SOLUTION: YMy question is; Lawrence High prevailed in Saturdays track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a

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Question 711700: YMy question is;
Lawrence High prevailed in Saturdays track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points, and third place earns 1 point. Lawrence had s strong second-place showing, with as many second-place finishes as first and third place finishers combined. (hint: the third equation will be as: y=x+z)

Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
Since you used XYZ in the write up I will continue that.
Let X = number of first place finishers
Let Y = number of second place finishers
Let Z = number of third place finishers
Below are the equations that you need to set up to solve the problem. Behind each equation is the part of the text where the equation was made from.
Equation 1: X+%2B+Y+%2B+Z+=+20 "with the help of 20 individual-event placers"
Equation 2: 5X+%2B+3Y+%2B+Z+=+68 "earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points, and third place earns 1 point."
Equation 3: Y+=+X+%2B+Z "with as many second-place finishes as first and third place finishers combined."
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Equation 3 is already solved for 1 varible.
Plug (X + Z) into equation 1 for Y. You can then solve for one of the other variables.
Equation 1: X+%2B+Y+%2B+Z+=+20
X+%2B+%28X+%2B+Z%29+%2B+Z+=+20
Combine like terms
2X+%2B+2Z+=+20
I will solve for Z.
Pull a 2 from the left sides of the equation.
2%2A%28X+%2B+Z%29+=+20
Divide both sides by 2
X+%2B+Z+=+10
Subtract X from both sides
Z+=+10+-+X
Now we have all the variables in terms of X.
Substitute these values into equation 2 and solve for X
Equation 2: 5X+%2B+3Y+%2B+Z+=+68
5X+%2B+3%2A%28Z+%2B+X%29+%2B+%2810+-+X%29+=+68
Multiply the 3 through.
5X+%2B+3Z+%2B+3X+%2B+10+-+X+=+68
Combine like terms.
7X+%2B+3Z+%2B+10+=+68
Subtract 10 from both sides
7X+%2B+3Z+=+58
We need to get rid of the Z, substitute the (10-X) in for Z.
7X+%2B+3%2A%2810+-+X%29+=+58
Multiply the 3 through
7X+%2B+30+-+3X+=+58
Combine like terms.
4X+%2B+30+=+58
Subtract 30 from both sides
4X+=+28
Divide both sides by 4
highlight%28X+=+7%29
Now solve for Y & Z by using 7 for X
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From earlier: Z+=+10+-+X
Z+=+10+-+7
highlight_green%28Z+=+3%29
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Equation 3: Y+=+X+%2B+Z
Y+=+%287%29+%2B+%283%29
highlight%28Y+=+10%29