Nancy and Ezra are both making bracelets with beads. Nancy's bracelet starts with 24 beads and she adds 2 more every minute. Ezra is taking apart his bracelet. Ezra's bracelet starts with 48 beads and he removes 4 every minute. Is there a time when their bracelets will have the same number of beads?

Question

The equation below represents the number of beads, y, that Ezra has on his bracelet after x minutes. Use the equation to complete the table with the number of beads on Ezra bracelet for the given number of minutes. You can solve on paper and then enter your answer on Zearn.
y=48−4x

Minutes (x) Beads (y)
2
4
8

Answer 1

To complete the table for the number of beads on Ezra's bracelet using the equation \( y = 48 - 4x \), we will substitute the values of \( x \) (representing minutes) into the equation.

1. For \( x = 2 \): \[ y = 48 - 4(2) = 48 - 8 = 40 \]

2. For \( x = 4 \): \[ y = 48 - 4(4) = 48 - 16 = 32 \]

3. For \( x = 8 \): \[ y = 48 - 4(8) = 48 - 32 = 16 \]

Now, we can fill in the table:

| Minutes (x) | Beads (y) | |--------------|-----------| | 2 | 40 | | 4 | 32 | | 8 | 16 |

Now, to determine if there is a time when Nancy's and Ezra's bracelets will have the same number of beads, we can set up the equations for both:

For Nancy: \[ N(x) = 24 + 2x \]

For Ezra: \[ E(x) = 48 - 4x \]

We want to find \( x \) when \( N(x) = E(x) \): \[ 24 + 2x = 48 - 4x \]

Now, we will solve for \( x \):

1. Add \( 4x \) to both sides: \[ 24 + 2x + 4x = 48 \] \[ 24 + 6x = 48 \]

2. Subtract \( 24 \) from both sides: \[ 6x = 48 - 24 \] \[ 6x = 24 \]

3. Divide by \( 6 \): \[ x = \frac{24}{6} = 4 \]

Now, let's check if they have the same number of beads at \( x = 4 \):

• Nancy's beads: \[ N(4) = 24 + 2(4) = 24 + 8 = 32 \]

• Ezra's beads: \[ E(4) = 48 - 4(4) = 48 - 16 = 32 \]

Yes, at \( x = 4 \) minutes, both Nancy and Ezra have 32 beads. Therefore, there is a time when their bracelets will have the same number of beads, and that time is 4 minutes.