Find the area bounded by the lines `y=4x+5, y=5-x" and "4y=x+5`.

Explanation

To find the area bounded by the lines \( y = 4x + 5 \), \( y = 5 - x \), and \( 4y = x + 5 \), we will follow these steps:

### Step 1: Find the Points of Intersection

1. **Intersection of \( y = 4x + 5 \) and \( y = 5 - x \)**:
\[ 4x + 5 = 5 - x \]
Rearranging gives:
\[ 4x + x = 5 - 5 \implies 5x = 0 \implies x = 0 \]
Substituting \( x = 0 \) into \( y = 5 - x \):
\[ y = 5 - 0 = 5 \]
So, the first point of intersection \( A \) is \( (0, 5) \).

2. **Intersection of \( y = 4x + 5 \) and \( 4y = x + 5 \)**:
First, rewrite \( 4y = x + 5 \) as \( y = \frac{x + 5}{4} \).
Setting \( 4x + 5 = \frac{x + 5}{4} \):
\[ 4(4x + 5) = x + 5 \implies 16x + 20 = x + 5 \implies 15x = -15 \implies x = -1 \]
Substituting \( x = -1 \) into \( y = 4x + 5 \):
\[ y = 4(-1) + 5 = 1 \]
So, the second point of intersection \( B \) is \( (-1, 1) \).

3. **Intersection of \( y = 5 - x \) and \( 4y = x + 5 \)**:
Setting \( 5 - x = \frac{x + 5}{4} \):
\[ 4(5 - x) = x + 5 \implies 20 - 4x = x + 5 \implies 20 - 5 = 5x \implies 15 = 5x \implies x = 3 \]
Substituting \( x = 3 \) into \( y = 5 - x \):
\[ y = 5 - 3 = 2 \]
So, the third point of intersection \( C \) is \( (3, 2) \).

### Step 2: Sketch the Region

We now have the points of intersection \( A(0, 5) \), \( B(-1, 1) \), and \( C(3, 2) \). The area we need to find is bounded by these points.

### Step 3: Set Up the Integrals

We will divide the area into two parts: \( \text{Area A} \) bounded by \( A \), \( B \), and the line \( 4y = x + 5 \), and \( \text{Area B} \) bounded by \( B \), \( C \), and the line \( y = 5 - x \).

1. **Area A**:
\[ \text{Area A} = \int_{-1}^{0} \left( (4x + 5) - \left(\frac{x + 5}{4}\right) \right) dx \]

2. **Area B**:
\[ \text{Area B} = \int_{0}^{3} \left( (5 - x) - \left(\frac{x + 5}{4}\right) \right) dx \]

### Step 4: Calculate the Integrals

1. **Calculating Area A**:
\[ \text{Area A} = \int_{-1}^{0} \left( 4x + 5 - \frac{x + 5}{4} \right) dx \]
Simplifying the integrand:
\[ 4x + 5 - \frac{x + 5}{4} = \frac{16x + 20 - x - 5}{4} = \frac{15x + 15}{4} = \frac{15(x + 1)}{4} \]
Now integrate:
\[ \text{Area A} = \frac{15}{4} \int_{-1}^{0} (x + 1) dx = \frac{15}{4} \left[ \frac{x^2}{2} + x \right]_{-1}^{0} = \frac{15}{4} \left(0 - \left(\frac{1}{2} - 1\right)\right) = \frac{15}{4} \left(\frac{1}{2}\right) = \frac{15}{8} \]

2. **Calculating Area B**:
\[ \text{Area B} = \int_{0}^{3} \left( 5 - x - \frac{x + 5}{4} \right) dx \]
Simplifying the integrand:
\[ 5 - x - \frac{x + 5}{4} = \frac{20 - 4x - x - 5}{4} = \frac{15 - 5x}{4} \]
Now integrate:
\[ \text{Area B} = \frac{1}{4} \int_{0}^{3} (15 - 5x) dx = \frac{1}{4} \left[ 15x - \frac{5x^2}{2} \right]_{0}^{3} = \frac{1}{4} \left( 45 - \frac{45}{2} \right) = \frac{1}{4} \left( \frac{90 - 45}{2} \right) = \frac{1}{4} \left( \frac{45}{2} \right) = \frac{45}{8} \]

### Step 5: Total Area

Now, we sum the areas:
\[ \text{Total Area} = \text{Area A} + \text{Area B} = \frac{15}{8} + \frac{45}{8} = \frac{60}{8} = \frac{15}{2} \]

### Final Answer
The area bounded by the lines is \( \frac{15}{2} \) square units.

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