Alegbra help again. 4 problems pls help me 🙁 i dont want to fail lool?

For all of these problems, make x one of the numbers. Then, just keep adding 1 to designate another “consecutive integer.” So if x is one of the integers, x + 1 is a consecutive integer, and so is x + 2. Use this strategy to complete all of the problems.

1. You have three consecutive integers, so it’s going to be x, x + 1, and x + 2. The sum of these integers is 69. So x + x + 1 + x + 2 = 69. Therefore 3x + 3 = 69, 3x = 66, x = 22. So the three consecutive integers are 22, 23, and 24.

2. This time you’ve got four consecutive integers, so your equation is x + x + 1 + x + 2 + x + 3 = -42. So 4x + 6 = -42. Then, 4x = -48, and x = -12. So the consecutive integers are -12, -11, -10, and -9.

3. This time, it’s consecutive odd integers. So instead of adding 1, you add 2 since you need to skip the even number. The equation is x + x + 2 + x + 4 = 147. So, 3x + 6 = 147, and 3x = 141. x = 47, so the consecutive integers are 47, 49, and 51.

4. If you are looking at the product of consecutive odd integers, just multiply instead of adding. The first integer is x, and the second integer is x + 2, so just multiply them. You get x(x + 2) = 195. Therefore, x^2 + 2x = 195 and x^2 + 2x – 195 = 0. Factor the equation, and you get (x + 15)(x – 13) = 0. So x equals either -15 or 13. This means that you have two possibilities for your answer. The two consecutive odd integers could be -15 and -13, or 13 and 15, since when you multiply them they reach 195.

Hope it helps!

All four questions are basically the same principal. Because of this I will do just the first problem to show the method. You should be able to figure out how to use the same method to answer the other three questions.

I’ll use the letter “a” to represent the unknown number.

The sum (the answer to addition) of three consecutive integers (three numbers that come one right after another) is 69.

Knowing this, if “a” represents the unknown number, then the next (consecutive) number would be “a+1” and the next (consecutive) number after that would be “a+2”

If the SUM of these three integers is 69, then:

a + (a+1) + (a+2) = 69

Add the like terms (add all the “a”s together, and all the digits together):

3a + 3 = 69

Now you need to move the integer “3” to the other side of the equation. It is being added on the left side of the equals sign, so it must be subtracted on the other side:

3a = 69 – 3

3a = 66

Now you need to get the variable “a” by itself on one side of the equation. Since (3a) means (3 times a), that means that the 3 is being multiplied. When you move the 3 to the other side of the equal sign it will have to be divided.:

a = 66 / 3 (66 divided by 3)

a = 22

Now check the work by substiting the answer for “a” into the original equation.

since we have figured out that “a” = 22, then:

22 + (22+1) + (22 + 2) = 69

22 + 23 + 24 = 69

69 = 69

Use the same principal for the next three equations. In number four, remember that product means the answer to multiplication, so 195 is the answer to two consecutive odd integers (“a”, and “a+2”) being multiplied.