If `(3y-1)(2y+k)=ay^(2)+by-5` for all values of y, what is the value of a+b?

Answer

To solve the equation \((3y-1)(2y+k)=ay^{2}+by-5\) for all values of \(y\) and find the value of \(a+b\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:

\[ (3y - 1)(2y + k) = 3y \cdot 2y + 3y \cdot k - 1 \cdot 2y - 1 \cdot k \]

Calculating each term gives:

\[ = 6y^2 + 3ky - 2y - k \]

### Step 2: Combine like terms
Now, we can combine the \(y\) terms:

\[ = 6y^2 + (3k - 2)y - k \]

### Step 3: Set the expanded equation equal to the right-hand side
Now, we set this equal to the right-hand side of the original equation:

\[ 6y^2 + (3k - 2)y - k = ay^2 + by - 5 \]

### Step 4: Compare coefficients
Since this equation holds for all values of \(y\), we can compare the coefficients of \(y^2\), \(y\), and the constant terms on both sides.

1. For \(y^2\):
\[ a = 6 \]

2. For \(y\):
\[ b = 3k - 2 \]

3. For the constant term:
\[ -k = -5 \implies k = 5 \]

### Step 5: Substitute \(k\) into the equation for \(b\)
Now that we have \(k\), we can substitute it back into the equation for \(b\):

\[ b = 3(5) - 2 = 15 - 2 = 13 \]

### Step 6: Calculate \(a + b\)
Now we can find \(a + b\):

\[ a + b = 6 + 13 = 19 \]

### Final Answer
Thus, the value of \(a + b\) is:

\[ \boxed{19} \]

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