a) Let's set up an equation to solve for x.
First, you share equally between you and your friend, which means you each get x/2 colored pencils.
Then, you add 3 more colored pencils to your own box, making the total number of pencils in your box (after sharing) x/2 + 3.
Finally, we know that the total number of pencils you have now after adding 3 more is 15. Therefore, the equation becomes:
x/2 + 3 = 15
To solve for x, we can start by subtracting 3 from both sides:
x/2 = 15 - 3
x/2 = 12
Next, we can multiply both sides by 2 to get rid of the fraction:
2 * (x/2) = 2 * 12
x = 24
So, the original number of colored pencils in your box was 24.
b) If you wanted to have at least 20 pencils left in your box, we can modify the equation from part (a).
After sharing equally with your friend, you each get x/2 colored pencils. If you want to have at least 20 pencils left in your box, you would have x/2 - 3 pencils left after sharing.
The modified equation becomes:
x/2 - 3 ≥ 20
To solve for x, we can start by adding 3 to both sides:
x/2 - 3 + 3 ≥ 20 + 3
x/2 ≥ 23
Next, we can multiply both sides by 2:
2 * (x/2) ≥ 23 * 2
x ≥ 46
Therefore, you would have needed originally at least 46 pencils in your box.
c) The inequality for this scenario would be:
x/2 + 3 ≥ 20
To solve for x, we can start by subtracting 3 from both sides:
x/2 ≥ 20 - 3
x/2 ≥ 17
Next, we can multiply both sides by 2:
2 * (x/2) ≥ 17 * 2
x ≥ 34
So the inequality solution is x ≥ 34, indicating that the original number of pencils in your box should be at least 34.
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 pencils in your box)
b) What if you wanted to have at least 20 pencils left in your box? How many pencils would you have needed originally?
c) Write an inequality and show all steps for solving for x.
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make a shorter awnswer
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