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Find three consecutive integers such that 4 times the first is 13 more than the third

Guest Sep 25, 2018
 #1
avatar+2424 
+1

let the first integer be n

 

the second is (n+1), the third (n+2)

 

4n = (n+2) + 13

 

4n = n + 15

 

3n = 15

 

n = 5

 

4 x 5 = 20 = 7 + 13

 

the 3 integers are 5, 6, 7

Rom  Sep 25, 2018
 #2
avatar+90053 
+1

Let the integers be   x, x + 1 and x + 2

 

So we have

 

4x  =  x+ 2 + 13    simplify

 

4x  = x + 15    subtract x from both sides

 

3x  = 15     divide both sides by 3

 

x  = 5

 

So...the integers are 5, 6, 7

 

 

cool cool cool

CPhill  Sep 25, 2018

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