16. 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 solve for x, the original number of colored pencils in your box.
According to the given information, you initially have x colored pencils.
After sharing them equally between you and your friend, both of you have x/2 colored pencils.
Then, you add 3 more colored pencils to your own box, which gives you x/2 + 3 colored pencils.
This gives you a total of 15 pencils, so we can set up the following equation:
x/2 + 3 = 15
To solve for x, we can subtract 3 from both sides of the equation:
x/2 = 15 - 3
x/2 = 12
Finally, to isolate x, we can multiply both sides of the equation by 2:
2*(x/2) = 2*12
x = 24
Therefore, the original number of colored pencils in your box was 24.
b. If you want to have at least 20 colored pencils left in your box, we can set up the following inequality:
x/2 + 3 ≥ 20
To solve for x, we can subtract 3 from both sides of the inequality:
x/2 ≥ 17
Next, we can multiply both sides of the inequality by 2 to get rid of the fraction:
2*(x/2) ≥ 2*17
x ≥ 34
Therefore, you would need at least 34 colored pencils originally to have at least 20 left in your box.
c. The inequality is x/2 + 3 ≥ 20.
Step 1: Subtract 3 from both sides of the inequality:
x/2 ≥ 20 - 3
x/2 ≥ 17
Step 2: Multiply both sides of the inequality by 2 to get rid of the fraction:
2*(x/2) ≥ 2*17
x ≥ 34
The solution to the inequality is x ≥ 34. This means that the original number of colored pencils in your box must be at least 34 in order to have at least 20 left.
a. Let's break down the steps:
Step 1: You share the colored pencils equally with your friend. So each of you gets x/2 colored pencils.
Step 2: You add 3 more colored pencils to your box, so you now have (x/2) + 3 colored pencils.
According to the problem, this gives you a total of 15 colored pencils. So, we can write the equation as:
(x/2) + 3 = 15
Step 3: To solve the equation, we need to isolate x. Let's start by subtracting 3 from both sides of the equation:
(x/2) = 15 - 3
(x/2) = 12
Step 4: To eliminate the fraction, we can multiply both sides of the equation by 2:
2 * (x/2) = 2 * 12
x = 24
So, the original number of colored pencils in your box was 24.
b. Now, let's solve for the original number of colored pencils if you wanted to have at least 20 pencils left in your box.
If you share the pencils equally with your friend, each of you will have (x/2) pencils. After adding 3 more pencils, you will have (x/2) + 3 pencils. And according to the problem, you want to have at least 20 pencils left, so we can write the inequality as:
(x/2) + 3 >= 20
Step 1: Let's isolate (x/2) by subtracting 3 from both sides:
(x/2) >= 17
Step 2: To eliminate the fraction, we multiply both sides by 2:
2 * (x/2) >= 2 * 17
x >= 34
So, the original number of colored pencils you needed to have to have at least 20 pencils left is 34.
c. The inequality we used in part b was:
(x/2) + 3 >= 20
Now, let's solve it step by step:
Step 1: Subtract 3 from both sides of the inequality:
(x/2) >= 20 - 3
(x/2) >= 17
Step 2: Multiply both sides of the inequality by 2 to eliminate the fraction:
2 * (x/2) >= 2 * 17
x >= 34
So, the solution to the inequality is x >= 34.