Question

The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 square inches. What are the dimensions of the rectangle? If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

A) 2x2 + 3x - 161 = 0
B)3x2 + 2x - 161 = 0
C) 6x2 - 161 = 0

Answer 1

Case b is correct but you have to write
$3 {x }^{2} + 2x - 161 = 0$

Answer 2

Answer:

Option B is correct

$3x^2+2x-161=0$

Step-by-step explanation:

Let the width of the rectangle be x units.

As per the statement:

The length of a rectangle is 2 more than three times the width.

⇒Length of rectangle = 3x+2 units

Recall the following result:

Area(A) of rectangle is given by:

$A = lw$

where, l is the length and w is the width of the rectangle respectively.

It is also given that the area of the rectangle is 161 square inches

⇒161 = (3x+2)x

By distributive property: $a \cdot (b+c) = a\cdot b+ a\cdot c$

then;

$161 = 3x^2+2x$

or

$3x^2+2x=161$

Subtract 161 from both sides we have;

$3x^2+2x-161=0$

Therefore, the following equations is used in the process of solving this problem is, $3x^2+2x-161=0$.