Question

Translate this problem to an equation and solve.
The width of a rectangle is 6 inches less than twice its length. Find the dimension of the rectangle if its area is 108 square inches. @johnweldon1993

Answer 1

let length be l
width= 2 * l - 6
let width be w
l * w = 108

I think you can solve this now

Answer 2

Just as @sampatho2 has shown

W = 2L - 6
A = L x W

We know area

108 = L x W

We know W = 2L - 6 so substitute that in for 'W'

108 = L x (2L - 6)

can you take it from here...?

Answer 3

na

Answer 4

where are you facing the problem. You just need to solve the equation for l

Answer 5

but i still dont get it

Answer 6

@johnweldon1993

Answer 7

Let x be the length and y the width.
Then :
\[2x-6=y\\xy=108\]
So from the 2nd equation we have :
\[x=\frac{108}{y}\]
And from the 1st equation :
\[2\frac{108}{y}-6=y\]
We multiply by y to get :
\[216-6y=y^2\]
And then :
\[y^2+6y-216=0\]
so :
\[\Delta=36-4\times(-216)=900\]
So :
\[y=\frac{-6+\sqrt{900}}{2}=\frac{-6+30}{2}=12 \text{ inches}\]
And then :
\[x=\frac{108}y=\frac{108}{12}=9 \text{ inches}\]

Answer 8

i dont get it still @johnweldon1993 @Noura11

Answer 9

Tell me what didn't understand in my solution and I will try to explain it to you ?

Answer 10

the whole thing tht u did its suppost to be like this type the answer in this from:6*7

Answer 11

Can you explain more , please ?

Answer 12

i got it

Answer 13

Glad to hear that !