GEOMETRY Find the value of x so the rectangles have the same area.

The area of a rectangle is given as $lw$, where $l$is the length of the rectangle and is the width of the rectangle.

From the given figure, the length of the first rectangle is $x$units and the width of the rectangle is $12$units. Then, the area of the first rectangle is $12x$units.

Similarly, from the given figure, the length of the second rectangle is $16$units and the width of the second rectangle is $x-2$units. Then, the area of the second rectangle is $16\left(x-2\right)$units.

It is given that the figures have the same area. Then, the areas of the first and the second rectangle are equal. So, “$12x=16\left(x-2\right)$”.

This is the required equation.

Solving,

$12x=16\left(x-2\right)$ (Given equation)

$12x=16x-32$ (Distributive property)

$12x-16x=16x-32-16x$ (Subtract $16x$from both sides)

$-4x=-32$ (Simplify)

$\frac{-4x}{-4}=\frac{-32}{-4}$ (Divide both sides by $-4$)

$x=8$ (Simplify)

So, $x=8$is the solution of the given equation.

It was assumed that the areas of both figures are equal. This assumption led to the obtained solution $x=8$.

Thus, the value of x so that the figures have the same area is 8.