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Solve the system of equations by substitution.\newliney=8y = 8\newline2x−2y+z=−132x - 2y + z = -13\newline−2x+y−3z=−1-2x + y - 3z = -1

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Q. Solve the system of equations by substitution.\newliney=8y = 8\newline2x−2y+z=−132x - 2y + z = -13\newline−2x+y−3z=−1-2x + y - 3z = -1
  1. Substitute y=8y = 8: First, let's substitute y=8y = 8 into the second and third equations.2x−2(8)+z=−132x - 2(8) + z = -13−2x+8−3z=−1-2x + 8 - 3z = -1
  2. Simplify the equations: Now, simplify the equations. 2x−16+z=−132x - 16 + z = -13 −2x+8−3z=−1-2x + 8 - 3z = -1
  3. Add constants: Add 1616 to both sides of the first simplified equation.\newline2x+z=32x + z = 3
  4. Eliminate xx: Add −8-8 to both sides of the second simplified equation.\newline−2x−3z=−9-2x - 3z = -9
  5. Combine equations: Now we have a system of two equations with two variables:\newline2x+z=32x + z = 3\newline−2x−3z=−9-2x - 3z = -9\newlineLet's add these two equations together to eliminate xx.
  6. Solve for z: Adding the equations gives us:\newline(2x−2x)+(z−3z)=3−9(2x - 2x) + (z - 3z) = 3 - 9\newline0x−2z=−60x - 2z = -6
  7. Substitute z=3z = 3: Divide both sides by −2-2 to solve for zz.z=3z = 3
  8. Solve for x: Now we'll substitute z=3z = 3 back into one of the two-variable equations to solve for xx.2x+3=32x + 3 = 3
  9. Find yy: Subtract 33 from both sides to solve for xx.2x=02x = 0
  10. Final solution: Divide both sides by 22 to find the value of xx.x=0x = 0
  11. Final solution: Divide both sides by 22 to find the value of xx.x=0x = 0We have found the values for xx and zz. Now we substitute these values into the original equation to find yy.y=8y = 8 (given)
  12. Final solution: Divide both sides by 22 to find the value of xx.\newlinex=0x = 0 We have found the values for xx and zz. Now we substitute these values into the original equation to find yy.\newliney=8y = 8 (given) We have the solution to the system of equations:\newlinex=0x = 0, y=8y = 8, z=3z = 3

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