The functions f and g are such that \( f(x)=5 x+3 \) and \( g(x)=a x+b \) where a and b are constants, \( \mathrm{g}(3)=20 \) and \( f^{-1}(33)=g(1) \). Find a and b \( \} \mathrm{a}=\square\{ \} \mathrm{b}=\square \)

To solve for constants \( a \) and \( b \), we start with the known functions. First, we calculate \( g(3) \) using the information given: Since \( g(3) = 20 \), we have: \[ g(3) = a(3) + b = 20 \implies 3a + b = 20 \quad \text{(1)} \] Next, we need to determine \( f^{-1}(33) \). The function \( f(x) = 5x + 3 \) can be inverted. To find the inverse, we set: \[ y = 5x + 3 \] Solving for \( x \): \[ y - 3 = 5x \implies x = \frac{y - 3}{5} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{x - 3}{5} \] Calculating \( f^{-1}(33) \): \[ f^{-1}(33) = \frac{33 - 3}{5} = \frac{30}{5} = 6 \] From the problem, we know: \[ f^{-1}(33) = g(1) \implies g(1) = 6 \] So: \[ g(1) = a(1) + b = 6 \implies a + b = 6 \quad \text{(2)} \] Now we have a system of two equations: 1. \( 3a + b = 20 \) 2. \( a + b = 6 \) We can solve these equations simultaneously. Subtract equation (2) from equation (1): \[ (3a + b) - (a + b) = 20 - 6 \] This simplifies to: \[ 2a = 14 \implies a = 7 \] Now, substitute \( a = 7 \) back into equation (2): \[ 7 + b = 6 \implies b = 6 - 7 = -1 \] Thus, the values of \( a \) and \( b \) are: \[ \mathrm{a} = 7 \quad \mathrm{b} = -1 \]