Please explain how to do this kind of problem
The length of a rectangle is 6 cm less than twice its width. Find the dimensions of this rectangle if its area is 108 cm^2.
The first step is always to assign letters to the variables. We can call length L and width W.
Since all units are in cm, unit conversion is unnecessary.
Next, set up some equations. L=2W-6 because of the first sentence of the problem. Length times width equals area, so L*W=108.
Now that there are two equations and two variables, the next step is to solve the system. I'm going to solve it by plugging in the right side of the first equation in for L in the second equation. This gives me (2W-6)*W=108. Distributing the W gives me 2W^2-6W=108, or W^2-3W-54=0. Using the quadratic formula, I get that W=9. Using the first equation, L=2*9-6=12. So the length is 12 and the width is 9.
Start with the basics: A=lw
Use your given information to substitute the variables: A=108, w=?, l=2w -6 <(twice the width & 6 less)
Now you have: A=lw -> 108=w(2w -6)
Distribute the w : 108=2w2 -6w
Simplify by factoring out 2: 108=2(w2 -3w) , Divide by 2: 54=w2 -3w
Set the equation equal to 0: 0=w2 -3w -54
You can now factor! 0=(w -9)(w +6)
Therefore, w= 9 or -6. Width can not be negative, so w=9. Substitute w=9 into your length equation to find the length. l=2w -6 -> l=2(9)-6=12
So, the length of the rectangle is 12 cm, as the width is 9 cm. (You may check your answer by multiplying 12 and 9 to see if you get an Area of 108 cm2)