Answer :
To solve this problem, we need to formulate and solve a system of linear equations based on the costs given.
Let's denote:
- [tex]\( r \)[/tex] as the cost of 1 kg of radishes
- [tex]\( c \)[/tex] as the cost of 1 kg of carrots
We are given two pieces of information which can be translated into two equations:
1. [tex]\( 4r + 1.5c = 14.80 \)[/tex]
2. [tex]\( 3r + 2c = 12.50 \)[/tex]
We need to solve this system of equations to find [tex]\( r \)[/tex] and [tex]\( c \)[/tex].
### Equation 1:
[tex]\[ 4r + 1.5c = 14.80 \][/tex]
### Equation 2:
[tex]\[ 3r + 2c = 12.50 \][/tex]
To solve these equations, we can use the method of substitution or elimination. In this case, let's use the method of elimination for simplicity.
#### Step 1: Make the coefficients of [tex]\( c \)[/tex] the same in both equations
Multiply Equation 1 by 2 to make the coefficient of [tex]\( c \)[/tex] equal to 3:
[tex]\[ 2(4r + 1.5c) = 2(14.80) \][/tex]
[tex]\[ 8r + 3c = 29.60 \][/tex] — [Equation 3]
Now we have:
[tex]\[ 8r + 3c = 29.60 \][/tex] — [Equation 3]
[tex]\[ 3r + 2c = 12.50 \][/tex] — [Equation 2]
#### Step 2: Eliminate [tex]\( c \)[/tex]
To eliminate [tex]\( c \)[/tex], we will multiply Equation 2 by 3 to match the coefficient of [tex]\( r \)[/tex] in Equation 3:
[tex]\[ 3(3r + 2c) = 3(12.50) \][/tex]
[tex]\[ 9r + 6c = 37.50 \][/tex] — [Equation 4]
Now we have:
[tex]\[ 8r + 3c = 29.60 \][/tex] — [Equation 3]
[tex]\[ 9r + 6c = 37.50 \][/tex] — [Equation 4]
#### Step 3: Align coefficients and subtract equations
To facilitate elimination, let's align Equation 3 and 4 by multiplying Equation 3 by 2:
[tex]\[ 2(8r + 3c) = 2(29.60) \][/tex]
[tex]\[ 16r + 6c = 59.20 \][/tex] — [Equation 5]
Now we have:
[tex]\[ 16r + 6c = 59.20 \][/tex] — [Equation 5]
[tex]\[ 9r + 6c = 37.50 \][/tex] — [Equation 4]
Subtract Equation 4 from Equation 5 to eliminate [tex]\( c \)[/tex]:
[tex]\[ (16r + 6c) - (9r + 6c) = 59.20 - 37.50 \][/tex]
[tex]\[ 7r = 21.70 \][/tex]
#### Step 4: Solve for [tex]\( r \)[/tex]
[tex]\[ r = \frac{21.70}{7} \][/tex]
[tex]\[ r = 3.10 \][/tex]
So, the cost of 1 kg of radishes is [tex]\( r = £3.10 \)[/tex].
#### Step 5: Substitute [tex]\( r \)[/tex] back into one of the original equations
Let's substitute [tex]\( r = 3.10 \)[/tex] into Equation 1 to find [tex]\( c \)[/tex]:
[tex]\[ 4(3.10) + 1.5c = 14.80 \][/tex]
[tex]\[ 12.40 + 1.5c = 14.80 \][/tex]
Subtract 12.40 from both sides:
[tex]\[ 1.5c = 2.40 \][/tex]
Divide both sides by 1.5:
[tex]\[ c = \frac{2.40}{1.5} \][/tex]
[tex]\[ c = 1.60 \][/tex]
So, the cost of 1 kg of carrots is [tex]\( c = £1.60 \)[/tex].
### Final Answers:
a) The cost of 1 kg of carrots is [tex]\( £1.60 \)[/tex].
b) The cost of 1 kg of radishes is [tex]\( £3.10 \)[/tex].
Let's denote:
- [tex]\( r \)[/tex] as the cost of 1 kg of radishes
- [tex]\( c \)[/tex] as the cost of 1 kg of carrots
We are given two pieces of information which can be translated into two equations:
1. [tex]\( 4r + 1.5c = 14.80 \)[/tex]
2. [tex]\( 3r + 2c = 12.50 \)[/tex]
We need to solve this system of equations to find [tex]\( r \)[/tex] and [tex]\( c \)[/tex].
### Equation 1:
[tex]\[ 4r + 1.5c = 14.80 \][/tex]
### Equation 2:
[tex]\[ 3r + 2c = 12.50 \][/tex]
To solve these equations, we can use the method of substitution or elimination. In this case, let's use the method of elimination for simplicity.
#### Step 1: Make the coefficients of [tex]\( c \)[/tex] the same in both equations
Multiply Equation 1 by 2 to make the coefficient of [tex]\( c \)[/tex] equal to 3:
[tex]\[ 2(4r + 1.5c) = 2(14.80) \][/tex]
[tex]\[ 8r + 3c = 29.60 \][/tex] — [Equation 3]
Now we have:
[tex]\[ 8r + 3c = 29.60 \][/tex] — [Equation 3]
[tex]\[ 3r + 2c = 12.50 \][/tex] — [Equation 2]
#### Step 2: Eliminate [tex]\( c \)[/tex]
To eliminate [tex]\( c \)[/tex], we will multiply Equation 2 by 3 to match the coefficient of [tex]\( r \)[/tex] in Equation 3:
[tex]\[ 3(3r + 2c) = 3(12.50) \][/tex]
[tex]\[ 9r + 6c = 37.50 \][/tex] — [Equation 4]
Now we have:
[tex]\[ 8r + 3c = 29.60 \][/tex] — [Equation 3]
[tex]\[ 9r + 6c = 37.50 \][/tex] — [Equation 4]
#### Step 3: Align coefficients and subtract equations
To facilitate elimination, let's align Equation 3 and 4 by multiplying Equation 3 by 2:
[tex]\[ 2(8r + 3c) = 2(29.60) \][/tex]
[tex]\[ 16r + 6c = 59.20 \][/tex] — [Equation 5]
Now we have:
[tex]\[ 16r + 6c = 59.20 \][/tex] — [Equation 5]
[tex]\[ 9r + 6c = 37.50 \][/tex] — [Equation 4]
Subtract Equation 4 from Equation 5 to eliminate [tex]\( c \)[/tex]:
[tex]\[ (16r + 6c) - (9r + 6c) = 59.20 - 37.50 \][/tex]
[tex]\[ 7r = 21.70 \][/tex]
#### Step 4: Solve for [tex]\( r \)[/tex]
[tex]\[ r = \frac{21.70}{7} \][/tex]
[tex]\[ r = 3.10 \][/tex]
So, the cost of 1 kg of radishes is [tex]\( r = £3.10 \)[/tex].
#### Step 5: Substitute [tex]\( r \)[/tex] back into one of the original equations
Let's substitute [tex]\( r = 3.10 \)[/tex] into Equation 1 to find [tex]\( c \)[/tex]:
[tex]\[ 4(3.10) + 1.5c = 14.80 \][/tex]
[tex]\[ 12.40 + 1.5c = 14.80 \][/tex]
Subtract 12.40 from both sides:
[tex]\[ 1.5c = 2.40 \][/tex]
Divide both sides by 1.5:
[tex]\[ c = \frac{2.40}{1.5} \][/tex]
[tex]\[ c = 1.60 \][/tex]
So, the cost of 1 kg of carrots is [tex]\( c = £1.60 \)[/tex].
### Final Answers:
a) The cost of 1 kg of carrots is [tex]\( £1.60 \)[/tex].
b) The cost of 1 kg of radishes is [tex]\( £3.10 \)[/tex].
