Amira wants to ride her bicycle 48 miles this week. She has already ridden 18 miles. If she rides for 5 more days, write and solve an equation which can be used to determine xx, the average number of miles she would have to ride each day to meet her goal.

Question

Equation: ______
Answer: x = _____

Answer 1

To determine the number of additional miles Amira needs to ride in order to reach her goal of 48 miles, we start by subtracting the miles she has already ridden from her total goal:

\[ 48 - 18 = 30 \]

This means Amira needs to ride 30 more miles over the next 5 days. We can write the equation to find the average number of miles she needs to ride each day, x, as follows:

\[ 5x = 30 \]

Now, we can solve the equation for x:

\[ x = \frac{30}{5} = 6 \]

So, the complete answers are:

Equation: \( 5x = 30 \)

Answer: \( x = 6 \)

This means Amira needs to ride an average of 6 miles each day for the next 5 days to meet her goal.

Answer 2

Brandon wants to ride his bicycle 21 miles this week. He has already ridden 6 miles. If he rides for 5 more days, write and solve an equation which can be used to determine x, the average number of miles he would have to ride each day to meet his goal.


Equation: ____

Answer x:______ (Do it for equation like for example cx + x = 60)

Answer 3

To determine how many more miles Brandon needs to ride to reach his goal of 21 miles, we first subtract the miles he has already ridden:

\[ 21 - 6 = 15 \]

This means Brandon needs to ride 15 more miles over the next 5 days. We can set up the equation to find the average number of miles he needs to ride each day, \(x\), as follows:

\[ 5x = 15 \]

This equation states that 5 times the average miles per day \(x\) is equal to the total miles he needs to ride (15 miles).

So, the complete answers are:

Equation: \( 5x = 15 \)

Now, solving for \(x\):

\[ x = \frac{15}{5} = 3 \]

Thus, the answer for \(x\) is:

Answer \(x\): \(3\)

This means Brandon would have to ride an average of 3 miles each day for the next 5 days to meet his goal.