We are given the following information.
Perimeter of rectangle = 8x + 12 units
Length of rectangle = 2x − 3 units
Recall that the perimeter of a rectangle is given by
[tex]P=2(L+W)[/tex]Where L is the length and W is the width of the rectangle.
Let us substitute the values of perimeter and length and solve for width.
[tex]\begin{gathered} P=2(L+W) \\ 8x+12=2(2x-3+W) \\ 8x+12=4x-6+2W \\ 8x-4x+12+6=2W \\ 4x+18=2W \\ 2W=4x+18 \\ W=\frac{4x+18}{2} \\ W=2x+9 \end{gathered}[/tex]Recall that the area of a rectangle is given by
[tex]A=L\cdot W[/tex]Substitute the values of length and width
[tex]\begin{gathered} A=L\cdot W \\ A=(2x-3)\cdot(2x+9) \\ A=2x\cdot2x+2x\cdot9-3\cdot2x-3\cdot9 \\ A=4x^2+18x-6x-27 \\ A=4x^2+12x-27 \end{gathered}[/tex]Therefore, the area of the rectangle is
[tex]A=4x^2+12x-27[/tex]