Question 1118569
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There are innumerable ways to solve a system like this algebraically.  Different methods will be appropriate for different systems of equations.<br>
You always want to look for ways to make the problem as easy as possible.  In your example, a quick look shows that adding the first two equations together eliminates both y and z, allowing you to find the value of x and thus immediately reduce the system of three equations and three unknowns to a system of two equations and two unknowns.<br>
4x+y+2z = 24; 2x-y-2z = -6  -->  6x = 18  -->  x = 3<br>
Substitute x=3 into either of the first two equations and the third. (Since you found x=3 using the first two equations, substituting x=3 in those same two equations won't get you anywhere.)<br>
6-y-2z = -6  -->  y+2z = 12;
-3+2y-z = -4  -->  2y-z = -1<br>
The solution from the other tutor then uses substitution to finish the problem -- solving one equation for one of the variables and substituting the resulting expression for that variable in the other equation.<br>
When the two equations are both in the form Ax+By=C, I think a solution using elimination is much easier.  Multiply the second equation by 2 and add the two equations to eliminate y:<br>
y+2z = 12; 4y-2z = -2  -->  5y = 10  -->  y = 2<br>
Then substitute y=2 in either of those last two equations to find z:<br>
2+2z = 12  -->  2z = 10  -->  z = 5<br>
Answer: x=3; y=2; z=5