Find three consecutive integers such that three times the sum of the first two integers is the same as thirteen more than four times the third integer. Please help

Question

I just need help with this since its my first time

Answer 1

Hello Mohammed, hopefully I can help!

So as we know, consecutive integers are integers that follow each other, and are separated by one whole number, such as say, the numbers 3, 4, and 5. This is not the answer, this is an example of three consecutive integers.

So let's say we're representing the three consecutive integers you'd like to find for this question by x, y, and z.

We could replace x, y, and z by x, x+1, and x+2, where

x=x

y=x+1

z=x+2

because the numbers follow each other.

Essentially what the question is saying, is that three times the sum of the first two integers (3(x+y)) is equal to 4 times the third integer, plus 13 (4z+13).

So we get the equation:

3(x+y) = 4z + 13

But as we know, x=x, y=x+1, and z=x+2

So by replacing y by (x+1) and z by (x+2), we get the following equation from the one above:

3(x+x+1) = 4(x+2) +13

All we have to do is solve for x, knowing that y is x+1 and z is x+2.

Continuing to expand and solve the equation above, we get:

3x+3x+3 = 4x+8+13

6x+3 = 4x+21

6x-4x = 21-3

2x = 18

x= 18/2 = 9

So therefore,

x=9

y= x+1=10

z=x+2=11

We can check our answer against our initial equation, 3(x+y) = 4z + 13

If we replace x, y, z by the values we got, we get:

3 x (9+10) = 4 x 11 + 13

3x19= 4x11 + 13

57= 44 + 13 = 57

So as you can see, our values are correct.

Let me know if this helped you!

All the best,

Iman