add Equation 22 and Equation 33 modified.(3x−3y−z)+(−12x+y−z)=−14+(−3)(3x - 3y - z) + (-\frac{1}{2}x + y - z) = -14 + (-3)52x−2y=−17\frac{5}{2}x - 2y = -17
  3. Substitute xx and yy into Equation: Now, let's solve the system formed by the two new equations we got.12x−8y=−4012x - 8y = -40 (From Step 11)52x−2y=−17\frac{5}{2}x - 2y = -17 (From Step 22)Multiply the second equation by 44 to make the coefficients of yy the same.4∗(52x−2y)=4∗(−17)4*(\frac{5}{2}x - 2y) = 4*(-17)10x−8y=−6810x - 8y = -68 (Equation 44)Now, subtract Equation 44 from the first new equation.(12x−8y)−(10x−8y)=−40−(−68)(12x - 8y) - (10x - 8y) = -40 - (-68)2x=282x = 28yy00
  4. Find zz: Substitute x=14x = 14 into Equation 44 to find yy.10x−8y=−6810x - 8y = -6810(14)−8y=−6810(14) - 8y = -68140−8y=−68140 - 8y = -68−8y=−68−140-8y = -68 - 140−8y=−208-8y = -208y=26y = 26
  5. Find zz: Substitute x=14x = 14 into Equation 44 to find yy.
    10x−8y=−6810x - 8y = -68
    10(14)−8y=−6810(14) - 8y = -68
    140−8y=−68140 - 8y = -68
    −8y=−68−140-8y = -68 - 140
    −8y=−208-8y = -208
    y=26y = 26Substitute x=14x = 14 and y=26y = 26 into one of the original equations to find zz.
    x=14x = 1422 (Equation 33)
    x=14x = 1433
    x=14x = 1444
    x=14x = 1455
    x=14x = 1466
    x=14x = 1477
    x=14x = 1488

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