To solve for the constants \( h \) and \( k \), we start by writing the vectors \( \mathbf{p} \), \( \mathbf{q} \), and \( \mathbf{y} \) using the equations provided:
- \( \mathbf{p} = 3\mathbf{t} - \mathbf{s} \)
- \( \mathbf{q} = 5\mathbf{t} + \mathbf{s} \)
- \( \mathbf{y} = h\mathbf{t} - (h+k)\mathbf{s} \)
We also know that \( \mathbf{y} = 3\mathbf{q} - 2\mathbf{p} \). Let's express \( \mathbf{y} \) using \( \mathbf{q} \) and \( \mathbf{p} \):
We first calculate \( 3\mathbf{q} \) and \( -2\mathbf{p} \):
Starting with \( 3\mathbf{q} \):
\[ 3\mathbf{q} = 3(5\mathbf{t} + \mathbf{s}) = 15\mathbf{t} + 3\mathbf{s} \]
Now, calculate \( -2\mathbf{p} \):
\[ -2\mathbf{p} = -2(3\mathbf{t} - \mathbf{s}) = -6\mathbf{t} + 2\mathbf{s} \]
Now, let's add these two results to find \( \mathbf{y} \):
\[ \mathbf{y} = 3\mathbf{q} - 2\mathbf{p} = (15\mathbf{t} + 3\mathbf{s}) + (-6\mathbf{t} + 2\mathbf{s}) \]
Combining the like terms:
\[ \mathbf{y} = (15\mathbf{t} - 6\mathbf{t}) + (3\mathbf{s} + 2\mathbf{s}) = 9\mathbf{t} + 5\mathbf{s} \]
Now we have two expressions for \( \mathbf{y} \):
- \( \mathbf{y} = h\mathbf{t} - (h+k)\mathbf{s} \)
- \( \mathbf{y} = 9\mathbf{t} + 5\mathbf{s} \)
Next, we can set the coefficients of \( \mathbf{t} \) and \( \mathbf{s} \) from both expressions equal to each other:
From the term with \( \mathbf{t} \): \[ h = 9 \]
From the term with \( \mathbf{s} \): \[ -(h + k) = 5 \]
Substituting \( h = 9 \) into the second equation:
\[ -(9 + k) = 5 \]
This can be simplified to:
\[ -9 - k = 5 \] \[ -k = 5 + 9 \] \[ -k = 14 \] \[ k = -14 \]
Finally, we have found the values for \( h \) and \( k \):
\[ h = 9, \quad k = -14 \]
This means the solution is:
\[ \boxed{h = 9, \quad k = -14} \]