Question

The sum of 3 times a number and 4 is between -8 and 10. Define a variable, write an inequality, and solve the problem.

Answer 1

The solution to the inequality where the sum of 3 times a number and 4 is between -8 and 10 is that the number, represented by the variable x, must be greater than -4 and less than 2. The inequality is -4 < x < 2.

Step-by-step explanation:

To solve the problem where the sum of 3 times a number and 4 is between -8 and 10, we start by defining a variable to represent the unknown number. Let's define the variable x to represent the number.

Given this variable, we can write the inequality as:

-8 < 3x + 4 < 10

We now need to solve this compound inequality, which is essentially two inequalities in one:

1. -8 < 3x + 4
2. 3x + 4 < 10

First, we subtract 4 from each part of the compound inequality:

• -8 - 4 < 3x + 4 - 4
• 3x + 4 - 4 < 10 - 4

Simplifying both inequalities, we get:

• -12 < 3x
• 3x < 6

Next, we divide each part by 3 to solve for x:

• -12 / 3 < 3x / 3
• 3x / 3 < 6 / 3

After division, the solution is:

• -4 < x
• x < 2

Therefore, the solution set for x is -4 < x < 2, meaning the number must be greater than -4 and less than 2.

Answer 2

Understanding:

The sum of 3 times a number (i'll define it as x) and 4 means
`3x+4` is between -8 and 10 which means the value of 3x+4 is less than 10 and greater than -8. With that said, the inequality for that problem would be
`-8<3x+4<10`.

Solving the problem:

Now we know that
`-8<3x+4<10`, you can solve the problem by subtracting 4 from each side then dividing by 3.

`-8-4<3x+4-4<10-4\\-12<3x<6\\(-12)/(3) <(3x)/(3)<(6)/(3)\\-4<x<2`

Final Answer:

`-4<x<2`