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 represent the situation:
\[x/2 + 3 = 15\]
To solve for x, we'll first subtract 3 from both sides of the equation:
\[x/2 = 12\]
Then, multiply both sides by 2 to get rid of the fraction:
\[x = 24\]
Therefore, the original number of colored pencils in your box was 24.
b. To have at least 20 colored pencils left in your box, we can set up the inequality:
\[x/2 + 3 \geq 20\]
Subtracting 3 from both sides gives:
\[x/2 \geq 17\]
Then, multiplying both sides by 2:
\[x \geq 34\]
Therefore, you would have needed at least 34 colored pencils originally.
c. The inequality for this situation would be:
\[x/2 + 3 \leq 15\]
Subtracting 3 from both sides:
\[x/2 \leq 12\]
Multiplying both sides by 2:
\[x \leq 24\]
Thus, the possible range for the original number of colored pencils is x ≤ 24.
are the [ and / nessesary?
No, the brackets [ ] and forward slash / are not necessary in this case. They were used to help separate and clarify the different parts of the equations, but the equations can be written without them. Here are the revised equations without the brackets and forward slash:
a. Equation: x/2 + 3 = 15
Solution: Subtract 3 from both sides: x/2 = 12
Multiply both sides by 2: x = 24
b. Inequality: x/2 + 3 ≥ 20
Solution: Subtract 3 from both sides: x/2 ≥ 17
Multiply both sides by 2: x ≥ 34
c. Inequality: x/2 + 3 ≤ 15
Solution: Subtract 3 from both sides: x/2 ≤ 12
Multiply both sides by 2: x ≤ 24
These equations and inequalities can be expressed without the brackets and forward slash, but they were used initially to make it clearer.
a. Let's use the information given to set up an equation and solve for x.
Let's call the original number of colored pencils in your pencil box as x.
According to the information given, you shared the pencils equally with your friend, so you each received x/2 colored pencils.
After sharing, you added 3 more colored pencils to your own box, making your total number of colored pencils (x/2) + 3.
According to the problem, this total is 15:
(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
To isolate x, we 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. If you want to have at least 20 colored pencils left in your box after sharing and adding 3 more, let's set up an equation.
Let's call the original number of colored pencils you needed as y.
After sharing equally with your friend, you each receive y/2 colored pencils.
After adding 3 more colored pencils, you have (y/2) + 3 colored pencils.
According to the problem, you want this total to be at least 20:
(y/2) + 3 ≥ 20
To solve for y, we can subtract 3 from both sides of the inequality:
(y/2) ≥ 20 - 3
(y/2) ≥ 17
To get rid of the fraction, we can multiply both sides of the inequality by 2:
2 * (y/2) ≥ 2 * 17
y ≥ 34
Therefore, you would have needed at least 34 colored pencils originally to have at least 20 left in your box after sharing and adding 3 more.
c. Here is an inequality to solve for x using the information given:
The original number of colored pencils in your pencil box is x.
The number of colored pencils you have after sharing equally with your friend and adding 3 more is (x/2) + 3.
According to the problem, this total is 15:
(x/2) + 3 ≤ 15
To solve for x, we can subtract 3 from both sides of the inequality:
(x/2) ≤ 15 - 3
(x/2) ≤ 12
To isolate x, we multiply both sides of the inequality by 2:
2 * (x/2) ≤ 2 * 12
x ≤ 24
Therefore, the original number of colored pencils in your box, x, is less than or equal to 24.