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Solve the system of equations by elimination.\newlinex−3y−2z=−20x - 3y - 2z = -20\newlinex−3y+z=−8x - 3y + z = -8\newline2x+3y−3z=92x + 3y - 3z = 9

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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.\newlinex−3y−2z=−20x - 3y - 2z = -20\newlinex−3y+z=−8x - 3y + z = -8\newline2x+3y−3z=92x + 3y - 3z = 9
  1. Combine Equations to Eliminate zz: First, let's eliminate zz by adding the first two equations.\newlinex−3y−2z=−20x - 3y - 2z = -20\newlinex−3y+z=−8x - 3y + z = -8\newlineAdd them up:\newline(x−3y−2z)+(x−3y+z)=−20+(−8)(x - 3y - 2z) + (x - 3y + z) = -20 + (-8)\newline2x−6y−z=−282x - 6y - z = -28
  2. Prepare Second Equation for Elimination: Now, let's multiply the second equation by 33 to prepare it for elimination with the third equation.\newline3(x−3y+z)=3(−8)3(x - 3y + z) = 3(-8)\newline3x−9y+3z=−243x - 9y + 3z = -24
  3. Eliminate z by Adding Equations: Add the new equation to the third equation to eliminate z.\newline3x−9y+3z=−243x - 9y + 3z = -24\newline2x+3y−3z=92x + 3y - 3z = 9\newlineAdd them up:\newline(3x−9y+3z)+(2x+3y−3z)=−24+9(3x - 9y + 3z) + (2x + 3y - 3z) = -24 + 9\newline5x−6y=−155x - 6y = -15
  4. Solve for x Using Elimination: Now we have two equations with two variables:\newline2x−6y−z=−282x - 6y - z = -28\newline5x−6y=−155x - 6y = -15\newlineLet's solve for x by multiplying the first equation by 55 and the second by 22.\newline5(2x−6y−z)=5(−28)5(2x - 6y - z) = 5(-28)\newline2(5x−6y)=2(−15)2(5x - 6y) = 2(-15)\newlineWe get:\newline10x−30y−5z=−14010x - 30y - 5z = -140\newline10x−12y=−3010x - 12y = -30
  5. Eliminate xx to Solve for yy: Subtract the second new equation from the first to eliminate xx.
    (10x−30y−5z)−(10x−12y)=−140−(−30)(10x - 30y - 5z) - (10x - 12y) = -140 - (-30)
    −18y−5z=−110-18y - 5z = -110
  6. Substitute zz into Equation to Solve for yy: Now let's solve for yy using the equation −18y−5z=−110-18y - 5z = -110. We can substitute zz from 2x−6y−z=−282x - 6y - z = -28. First, solve for zz in terms of xx and yy: 2x−6y−z=−282x - 6y - z = -28 yy00 Now substitute zz into −18y−5z=−110-18y - 5z = -110: yy33 yy44 yy55

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