Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the system of equations by elimination.\newline2x−3y−z=12x - 3y - z = 1\newline3x+3y−2z=−123x + 3y - 2z = -12\newline−2x−y−2z=16-2x - y - 2z = 16

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Solve a system of equations in three variables using eliminationright chevron icon

Full solution

Q. Solve the system of equations by elimination.\newline2x−3y−z=12x - 3y - z = 1\newline3x+3y−2z=−123x + 3y - 2z = -12\newline−2x−y−2z=16-2x - y - 2z = 16
  1. Eliminate xx by adding equations: Add the first and third equations to eliminate xx.
    (2x−3y−z)+(−2x−y−2z)=1+16(2x - 3y - z) + (-2x - y - 2z) = 1 + 16
    2x−3y−z−2x−y−2z=172x - 3y - z - 2x - y - 2z = 17
    −4y−3z=17-4y - 3z = 17
  2. Eliminate xx by multiplying and adding equations: Multiply the second equation by 22 and add it to the third equation to eliminate xx.
    (2×(3x+3y−2z))+(−2x−y−2z)=2×(−12)+16(2 \times (3x + 3y - 2z)) + (-2x - y - 2z) = 2 \times (-12) + 16
    6x+6y−4z−2x−y−2z=−24+166x + 6y - 4z - 2x - y - 2z = -24 + 16
    4x+5y−6z=−84x + 5y - 6z = -8
  3. Solve for y from new equations: Now we have two new equations:\newline−4y−3z=17-4y - 3z = 17\newline4x+5y−6z=−84x + 5y - 6z = -8\newlineLet's solve for y from the first new equation.\newline−4y=17+3z-4y = 17 + 3z\newliney=−17+3z4y = -\frac{17 + 3z}{4}
  4. Substitute y in second new equation: Substitute y in the second new equation.\newline4x+5(−17+3z4)−6z=−84x + 5\left(-\frac{17 + 3z}{4}\right) - 6z = -8\newline4x−(854+15z4)−6z=−84x - \left(\frac{85}{4} + \frac{15z}{4}\right) - 6z = -8\newline4x−854−15z4−24z4=−84x - \frac{85}{4} - \frac{15z}{4} - \frac{24z}{4} = -8\newline4x−854−39z4=−84x - \frac{85}{4} - \frac{39z}{4} = -8
  5. Clear fractions by multiplying: Multiply everything by 44 to clear the fractions.\newline4×(4x−854−39z4)=4×(−8)4 \times (4x - \frac{85}{4} - \frac{39z}{4}) = 4 \times (-8)\newline16x−85−39z=−3216x - 85 - 39z = -32
  6. Solve for xx from first equation: Add 8585 to both sides.\newline16x−39z=−32+8516x - 39z = -32 + 85\newline16x−39z=5316x - 39z = 53
  7. Substitute xx in original equation: Now we have two equations with two variables:\newline16x−39z=5316x - 39z = 53\newline−4y−3z=17-4y - 3z = 17\newlineLet's solve for xx from the first equation.\newline16x=53+39z16x = 53 + 39z\newlinex=53+39z16x = \frac{53 + 39z}{16}
  8. Clear fractions by multiplying: Substitute xx in the first original equation.2(53+39z16)−3y−z=12\left(\frac{53 + 39z}{16}\right) - 3y - z = 1106+78z16−3y−z=1\frac{106 + 78z}{16} - 3y - z = 1
  9. Combine like terms: Multiply everything by 1616 to clear the fractions.\newline16×(106+78z16)−16×(3y)−16×(z)=16×(1)16 \times \left(\frac{106 + 78z}{16}\right) - 16 \times (3y) - 16 \times (z) = 16 \times (1)\newline106+78z−48y−16z=16106 + 78z - 48y - 16z = 16
  10. Solve for zz from second equation: Combine like terms.106+62z−48y=16106 + 62z - 48y = 16
  11. Substitute zz in equation: Subtract 106106 from both sides.\newline62z−48y=16−10662z - 48y = 16 - 106\newline62z−48y=−9062z - 48y = -90
  12. Clear fractions by multiplying: Now we have two equations with two variables:\newline62z−48y=−9062z - 48y = -90\newline−4y−3z=17-4y - 3z = 17\newlineLet's solve for zz from the second equation.\newline−3z=17+4y-3z = 17 + 4y\newlinez=−17+4y3z = -\frac{17 + 4y}{3}
  13. Distribute −62-62: Substitute zz in the equation 62z−48y=−9062z - 48y = -90.62(−17+4y3)−48y=−9062\left(-\frac{17 + 4y}{3}\right) - 48y = -90−62(17+4y)/3−48y=−90-62\left(17 + 4y\right) / 3 - 48y = -90
  14. Combine like terms: Multiply everything by 33 to clear the fractions.3×(−62(17+4y)/3)−3×(48y)=3×(−90)3 \times (-62(17 + 4y) / 3) - 3 \times (48y) = 3 \times (-90)−62(17+4y)−144y=−270-62(17 + 4y) - 144y = -270
  15. Solve for yy: Distribute −62-62.
    −62×17−62×4y−144y=−270-62 \times 17 - 62 \times 4y - 144y = -270
    −1054−248y−144y=−270-1054 - 248y - 144y = -270
  16. Substitute yy in equation: Combine like terms.−1054−392y=−270-1054 - 392y = -270
  17. Solve for zz: Add 10541054 to both sides.\newline−392y=−270+1054-392y = -270 + 1054\newline−392y=784-392y = 784
  18. Substitute zz in equation: Divide by −392-392 to solve for yy.y=784−392y = \frac{784}{-392}y=−2y = -2
  19. Substitute zz in equation: Substitute yy back into the equation z=−17+4y3z = -\frac{17 + 4y}{3}.z=−17+4(−2)3z = -\frac{17 + 4(-2)}{3}z=−17−83z = -\frac{17 - 8}{3}z=−93z = -\frac{9}{3}z=−3z = -3
  20. Substitute zz in equation: Substitute yy back into the equation z=−17+4y3z = -\frac{17 + 4y}{3}.
    z=−17+4(−2)3z = -\frac{17 + 4(-2)}{3}
    z=−17−83z = -\frac{17 - 8}{3}
    z=−93z = -\frac{9}{3}
    z=−3z = -3Substitute zz back into the equation x=53+39z16x = \frac{53 + 39z}{16}.
    x=53+39(−3)16x = \frac{53 + 39(-3)}{16}
    yy00
    yy11
    yy22

More problems from Solve a system of equations in three variables using elimination

Question
Solve using substitution. 5x−2y=−7 5x - 2y = -7 x=−5 x = -5 (_,_)
Get tutor helpright-arrow

Posted 6 months ago

Question
Is (1,1)(1,1) a solution to this system of equations? \newline4x+10y=144x + 10y = 14\newlinex+6y=7x + 6y = 7\newlineChoices:\newline(A) yes \newline(B) no
Get tutor helpright-arrow

Posted 6 months ago

Question
Which describes the system of equations below?\newliney=–3x+9y = –3x + 9 \newliney=–3x+9y = –3x + 9 \newlineChoices:\newline(A) consistent and independent\text{(A) consistent and independent} \newline(B) consistent and dependent\text{(B) consistent and dependent} \newline(C) inconsistent\text{(C) inconsistent}
Get tutor helpright-arrow

Posted 6 months ago

Question
Solve using elimination. \newline7x−8y=−17 7x - 8y = -17 \newline −7x+3y=2 -7x + 3y = 2 \newline (____,____)(\_\_\_\_, \_\_\_\_)
Get tutor helpright-arrow

Posted 6 months ago

Question
Solve. \newlinex=−2 x = -2 \newline−2x+2y=−8 -2x + 2y = -8 \newline(____,____)(\_\_\_\_, \_\_\_\_)
Get tutor helpright-arrow

Posted 10 months ago

Question
Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineAt a community barbecue, Mrs. Wilkerson and Mr. Hogan are buying dinner for their families. Mrs. Wilkerson purchases 33 hot dog meals and 33 hamburger meals, paying a total of $36\$36. Mr. Hogan buys 11 hot dog meal and 33 hamburger meals, spending $26\$26 in all. How much do the meals cost?\newlineHot dog meals cost $\$_______ each, and hamburger meals cost $\$________ each.
Get tutor helpright-arrow

Posted 10 months ago

Question
Solve the system of equations by substitution.\newline−3x−y−3z=−11-3x - y - 3z = -11\newlinez=5z = 5\newlinex−y+3z=19x - y + 3z = 19\newline(____.____,____)
Get tutor helpright-arrow

Posted 6 months ago

Question
Solve the system of equations by elimination.\newlinex−3y−2z=10 x - 3y - 2z = 10 \newline3x+2y+2z=14 3x + 2y + 2z = 14 \newline2x−3y−2z=16 2x - 3y - 2z = 16 \newline(_,_,_)(\_,\_,\_)
Get tutor helpright-arrow

Posted 6 months ago

Question
Solve the system of equations.\newliney=x2+36x+3 y = x^2 + 36x + 3 \newliney=22x−37 y = 22x - 37 \newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(_,_)(\_,\_)\newline(_,_)(\_,\_)
Get tutor helpright-arrow

Posted 6 months ago

Question
Solve the system of equations.\newliney=−x−24 y = -x - 24 \newlinex2+y2=488 x^2 + y^2 = 488 \newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(_,_) (\_,\_) \newline(_,_) (\_,\_)
Get tutor helpright-arrow

Posted 6 months ago

View all (39)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant