SOLUTION: Help needed please Find X, Y, and Z X + Y + Z =1150 X = 4Z + 100 X = 6Y + 50 thank you

Question 110385: Help needed please
Find X, Y, and Z
X + Y + Z =1150
X = 4Z + 100
X = 6Y + 50
thank you
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!
Help needed please
Find X, Y, and Z
X + Y + Z = 1150
X = 4Z + 100
X = 6Y + 50
Solve the 2nd eq for Z:
X = 4Z + 100
X - 100 = 4Z
Divide every term by 4
- =
- =
= -
Solve the 3rd eq for Y:
X = 6Y + 50
X - 50 = 6Y
Divide every term by 6
- =
- =
= -
Substitute = -
and = -
into the first equation:
X + Y + Z = 1150
X + ( - ) + ( - ) = 1150
X + - + - = 1150
Nultiply every term by 12
12�X + 12� - 12� + 12� - 12� = 12�1150
12X + 2X - 100 + 3X - 300 = 13800
17X - 400 = 13800
17X = 13800 + 400
17X = 14200
X =
Substitute X = in X = 6Y + 50
X = 6Y + 50
= 6Y + 50
Multiply through by 17
17� = 17�6Y + 17�50
14200 = 102Y + 850
14200 - 850 = 102Y
13350 = 102Y
= Y
That fraction reduces by dividing top and bottom
by 6:
= Y
800 - 50 = 6Y
750 = 6Y
125 = Y
Substitute X = in X = 4Z + 100
X = 4Z + 100
= 4Z + 100
Multiply through by 17
17� = 17�4Z + 17�100
14200 = 68Z + 1700
14200 - 1700 = 68Z
12500 = 68Z
= Z
That fraction reduces by dividing top and bottom
by 4:
= Z
So the solution is
(X, Y, Z) = (, , )
Are you sure your teacher asked you to solve such a
horrible problem with such nasty fractions? This solution
is correct for what you gave me. But I'll bet anything that
there is a sign wrong somewhere in the original problem.
Edwin

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