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Let's define the variables:
- Let \( f \) be the cost of one football.
- Let \( t \) be the cost of one tennis ball.

We have two pieces of information from the problem:
1. \( 2f + 3t = 21.73 \) (the cost of 2 footballs and 3 tennis balls is £21.73)
2. \( 5f + 7t = 53.20 \) (the cost of 5 footballs and 7 tennis balls is £53.20)

We can solve these two simultaneous equations to find the values of \( f \) (the cost of a football) and \( t \) (the cost of a tennis ball).

### Step 1: Multiply to eliminate one variable
To eliminate \( t \), let's first make the coefficients of \( t \) the same in both equations. Multiply the first equation by 7 and the second equation by 3:

\[
7(2f + 3t) = 7(21.73) \quad \Rightarrow \quad 14f + 21t = 152.11
\]
\[
3(5f + 7t) = 3(53.20) \quad \Rightarrow \quad 15f + 21t = 159.60
\]

### Step 2: Subtract the two equations
Now subtract the first new equation from the second:

\[
(15f + 21t) - (14f + 21t) = 159.60 - 152.11
\]
\[
f = 7.49
\]

So, the cost of a football is \( f = 7.49 \) pounds.

### Step 3: Substitute back to find \( t \)
Now that we know \( f = 7.49 \), substitute this into one of the original equations to find \( t \). Let's use the first equation:

\[
2(7.49) + 3t = 21.73
\]
\[
14.98 + 3t = 21.73
\]
\[
3t = 21.73 - 14.98
\]
\[
3t = 6.75
\]
\[
t = \frac{6.75}{3} = 2.25
\]

So, the cost of a tennis ball is \( t = 2.25 \) pounds.

### Final Answer:
- The cost of a football is £7.49.
- The cost of a tennis ball is £2.25.

Would you like further details on any step or have any questions?

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Here are 5 questions to deepen understanding:
1. How can simultaneous equations be solved using substitution instead of elimination?
2. What happens if we get negative prices when solving this system?
3. Could we solve the system graphically, and how would that look?
4. Why do we multiply the equations to align coefficients for elimination?
5. What if there were three variables? How would the solution process change?

### Tip:
When solving simultaneous equations, always verify your solution by plugging the values back into the original equations to ensure they satisfy both conditions.

The cost of 2 footballs and 3 tennis balls is £21.73. The cost of 5 footballs and 7 tennis balls is £53.20. Work out the cost of a football and a tennis ball.

Learn how to solve the system of linear equations to find the cost of footballs and tennis balls. This problem involves algebraic methods such as elimination and substitution. Perfect for students in Grades 8-10.

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