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Sagot :
Let's solve the given problem step by step:
### 1. Formulating the Linear Equation
Given:
- The cost of 5 kg of apples and 2 kg of oranges is Rs. 330.
Let's denote:
- [tex]\( x \)[/tex] as the cost of 1 kg of apples (in Rs).
- [tex]\( y \)[/tex] as the cost of 1 kg of oranges (in Rs).
The relationship provided in the problem can be written as a linear equation:
[tex]\[ 5x + 2y = 330 \][/tex]
### 2. Representing the Equation Graphically
To graph this equation, we need to express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ 2y = 330 - 5x \][/tex]
[tex]\[ y = \frac{330 - 5x}{2} \][/tex]
Now, let's find some points that satisfy this equation. By setting different values for [tex]\( x \)[/tex], we can calculate the corresponding values of [tex]\( y \)[/tex]:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \frac{330 - 5 \cdot 0}{2} = \frac{330}{2} = 165 \][/tex]
Point: (0, 165)
- When [tex]\( x = 10 \)[/tex]:
[tex]\[ y = \frac{330 - 5 \cdot 10}{2} = \frac{330 - 50}{2} = \frac{280}{2} = 140 \][/tex]
Point: (10, 140)
- When [tex]\( x = 50 \)[/tex]:
[tex]\[ y = \frac{330 - 5 \cdot 50}{2} = \frac{330 - 250}{2} = \frac{80}{2} = 40 \][/tex]
Point: (50, 40)
Additionally, for a thorough representation, we can use many more points. For instance, calculate [tex]\( y \)[/tex] for [tex]\( x \)[/tex] from 0 to 100 with a fine increment. This ensures a smooth continuous curve on the graph.
### 3. Plotting the Graph
Let's plot the equation using the points we calculated. Here is the plotting process:
1. X-axis (Horizontal): Represents the cost in Rs of 1 kg of apples ([tex]\( x \)[/tex]).
2. Y-axis (Vertical): Represents the cost in Rs of 1 kg of oranges ([tex]\( y \)[/tex]).
The plot should clearly show the line passing through points such as (0, 165), (10, 140), and (50, 40). The line represents all possible combinations of costs for apples and oranges that sum up to Rs. 330 when taken in the given proportions.
### Graphical Representation:
Using the data previously processed:
- X-axis range: from 0 to around 100 for [tex]\( x \)[/tex].
- Y-axis range: calculated accordingly using the equation.
Here is a simplified illustration of how the graph should look:
```
|
165 | (0,165)
|
140 | (10,140)
|
|
40 | * (50,40)
|
|
0 ---------+------------------------x (cost of apples)
0 10 50 100
```
The line will start from the point (0, 165) and extend through other calculated points like (10, 140) and (50, 40), continuing in a straight path as the equation is linear. This implies any other combinations lying on this line will maintain the given cost relationship between apples and oranges.
Summary:
We formulated a linear equation from the given problem and represented it graphically by plotting calculated points. This shows how the costs of apples and oranges sum up to Rs. 330, representing possible cost combinations graphically.
### 1. Formulating the Linear Equation
Given:
- The cost of 5 kg of apples and 2 kg of oranges is Rs. 330.
Let's denote:
- [tex]\( x \)[/tex] as the cost of 1 kg of apples (in Rs).
- [tex]\( y \)[/tex] as the cost of 1 kg of oranges (in Rs).
The relationship provided in the problem can be written as a linear equation:
[tex]\[ 5x + 2y = 330 \][/tex]
### 2. Representing the Equation Graphically
To graph this equation, we need to express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ 2y = 330 - 5x \][/tex]
[tex]\[ y = \frac{330 - 5x}{2} \][/tex]
Now, let's find some points that satisfy this equation. By setting different values for [tex]\( x \)[/tex], we can calculate the corresponding values of [tex]\( y \)[/tex]:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \frac{330 - 5 \cdot 0}{2} = \frac{330}{2} = 165 \][/tex]
Point: (0, 165)
- When [tex]\( x = 10 \)[/tex]:
[tex]\[ y = \frac{330 - 5 \cdot 10}{2} = \frac{330 - 50}{2} = \frac{280}{2} = 140 \][/tex]
Point: (10, 140)
- When [tex]\( x = 50 \)[/tex]:
[tex]\[ y = \frac{330 - 5 \cdot 50}{2} = \frac{330 - 250}{2} = \frac{80}{2} = 40 \][/tex]
Point: (50, 40)
Additionally, for a thorough representation, we can use many more points. For instance, calculate [tex]\( y \)[/tex] for [tex]\( x \)[/tex] from 0 to 100 with a fine increment. This ensures a smooth continuous curve on the graph.
### 3. Plotting the Graph
Let's plot the equation using the points we calculated. Here is the plotting process:
1. X-axis (Horizontal): Represents the cost in Rs of 1 kg of apples ([tex]\( x \)[/tex]).
2. Y-axis (Vertical): Represents the cost in Rs of 1 kg of oranges ([tex]\( y \)[/tex]).
The plot should clearly show the line passing through points such as (0, 165), (10, 140), and (50, 40). The line represents all possible combinations of costs for apples and oranges that sum up to Rs. 330 when taken in the given proportions.
### Graphical Representation:
Using the data previously processed:
- X-axis range: from 0 to around 100 for [tex]\( x \)[/tex].
- Y-axis range: calculated accordingly using the equation.
Here is a simplified illustration of how the graph should look:
```
|
165 | (0,165)
|
140 | (10,140)
|
|
40 | * (50,40)
|
|
0 ---------+------------------------x (cost of apples)
0 10 50 100
```
The line will start from the point (0, 165) and extend through other calculated points like (10, 140) and (50, 40), continuing in a straight path as the equation is linear. This implies any other combinations lying on this line will maintain the given cost relationship between apples and oranges.
Summary:
We formulated a linear equation from the given problem and represented it graphically by plotting calculated points. This shows how the costs of apples and oranges sum up to Rs. 330, representing possible cost combinations graphically.
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