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Solve the system of equations by substitution.\newline2x+y3z=16-2x + y - 3z = 16\newline3xy+3z=103x - y + 3z = -10\newliney=7y = 7

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Q. Solve the system of equations by substitution.\newline2x+y3z=16-2x + y - 3z = 16\newline3xy+3z=103x - y + 3z = -10\newliney=7y = 7
  1. Substitute y=7y = 7: Substitute y=7y = 7 into the first equation 2x+y3z=16-2x + y - 3z = 16.\newline2x+73z=16-2x + 7 - 3z = 16\newline2x3z=9-2x - 3z = 9
  2. Substitute y=7y = 7: Substitute y=7y = 7 into the second equation 3xy+3z=103x - y + 3z = -10.\newline3x7+3z=103x - 7 + 3z = -10\newline3x+3z=33x + 3z = -3
  3. Divide and simplify: Divide the second equation by 33 to simplify.\newlinex+z=1x + z = -1
  4. Solve for z: Solve for zz from x+z=1x + z = -1.\newlinez=1xz = -1 - x
  5. Substitute zz into equation: Substitute z=1xz = -1 - x into 2x3z=9.-2x - 3z = 9.\newline2x3(1x)=9-2x - 3(-1 - x) = 9\newline2x+3+3x=9-2x + 3 + 3x = 9\newlinex=6x = 6

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