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Question 78028: Find two consecutive integers such that the sum of 3 times the first integer and 7 times second integer is 47.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=first integer
Three times the first integer=3x
Then x+1=the second integer
Seven times the second integer=7(x+1)
Now we are told that if we add three times the first and seven times the second we end up with 47, so:
3x+7(x+1)=47 get rid of parens
3x+7x+7=47 subtract 7 from both sides
3x+7x+7-7=47-7 collect like terms
10x=40 divide both sides by 10
x=4-------------------------------------first integer
4+1=5--------------------------------------second integer
CK
3*4+7*5=47
12+35=47
47=47
Hope this helps----ptaylor
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