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When four consecutive integers are added ,the sum is 46. Find the integers

Let the first integer be x
Therefore , 
                 x +(x+1)+(x+2)+(x+3) = 46
                 x + x +1 +x +2 +x +3 =46
                 4x +6 =46
                 4x = 46 -6
                 4x = 40 
                 x = 40/4
                 x=10
1st integer = x =10
2nd integer = x+1= 11
3rd integer = x+2 = 12
4th integer = x+3 = 13


 
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let the four consecutive integers be n, n+1, n+2 and n+3
According to the question,
n+n+1+n+2+n+3=46
4n+6=46
4n=46-6
4n=40
n=40/4
n=10
So the four consecutive integers are:
n=10
n+1=10+1=11
n+2=10+2=12
n+3=10+3=13
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10,11,12,13
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