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How many solutions does the system of equations below have?\newline−3x+y−z=−10-3x + y - z = -10\newline3x−3y+2z=−183x - 3y + 2z = -18\newline3x+3y−z=−143x + 3y - z = -14\newlineChoices:\newline(A)no solution\newline(B)one solution\newline(C)infinitely many solutions

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Q. How many solutions does the system of equations below have?\newline−3x+y−z=−10-3x + y - z = -10\newline3x−3y+2z=−183x - 3y + 2z = -18\newline3x+3y−z=−143x + 3y - z = -14\newlineChoices:\newline(A)no solution\newline(B)one solution\newline(C)infinitely many solutions
  1. Combine equations to eliminate yy: First, let's add the first and the third equation to eliminate yy.
    −3x+y−z=−10-3x + y - z = -10
    3x+3y−z=−143x + 3y - z = -14
    This gives us: 4y−2z=−244y - 2z = -24
  2. Eliminate xx by multiplying and adding equations: Now, let's multiply the first equation by 33 and add it to the second equation to eliminate xx.−9x+3y−3z=−30-9x + 3y - 3z = -303x−3y+2z=−183x - 3y + 2z = -18This gives us: −6z=−48-6z = -48
  3. Solve for z: Solve for z in the equation −6z=−48-6z = -48.z=−48−6z = \frac{-48}{-6}z=8z = 8
  4. Find yy by substituting zz: Substitute z=8z = 8 into the equation 4y−2z=−244y - 2z = -24 to find yy.
    4y−2(8)=−244y - 2(8) = -24
    4y−16=−244y - 16 = -24
    4y=−24+164y = -24 + 16
    4y=−84y = -8
    y=−8/4y = -8 / 4
    zz00
  5. Find xx by substituting yy and zz: Now, substitute y=−2y = -2 and z=8z = 8 into the first original equation to find xx.
    −3x+(−2)−8=−10-3x + (-2) - 8 = -10
    −3x−2−8=−10-3x - 2 - 8 = -10
    −3x−10=−10-3x - 10 = -10
    −3x=−10+10-3x = -10 + 10
    yy00
    yy11
    yy22
  6. Final solution: Since we found a single solution for xx, yy, and zz (x=0x=0, y=−2y=-2, z=8z=8), the system of equations has one solution.

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