Question

Equation for the length of a rectangle is two more than three times the width area is 161

Answer 1

Length = 3w + 2 = 23.

Explanation:

Given :-

• The length of a rectangle is two more than three times the width = ( 3w + 2 )
• Let width of rectangle be w.
• Area of rectangle = 161.

Find :-

• Length of rectangle .

Solution :-

We know that ,

Area of rectangle = Length × Width.

• Substitute the values

161 = (3w + 2) × w

• Distribute w .

161 = 3w² + 2w

• Move constant to the left-hand side and change their signs.

0 = 3w² + 2w - 161

• Split the middle term- 2w as 23w - 21w.

0 = 3w² + 23w - 21w - 161.

• Factor out w from first pair and -7 from second pair.

0 = w ( 3w + 23 ) -7 ( 3w + 23)

• Factor out 3w + 23.

0 = (3w + 23 ) ( w - 7)

• Using zero product property.

w - 7 = 0 , 3w + 23 = 0

• Solve for w.

w = 7 , w = -23/3

Since, width of rectangle can't be negative.

so, w = 7 is right value of width.

Calculate for Length.

length = 3w + 2 ( plug the value of width)

• length = 3 ( 7 ) + 2
• Length = 21 + 2

Hence, length = 23

Answer 2

l = 3w+2

l = 23

Explanation:

The area of a rectangle is

A = lw where l is the length and w is the width

l = 3w+2

A = 161

161 = (3w+2) *w

Distribute

161 = 3w^2 +2w

Subtract 161 from each side

0=3w^2 +2w -161

Factor

0=(w-7)(3w+23)

Using the zero product property

w-7 =0 3w+23 =0

w=7 w = -23/3

Since the width cannot be negative

w=7

l = 3w+2

l = 3(7)+2

l = 21+2 = 23