You originally have x colored pencils in your pencil box. You share them equally between you and your friend. You add in 3 more colored pencils to your own box after sharing them. This gives you a total of 15 pencils.
a. Write an equation and show all of your steps for solving x (the original number of colored pencils in your box).
b. If you wanted to have at least 20 colored pencils left in your box, how many colored pencils would you have needed originally?
c. Write an inequality and show all steps for solving for x.
a. Let's set up the equation to solve for x.
After sharing equally with your friend, you have x/2 colored pencils left in your box. Then, you add 3 more colored pencils to your box, resulting in x/2 + 3.
This sum gives you a total of 15 colored pencils, so we can write the equation:
x/2 + 3 = 15
To solve for x, we need to isolate it on one side of the equation. Here are the steps:
x/2 + 3 = 15
Subtract 3 from both sides:
x/2 = 15 - 3
x/2 = 12
Multiply both sides by 2 to remove the fraction:
2 * (x/2) = 2 * 12
x = 24
Therefore, the original number of colored pencils in your box was 24.
b. If you wanted to have at least 20 colored pencils left in your box, we need to modify the equation.
After sharing equally with your friend, you have x/2 colored pencils left in your box. Then, you add 3 more colored pencils to your box, resulting in x/2 + 3.
To determine the original number of colored pencils needed to have at least 20, we can write the inequality:
x/2 + 3 ≥ 20
To solve for x, we need to isolate it on one side of the inequality. Here are the steps:
x/2 + 3 ≥ 20
Subtract 3 from both sides:
x/2 ≥ 20 - 3
x/2 ≥ 17
Multiply both sides by 2 to remove the fraction (remember to flip the inequality when multiplying by a negative number):
2 * (x/2) ≤ 2 * 17
x ≤ 34
Therefore, you would need at least 34 colored pencils originally to have 20 left in your box.
c. The inequality we wrote in part b is:
x/2 + 3 ≥ 20
Alternatively, since we want to find the original number of colored pencils (x) and we know the total left should be at least 20, we can set up another inequality:
x/2 + 3 ≥ 20
To solve for x, we need to isolate it on one side of the inequality. Here are the steps:
x/2 + 3 ≥ 20
Subtract 3 from both sides:
x/2 ≥ 20 - 3
x/2 ≥ 17
Multiply both sides by 2 to remove the fraction (remember to flip the inequality when multiplying by a negative number):
2 * (x/2) ≤ 2 * 17
x ≤ 34
Therefore, the original number of colored pencils in your box (x) should be less than or equal to 34.
a. Let's solve for x step-by-step.
Let's start with the number of colored pencils you initially had in your box, which is x. After sharing them equally with your friend, you would have x/2 pencils left in your box.
Next, you added 3 more colored pencils to your box, increasing the total number of pencils to x/2 + 3.
According to the problem, the total number of pencils after adding 3 more is 15. We can write this as an equation:
x/2 + 3 = 15
To solve for x, we need to isolate the x term. First, subtract 3 from both sides of the equation:
x/2 = 15 - 3
x/2 = 12
To remove the fraction, multiply both sides of the equation by 2:
2*(x/2) = 12 * 2
x = 24
So, the original number of colored pencils in your pencil box was 24.
b. Now, let's find out how many colored pencils you would have needed originally to have at least 20 pencils left in your box.
We can set up the following inequality:
x/2 + 3 >= 20
To isolate x, we need to subtract 3 from both sides of the inequality:
x/2 >= 20 - 3
x/2 >= 17
To remove the fraction, multiply both sides of the inequality by 2:
(x/2)*2 >= 17*2
x >= 34
So, you would have needed at least 34 colored pencils originally to have at least 20 pencils left in your box.
c. The inequality we derived in part b is:
x/2 + 3 >= 20
Alternatively, we can write it as:
x/2 >= 17
This inequality states that the original number of colored pencils divided by 2 should be greater than or equal to 17.