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Use the substitution method to solve the following system of equations. 4x – y – 3z = –3 x – 2y – 2z = –3 x + y + 3z = –2
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Hint: Solve the second equation for x to get x – 2y – 2z = –3 x = 2y+2z–3 Then plug this into the first equation to get 4x – y – 3z = –3 4(2y+2z–3) – y – 3z = –3 8y + 8z - 12 - y - 3z = -3 7y + 5z - 12 = -3 7y + 5z = -3+12 7y + 5z = 9 Then plug x = 2y+2z–3 into the third equation x + y + 3z = –2 2y+2z–3 + y + 3z = –2 3y + 5z - 3 = -2 3y + 5z = -2+3 3y + 5z = 1
So you should end up with this new system of equations 7y + 5z = 9 3y + 5z = 1 Solve this system (ignore the other equations or previous system) for y and z. Once you have y and z, use them to find x.
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