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Solve the system of equations by substitution.\newliney=−4y = -4\newline−x+3y−2z=4-x + 3y - 2z = 4\newlinex+2y+z=−19x + 2y + z = -19

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Q. Solve the system of equations by substitution.\newliney=−4y = -4\newline−x+3y−2z=4-x + 3y - 2z = 4\newlinex+2y+z=−19x + 2y + z = -19
  1. Substitute y=−4y = -4: Substitute y=−4y = -4 into the second equation −x+3y−2z=4-x + 3y - 2z = 4.
    −x+3(−4)−2z=4-x + 3(-4) - 2z = 4
    −x−12−2z=4-x - 12 - 2z = 4
  2. Add 1212 to isolate: Add 1212 to both sides to isolate terms with variables on one side.\newline−x−2z=4+12-x - 2z = 4 + 12\newline−x−2z=16-x - 2z = 16
  3. Substitute y=−4y = -4: Now, substitute y=−4y = -4 into the third equation x+2y+z=−19x + 2y + z = -19.\newlinex+2(−4)+z=−19x + 2(-4) + z = -19\newlinex−8+z=−19x - 8 + z = -19
  4. Add 88 to isolate: Add 88 to both sides to isolate terms with variables on one side.\newlinex+z=−19+8x + z = -19 + 8\newlinex+z=−11x + z = -11
  5. Two equations with x and z: Now we have two equations with x and z:\newline11) −x−2z=16-x - 2z = 16\newline22) x+z=−11x + z = -11\newlineLet's solve for z by adding the two equations together.\newline(−x−2z)+(x+z)=16−11(-x - 2z) + (x + z) = 16 - 11\newline−x+x−2z+z=5-x + x - 2z + z = 5\newline−z=5-z = 5
  6. Divide both sides: Divide both sides by −1-1 to solve for zz.z=−5z = -5
  7. Substitute z=−5z = -5: Substitute z=−5z = -5 back into the equation x+z=−11x + z = -11 to find xx.
    x+(−5)=−11x + (-5) = -11
    x−5=−11x - 5 = -11
  8. Add 55 to solve: Add 55 to both sides to solve for xx.x=−11+5x = -11 + 5x=−6x = -6
  9. Solution is (−6,−4,−5)(-6, -4, -5): We have found the values for xx and zz, and we were given y=−4y = -4.\newlineSo the solution is (x,y,z)=(−6,−4,−5)(x, y, z) = (-6, -4, -5).

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