The entry ticket to a local football game costs $10 for an adult and $5 for a child. A group of 100 people paid an entry fee of
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The first step in any problem is to look at what you are given. When solving systems of linear equations, it is often helpful to eliminate one or more of the variables by adding or subtracting a multiple of one equation with a multiple of another. It is convenient when at least one of the multipliers is 1.
Here, we can cancel the y-terms in the 2nd and 3rd equations simply by adding them together. This gives
... (5x +y -4z) +(-3x -y +5z) = (41) +(-45)
... 2x +z = -4 . . . . simplified
Likewise, we can add 3 times the second equation to the first to cancel y in that sum.
... 3(5x +y -4z) +(2x -3y +z) = 3(41) + (-1)
...15x +3y -12z +2x -3y +z = 123 -1
...17x -11z = 122 . . . . simplified
Now that we have 2 equations in x and z, we can go through the same process. We observe that the coefficient of z is +1 in the first equation and -11 in the second. The means we can cancel the z terms by adding 11 times the first equation to the second:
... 11(2x +z) +(17x -11z) = 11(-4) +122
...22x +17x = 78 . . . . . . simplify a little bit
... x = 78/39 = 2 . . . . . . divide by 39
From above, we find
... z = -4 -2x = -4 -2·2 = -8
... y = 41 -5x +4z = 41-5(2) +4(-8) = 41 -42 = -1
The solution is (x, y, z) = (2, -1, -8).
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The method of elimination used here will vary with the system of equations. If you want to employ a consistent method, you can use Cramer's Rule, Gaussian elimination, or matrix methods. Since you apparently don't mind help from technology, learning to do this on your graphing calculator can also be a good idea.
