Question

By what like amount does the length and width of 6 by 4 rectangle need to be increased for its area to be doubled

Answer 1

Answer is to add “2”
to 6 and 4

Step by step

We know the sides are 6 x 4
We know the current Area is 24
We know we want the area to double to 48

So what do we add to 6 and 4 is the question

Let “x” be the unknown amount to add

So 6 + x and 4 + x equal 24

( x + 6) ( x + 4)
Distribute

x^2 + 4x + 6x + 24
Simplify

x^2 + 10x + 24
Factor

* what two numbers multiplied equal 24 that also add up to 10. Use a factor tree or search online for factors

(x + 12) or ( x - 2 ) = 0

find each solution by = 0

x + 12 = 0
Subtract 12 from each side to solve for x

x + 12 -12 = 0 -12
Simplify

x = -12

x - 2 = 0
add 2 to both sides to solve for x

x -2 +2 = 0 + 2
Simplify

x = 2

So x = -12 or x = 2

I assume a negative number here is not the solution because we are getting a bigger area so it should be a positive number, we don’t use -12.

Check your work with +2

2 + 6 = 8
2 + 4 = 6

L x W = area
8 x 6 = 48

This is correct, we were looking for double the area of 24, so 48 was our goal, problem solved!

Answer 2

Explanation:

Area = 8 x 4 =24

Area doubled = 48

Let x be the amount we increase width and length to get area =48.

`(6+x)* (4+x)=48`

`(x+6)(x+4)=48`

`x^2+10x+24=48`

`x^2+10x-24=0`

`(x+12)(x-2)=0`

`\text{gives }x=-12,2`

But
`x=-12` is not a practical solution.

So
`x=2` is the required solution.

We must increase the length and width by 2.