Question
Complete the steps to solve the bike-shop problem.
Flower Simpl..
(x)
Quizzes 2
今 56%口
+
口
Complete th...
The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. One day, she counts 18 seats and 28 wheels. How many of each type of cycle are there?
Graph & Come to an answer
This is the graph of the two equations.
30
20
20
30
6+ u = 18 26 + u = 28
At what point do the two lines intersect?
choose your answer...
Flower Simpl..
(x)
Quizzes 2
今 56%口
+
口
Complete th...
The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. One day, she counts 18 seats and 28 wheels. How many of each type of cycle are there?
Graph & Come to an answer
This is the graph of the two equations.
30
20
20
30
6+ u = 18 26 + u = 28
At what point do the two lines intersect?
choose your answer...
Answers
Answered by
GPT-4o mini
To solve the bike shop problem, we need to establish a system of equations based on the inventory counts of seats and wheels.
Let's define variables:
- Let \( u \) be the number of unicycles.
- Let \( b \) be the number of bicycles.
We have the following information:
1. Each unicycle has 1 seat and 1 wheel.
2. Each bicycle has 1 seat and 2 wheels.
From the problem, we can create two equations:
1. For seats:
\[
u + b = 18
\]
2. For wheels:
\[
u + 2b = 28
\]
Now we can solve this system of equations.
### Step 1: Solve the first equation for \( b \)
From the first equation:
\[
b = 18 - u
\]
### Step 2: Substitute \( b \) into the second equation
Substituting \( b \) in the second equation:
\[
u + 2(18 - u) = 28
\]
### Step 3: Simplify
This simplifies to:
\[
u + 36 - 2u = 28
\]
Combining like terms:
\[
36 - u = 28
\]
### Step 4: Solve for \( u \)
Rearranging gives:
\[
-u = 28 - 36 \\
-u = -8 \\
u = 8
\]
### Step 5: Substitute \( u \) back to find \( b \)
Using \( u = 8 \) in the first equation:
\[
8 + b = 18 \\
b = 18 - 8 \\
b = 10
\]
### Answer
There are 8 unicycles and 10 bicycles.
### Intersection of the Graphs
The two equations can be graphically represented as lines in a coordinate plane, where you identified the equations:
- The first equation \( u + b = 18 \).
- The second equation \( u + 2b = 28 \).
The lines will intersect at the point (8, 10), which corresponds to our solution.
**Choose the coordinates for the intersection:**
The point where the two lines intersect is \((u, b) = (8, 10)\).
Let's define variables:
- Let \( u \) be the number of unicycles.
- Let \( b \) be the number of bicycles.
We have the following information:
1. Each unicycle has 1 seat and 1 wheel.
2. Each bicycle has 1 seat and 2 wheels.
From the problem, we can create two equations:
1. For seats:
\[
u + b = 18
\]
2. For wheels:
\[
u + 2b = 28
\]
Now we can solve this system of equations.
### Step 1: Solve the first equation for \( b \)
From the first equation:
\[
b = 18 - u
\]
### Step 2: Substitute \( b \) into the second equation
Substituting \( b \) in the second equation:
\[
u + 2(18 - u) = 28
\]
### Step 3: Simplify
This simplifies to:
\[
u + 36 - 2u = 28
\]
Combining like terms:
\[
36 - u = 28
\]
### Step 4: Solve for \( u \)
Rearranging gives:
\[
-u = 28 - 36 \\
-u = -8 \\
u = 8
\]
### Step 5: Substitute \( u \) back to find \( b \)
Using \( u = 8 \) in the first equation:
\[
8 + b = 18 \\
b = 18 - 8 \\
b = 10
\]
### Answer
There are 8 unicycles and 10 bicycles.
### Intersection of the Graphs
The two equations can be graphically represented as lines in a coordinate plane, where you identified the equations:
- The first equation \( u + b = 18 \).
- The second equation \( u + 2b = 28 \).
The lines will intersect at the point (8, 10), which corresponds to our solution.
**Choose the coordinates for the intersection:**
The point where the two lines intersect is \((u, b) = (8, 10)\).