Answer :
To solve for \( d \) (the cost of the dance lesson) and \( v \) (the cost of the violin lesson) given the conditions, we will set up a system of equations based on the problem statement.
1. Define the relationship between the costs of the dance and violin lessons:
- We are told that the dance lesson costs \(\frac{2}{3}\) as much as the violin lesson. This can be written as:
[tex]\[ d = \frac{2}{3} v \][/tex]
2. Sum of the costs:
- We are also told the total cost of both lessons combined is $75. This gives us:
[tex]\[ d + v = 75 \][/tex]
Now we have the two equations:
[tex]\[ \begin{cases} d = \frac{2}{3} v & \text{(1)} \\ d + v = 75 & \text{(2)} \end{cases} \][/tex]
This system of equations can be used to find \( d \) and \( v \).
Solving the system of equations:
First, substitute equation (1) into equation (2):
[tex]\[ \left( \frac{2}{3} v \right) + v = 75 \][/tex]
Combine the terms involving \( v \):
[tex]\[ \frac{2}{3} v + v = 75 \][/tex]
Convert \( v \) to a common fraction:
[tex]\[ \frac{2}{3} v + \frac{3}{3} v = 75 \][/tex]
Simplify:
[tex]\[ \frac{5}{3} v = 75 \][/tex]
To isolate \( v \), multiply both sides of the equation by \( \frac{3}{5} \):
[tex]\[ v = 75 \times \frac{3}{5} \][/tex]
So, we get:
[tex]\[ v = 45 \][/tex]
Next, use this value to find \( d \). From equation (1):
[tex]\[ d = \frac{2}{3} \times 45 \][/tex]
Thus,
[tex]\[ d = 30 \][/tex]
Conclusion:
The system of equations to determine the costs of the dance lesson (\( d \)) and the violin lesson (\( v \)) is:
[tex]\[ \begin{cases} d = \frac{2}{3} v \\ d + v = 75 \end{cases} \][/tex]
And the solution to this system is:
[tex]\[ v = 45, \quad d = 30 \][/tex]
1. Define the relationship between the costs of the dance and violin lessons:
- We are told that the dance lesson costs \(\frac{2}{3}\) as much as the violin lesson. This can be written as:
[tex]\[ d = \frac{2}{3} v \][/tex]
2. Sum of the costs:
- We are also told the total cost of both lessons combined is $75. This gives us:
[tex]\[ d + v = 75 \][/tex]
Now we have the two equations:
[tex]\[ \begin{cases} d = \frac{2}{3} v & \text{(1)} \\ d + v = 75 & \text{(2)} \end{cases} \][/tex]
This system of equations can be used to find \( d \) and \( v \).
Solving the system of equations:
First, substitute equation (1) into equation (2):
[tex]\[ \left( \frac{2}{3} v \right) + v = 75 \][/tex]
Combine the terms involving \( v \):
[tex]\[ \frac{2}{3} v + v = 75 \][/tex]
Convert \( v \) to a common fraction:
[tex]\[ \frac{2}{3} v + \frac{3}{3} v = 75 \][/tex]
Simplify:
[tex]\[ \frac{5}{3} v = 75 \][/tex]
To isolate \( v \), multiply both sides of the equation by \( \frac{3}{5} \):
[tex]\[ v = 75 \times \frac{3}{5} \][/tex]
So, we get:
[tex]\[ v = 45 \][/tex]
Next, use this value to find \( d \). From equation (1):
[tex]\[ d = \frac{2}{3} \times 45 \][/tex]
Thus,
[tex]\[ d = 30 \][/tex]
Conclusion:
The system of equations to determine the costs of the dance lesson (\( d \)) and the violin lesson (\( v \)) is:
[tex]\[ \begin{cases} d = \frac{2}{3} v \\ d + v = 75 \end{cases} \][/tex]
And the solution to this system is:
[tex]\[ v = 45, \quad d = 30 \][/tex]
