15x + 5y =-8
y+4 =-3x
How many solutions does the system above have ?

Options

A. 0
B. 1
C. 2
D. An infinite number

Answer

To determine how many solutions the system of equations has, we will analyze the given equations step by step.

### Given Equations:
1. \( 15x + 5y = -8 \) (Equation 1)
2. \( y + 4 = -3x \) (Equation 2)

### Step 1: Rewrite the second equation in standard form
We can rearrange Equation 2 to express it in the standard form \( Ax + By + C = 0 \).

Starting with:
\[ y + 4 = -3x \]

We can move \( 3x \) to the left side:
\[ 3x + y + 4 = 0 \]

### Step 2: Identify coefficients
Now we have two equations in standard form:
1. \( 15x + 5y + 8 = 0 \) (from Equation 1)
2. \( 3x + y + 4 = 0 \) (from Equation 2)

From these equations, we can identify:
- For Equation 1: \( A_1 = 15, B_1 = 5, C_1 = 8 \)
- For Equation 2: \( A_2 = 3, B_2 = 1, C_2 = 4 \)

### Step 3: Calculate the ratios of coefficients
Next, we will calculate the ratios of the coefficients \( A_1/A_2 \), \( B_1/B_2 \), and \( C_1/C_2 \):

1. \( \frac{A_1}{A_2} = \frac{15}{3} = 5 \)
2. \( \frac{B_1}{B_2} = \frac{5}{1} = 5 \)
3. \( \frac{C_1}{C_2} = \frac{8}{4} = 2 \)

### Step 4: Analyze the ratios
Now we compare the ratios:
- \( \frac{A_1}{A_2} = 5 \)
- \( \frac{B_1}{B_2} = 5 \)
- \( \frac{C_1}{C_2} = 2 \)

We see that:
- \( \frac{A_1}{A_2} = \frac{B_1}{B_2} \) (both equal to 5)
- \( \frac{C_1}{C_2} \) is not equal to \( \frac{A_1}{A_2} \) or \( \frac{B_1}{B_2} \) (it equals 2)

### Conclusion
Since \( \frac{A_1}{A_2} = \frac{B_1}{B_2} \) but \( \frac{C_1}{C_2} \) is not equal to these ratios, the system of equations has no solutions. This means the lines represented by these equations are parallel and do not intersect.

### Final Answer
The system has **0 solutions**.

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