## shortie1515 2 years ago 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

1. sampatho2

let length be l width= 2 * l - 6 let width be w l * w = 108 I think you can solve this now

2. johnweldon1993

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...?

3. shortie1515

na

4. sampatho2

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

5. shortie1515

but i still dont get it

6. shortie1515

@johnweldon1993

7. Noura11

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 : $x=\frac{108}y=\frac{108}{12}=9 \text{ inches}$

8. shortie1515

i dont get it still @johnweldon1993 @Noura11

9. Noura11

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

10. shortie1515

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

11. Noura11

Can you explain more , please ?

12. shortie1515

i got it

13. Noura11