If `f(x)=ax+b` , where a and b are integers, `f(-1)=-5` and `f(3)=3` then a and b are equal to

Options

A. `a=-3,b=-1`
B. `a=2,b=-3`
C. `a=0,b=2`
D. `a=2,b=3`

Explanation

To solve the problem where \( f(x) = ax + b \) and we are given \( f(-1) = -5 \) and \( f(3) = 3 \), we will follow these steps:

### Step 1: Set up the equations based on the given conditions
Using the function \( f(x) = ax + b \):

1. For \( f(-1) = -5 \):
\[ f(-1) = a(-1) + b = -a + b = -5 \quad \text{(Equation 1)} \]

2. For \( f(3) = 3 \):
\[ f(3) = a(3) + b = 3a + b = 3 \quad \text{(Equation 2)} \]

### Step 2: Write down the equations
We have the following two equations:
1. \(-a + b = -5\) (Equation 1)
2. \(3a + b = 3\) (Equation 2)

### Step 3: Eliminate \( b \) by subtracting the equations
To eliminate \( b \), we can subtract Equation 1 from Equation 2:
\[ (3a + b) - (-a + b) = 3 - (-5) \]
This simplifies to:
\[ 3a + b + a - b = 3 + 5 \]
\[ 4a = 8 \]

### Step 4: Solve for \( a \)
Now, we can solve for \( a \):
\[ 4a = 8 \implies a = \frac{8}{4} = 2 \]

### Step 5: Substitute \( a \) back to find \( b \)
Now that we have \( a = 2 \), we can substitute this value back into either Equation 1 or Equation 2 to find \( b \). We will use Equation 1:
\[ -a + b = -5 \]
Substituting \( a = 2 \):
\[ -2 + b = -5 \]
\[ b = -5 + 2 = -3 \]

### Step 6: State the final values of \( a \) and \( b \)
Thus, the values of \( a \) and \( b \) are:
\[ a = 2, \quad b = -3 \]

### Final Answer:
\[ a = 2, \quad b = -3 \]

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