Find four consecutive integers such that if the sum of the first and third is increased by 10, the result is 6 greater than 4 times the fourth

Four consecutive integers are four numbers in a row, with a difference of one between each of them.

x= first integer
x+1= second integer
x+2= third integer
x+3= fourth integer

sum= add
increased by 10= add 10
the result is= equal sign
6 greater than= add 6
4 times= multiply

EQUATION
(x + x + 2) + 10= (4(x + 3)) + 6
combine like terms on the left side

2x + 12= (4(x + 3)) + 6
multiply 4 by terms in parentheses

2x + 12= 4x + 12 + 6
combine like terms on right side

2x + 12= 4x + 18
subtract 4x from both sides

-2x + 12= 18
subtract 12 from both sides

-2x= 6
divide both sides by -2

x= -3 first integer

x+1= -2 second integer
x+2= -1 third integer
x+3= 0 fourth integer

ANSWER: The four consecutive integers are -3, -2, -1 and 0.

Hope this helps! :)

The year before the 10s so it was like when caveman were alive