Question

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 painted
Help me please

Answer 1

Answer:

The area that was not painted is $(2x^{2}-6x+6)\ units^{2}$

Step-by-step explanation:

step 1

Find the area of the rectangle

we know that

The area of a rectangle is equal to

$A=LW$

In this problem we have

$L=2x-3$

$W=x$

substitute

$A=(2x-3)x\\A=(2x^{2}-3x)\ units^{2}$

step 2

Find the area that was painted

$A=(3)(x-2)\\A=(3x-6)\ units^{2}$

step 3

Find the area that was not painted

Subtract the area that was painted from the total area of rectangle

so

$A=(2x^{2}-3x)-(3x-6)=(2x^{2}-6x+6)\ units^{2}$