How do i solve the sum of three consecutive integers is 48 less than twice the smallest of the three integers?

So let's say the three consecutive integers are x-1,x,x+1.

2(x-1)=x-1+x+x+1+48

2x-2=3x+48

x=-50.

The three integers will be -49,-50,-51

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b