Let the width of the rectangle be $w$ meters. Since the length is 1 meter less than the width, the length can be represented as $l = w - 1$ meters. The area of a rectangle is given by the formula:

$\text{Area} = \text{Length} \times \text{Width}$

Substitute the given values into this equation:

$42 = (w - 1) \times w$

Now solve for $w$:

$42 = w^2 - w$

Rearrange the equation to form a quadratic equation:

$w^2 - w - 42 = 0$

Solve this quadratic equation using the quadratic formula:

$w = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-42)}}{2(1)}$

Simplify:

$w = \frac{1 \pm \sqrt{1 + 168}}{2}$ $w = \frac{1 \pm \sqrt{169}}{2}$ $w = \frac{1 \pm 13}{2}$

The two possible solutions are:

$w = \frac{1 + 13}{2} = 7 \quad \text{or} \quad w = \frac{1 - 13}{2} = -6$

Since the width cannot be negative, the width of the rectangle is $w = 7$ meters. Therefore, the length is:

$l = w - 1 = 7 - 1 = 6 \text{ meters}$

Thus, the dimensions of the rectangle are:

$\text{Length} = 6 \text{ meters}, \text{Width} = 7 \text{ meters}$

Would you like a more detailed explanation, or do you have any questions?

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Here are 5 related questions:

1. How would the problem change if the area were 56 square meters instead of 42?
2. What would happen if the length were 2 meters less than the width?
3. How would you solve the problem if the width was instead 1 meter more than the length?
4. Can you find the perimeter of the rectangle with the given dimensions?
5. What would the dimensions be if the area was doubled to 84 square meters?

Tip: When solving quadratic equations, always check if the problem constraints (like positive length or width) allow for both roots.