1. find three consecutive integers such that the sum of the first and twice the third
integer is 93 more than the second integer

Step-by-step explanation:

Let (x), (x+1), and (x+2) be the three consecutive integers.

Put the words into expressions

Sum of the 3rd and twice the 3rd:  (x)+2(x+2)

93 more than the second integer: 93+(x+1)

Now you can set up the equation and solve:

(x)+2(x+2)=93+(x+1)

Distribute and combine like terms

x+2x+4=94+x

3x+4=94+x

2x=90

x=45

So the 3 consecutive integers are 45, 46, 47.

Plug back in to check.

answer: idk

step-by-step explanation: