You have the system of equations:

1. $40 = a \cdot b$
2. $a = b + 5$

To solve this system, we can substitute the expression for $a$ from the second equation into the first one.

Substituting $a = b + 5$ into $40 = a \cdot b$, we get:

$40 = (b + 5) \cdot b$

Now expand the equation:

$40 = b^2 + 5b$

Rearrange the equation into standard quadratic form:

$b^2 + 5b - 40 = 0$

Next, we can solve this quadratic equation using the quadratic formula. The quadratic formula is given by:

$b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

For $b^2 + 5b - 40 = 0$, $A = 1$, $B = 5$, and $C = -40$.

Substitute the values into the formula:

$b = \frac{-5 \pm \sqrt{5^2 - 4(1)(-40)}}{2(1)}$

Simplify:

$b = \frac{-5 \pm \sqrt{25 + 160}}{2}$

$b = \frac{-5 \pm \sqrt{185}}{2}$

Now, calculate the two possible values for $b$:

$b = \frac{-5 + \sqrt{185}}{2} \quad \text{or} \quad b = \frac{-5 - \sqrt{185}}{2}$

Approximating the square root of 185:

$b = \frac{-5 + 13.6}{2} \quad \text{or} \quad b = \frac{-5 - 13.6}{2}$

$b \approx \frac{8.6}{2} = 4.3 \quad \text{or} \quad b \approx \frac{-18.6}{2} = -9.3$

So, the two possible values for $b$ are approximately $4.3$ and $-9.3$.

Now, substitute these back into the second equation $a = b + 5$:

• If $b \approx 4.3$, then $a \approx 4.3 + 5 = 9.3$
• If $b \approx -9.3$, then $a \approx -9.3 + 5 = -4.3$

Thus, the solutions are:

• $a \approx 9.3$, $b \approx 4.3$
• $a \approx -4.3$, $b \approx -9.3$

Would you like further clarification or a more detailed breakdown of these steps?

Here are five related questions to explore:

1. How do you solve quadratic equations by factoring?
2. What are the other methods for solving quadratic equations besides the quadratic formula?
3. What does the discriminant of a quadratic equation tell you about the number of solutions?
4. How can you verify the solutions of a system of equations?
5. How would you solve a system of equations with two variables but with a different relationship, like $a = 2b - 3$?

Tip: When solving quadratic equations, always check the solutions by substituting them back into the original system to ensure they satisfy both equations.