Question 110385
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

        {{{X/4}}} - {{{100/4}}} = {{{(4Z)/4}}}     
        {{{X/4}}} - {{{25}}} = {{{Z}}}
        {{{Z}}} = {{{X/4}}} - {{{25}}}
Solve the 3rd eq for Y:

               X = 6Y + 50
          X - 50 = 6Y

Divide every term by 6

        {{{X/6}}} - {{{50/6}}} = {{{(6Y)/6}}}     
        {{{X/6}}} - {{{25/3}}} = {{{Y}}}
        {{{Y}}} = {{{X/6}}} - {{{25/3}}}

Substitute {{{Y}}} = {{{X/6}}} - {{{25/3}}}
and {{{Z}}} = {{{X/4}}} - {{{25}}}
into the first equation:

X + Y + Z = 1150
X + ({{{X/6}}} - {{{25/3}}}) + ({{{X/4}}} - {{{25}}}) = 1150
X + {{{X/6}}} - {{{25/3}}} + {{{X/4}}} - {{{25}}} = 1150
Nultiply every term by 12

12·X + 12·{{{X/6}}} - 12·{{{25/3}}} + 12·{{{X/4}}} - 12·{{{25}}} = 12·1150

12X + 2X - 100 + 3X - 300 = 13800
                17X - 400 = 13800
                      17X = 13800 + 400
                      17X = 14200
                        X = {{{14200/17}}}

Substitute X = {{{14200/17}}} in X = 6Y + 50

       X = 6Y + 50
     {{{14200/17}}} = 6Y + 50

Multiply through by 17

    17·{{{14200/17}}} = 17·6Y + 17·50
              14200 = 102Y + 850
        14200 - 850 = 102Y
              13350 = 102Y
    {{{13350/102}}} = Y     

That fraction reduces by dividing top and bottom
by 6:

      {{{2225/17}}} = Y          

800 - 50 = 6Y
     750 = 6Y
     125 = Y 

Substitute X = {{{14200/17}}} in X = 4Z + 100

             X = 4Z + 100
{{{14200/17}}} = 4Z + 100

Multiply through by 17

    17·{{{14200/17}}} = 17·4Z + 17·100
                14200 = 68Z + 1700
         14200 - 1700 = 68Z
                12500 = 68Z
       {{{12500/68}}} = Z

That fraction reduces by dividing top and bottom
by 4:

      {{{3125/17}}} = Z

So the solution is

(X, Y, Z) = ({{{14200/17}}}, {{{2225/17}}}, {{{3125/17}}})

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</pre>