Let's solve the problem step by step.

Step 1: Define variables

Let:

• $k$ = cost of 1 kg of kale (in pounds)
• $b$ = cost of 1 kg of beetroot (in pounds)

Step 2: Form the equations

From the given information:

1. The cost of 4 kg of kale and 1.5 kg of beetroot is £12.20: $4k + 1.5b = 12.20$
2. The cost of 5 kg of kale and 2 kg of beetroot is £15.40: $5k + 2b = 15.40$

Step 3: Solve the system of equations

Equation 1:

$4k + 1.5b = 12.20$

Equation 2:

$5k + 2b = 15.40$

Multiply both equations to eliminate one variable:

• Multiply Equation 1 by 2: $8k + 3b = 24.40$
• Multiply Equation 2 by 1.5: $7.5k + 3b = 23.10$

Subtract the second equation from the first:

$(8k + 3b) - (7.5k + 3b) = 24.40 - 23.10$ $0.5k = 1.30$ $k = \frac{1.30}{0.5} = 2.60$

Step 4: Substitute $k = 2.60$ into one of the original equations

Substitute $k = 2.60$ into Equation 1: $4(2.60) + 1.5b = 12.20$ $10.40 + 1.5b = 12.20$ $1.5b = 12.20 - 10.40$ $1.5b = 1.80$ $b = \frac{1.80}{1.5} = 1.20$

Final Answer:

• Cost of 1 kg of kale ($k$) = £2.60
• Cost of 1 kg of beetroot ($b$) = £1.20

Would you like me to verify this or provide further explanation?

Related Questions:

1. How can we check if these results satisfy both equations?
2. How would the equations change if the weights of kale and beetroot were increased proportionally?
3. Can we solve these equations graphically? If yes, how?
4. What are the advantages of elimination vs. substitution methods in this case?
5. How would the problem change if kale or beetroot had a bulk discount rate?

Tip:

When solving systems of linear equations, elimination is a powerful tool to eliminate one variable and simplify calculations.