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Question 173433: Four numbers have a total of 92. The first number is twice the second number and the third number is 4 more than the sum of the first two. The fourth number is 3 less than the difference of the first two numbers. What are the numbers?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let the four numbers be a, b, c, and d.
From the problem description you can write:
1) a+b+c+d = 92 "The four numbers have a total of 92"
2) a = 2b "The first number is twice the second number..."
3) c = (a+b)+4 "...the third number is 4 more than the sum of the first two."
4) d = (a-b)-3 "The fourth number is 3 less than the difference of the first two numbers."
Substitute equation 2) into equation 1) to get:
1a) 2b+b+c+d = 92 Simplify.
2a) 3b+c+d = 92 Now substitute equation 3 into this equation.
2b) 3b+(a+b)+4+d = 92 Simplify this.
2b) a+4b+d+4 = 92 Substitute equation 2) into this equation.
2c) 2b+4b+d+4 = 94 Simplify.
2c) 6b+d+4 = 92 Substitute equation 4) into this equation.
2d) 6b+((a-b)-3)+4 = 92 Substitute equation 2) into this equation.
2e) 6b+((2b-b)-3)+4 = 92 Simplify.
2e) 7b+1 = 92 Subtract 1 from both sides.
7b = 91 Divide both sides by 7.
b = 13
a = 2b = 2(13) = 26
c = (a+b)+4 = 26+13+4 = 43
d = (a-b)-3 = 26-13-3 = 10
The numbers are:
a = 26
b = 13
c = 43
d = 10
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