Find the area of a rectangle that has the length of 2x-3 and the width of x .Someone painted an interior area of the rectangle and it has a length of 3 and a width of x-2 .Find the size of the area that was not paintedHelp me please

Answer:The area that was not painted is [tex](2x^{2}-6x+6)\ units^{2}[/tex]Step-by-step explanation:step 1Find the area of the rectanglewe know thatThe area of a rectangle is equal to[tex]A=LW[/tex]In this problem we have[tex]L=2x-3[/tex][tex]W=x[/tex]substitute[tex]A=(2x-3)x\\A=(2x^{2}-3x)\ units^{2}[/tex]step 2Find the area that was painted[tex]A=(3)(x-2)\\A=(3x-6)\ units^{2}[/tex]step 3Find the area that was not paintedSubtract the area that was painted from the total area of rectangleso[tex]A=(2x^{2}-3x)-(3x-6)=(2x^{2}-6x+6)\ units^{2}[/tex]