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Algebra 2 question! Find 4 consecutive integers such that when twice the 2nd is added to 3 times the 4th, the result is 31.
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let first integer = x second integer = x + 1 third integer = x + 2 fourth integer = x + 3 you know why right? when twice the 2nd means 2(x+1) 3 times the fourth is 3(x+3) it says the sum is 31 so... 2(x+1) + 3(x+3) = 31 do you know how to solve for x now? or do you need more help?
yes i do! I actually had it just like you, the 1= x 2=x+1 3=x+3 4= x+4
I just wasnt sure how to put it into an equation, thanks!
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