# The width of a rectangle is 3 less than twice the length of x. if the area of a rectangle is 43...

## Question:

The width of a rectangle is 3 less than twice the length of x. if the area of a rectangle is 43 in sq ft. which equation can be used to find length in feet

{eq}a. 2x(x-3) = 43 \\b.2x + 2(2x-3) = 43 \\c.x(3 - 2x) = 43 \\d.x(2x-3) = 43 {/eq}

## Area of a Rectangle as Algebraic Expression:

An algebraic expression can be written for the area of a rectangle if an independent value is unknown in the formula. We can do this as long as we can express properly the equivalency of the length and width of the rectangle regardless if there is one or more unknown value.

The width of a rectangle is {eq}3{/eq} less than twice the length of {eq}x{/eq}. Thus, We can write the width of rectangle as:

\begin{align} Rectangle_{width}&=2Rectangle_{length}-3\\[0.2cm] Rectangle_{width}&=2x-3 \end{align}

Wherein, we have {eq}Rectangle_{length}=x {/eq}.

The area of a rectangle is {eq}43{/eq} in sq ft and the formula is given by:

\begin{align} Rectangle_{area}&=Rectangle_{length}*Rectangle_{width}\\[0.2cm] Rectangle_{area}&=\left ( x \right ) \left ( 2Rectangle_{length}-3\right ) \\[0.2cm] 43 \ \text{ft}^2&=\left ( x \right ) \left ( 2x-3\right ) \end{align}

The area of the rectangle can be expressed by letter d. {eq}x(2x-3) = 43{/eq}