the perimeter of a rectangle is twice the sum of its length and its width. the perimeter is 40 meters and its length is 2 meters more than twice its width. what is its length?
Perimeter = 2*(length +width) Can you write a relationship between the length and the width? length = (how are the length and width related)
idk, i could never understand this question
Okay, this is just like the roses and daisies problem. If the width is 1, what is the length? The length is 2 meters more than twice the width so 2 + 2(1) If the width is 2, what is the length? If the width is w, what is the length?
Width Length 1 2 + 2(1) 2 2 + 2(2) 3 2 + 2(3) w ??????
Do you see a pattern?
kinda, but i dont kno where i wud start
Well you need to write a general rule that relates the length and the width. Since we know the circumference is 2(l + w) we need to somehow write one of those variables in terms of the other so we can solve for one of the variables. If we have two unknowns, we need two equations relating those unknowns. (1) P = 2(l + w) (2) l = ???? The length equation has to do with the length being 2 more than twice the width.
Not circumference.. perimeter. Sorry. :)
So again, what pattern do you see when you look at the width and length in this table? Width Length 1 2 + 2(1) 2 2 + 2(2) 3 2 + 2(3) w ??????
for every 2+2 you multiply going up by 1?
Right, so if you wrote a general rule relating the two.. l = 2 + 2w So whenever you have a width, you multiply it by 2 and then add 2.
Now I have two equations with two unknowns. (1) P = 2(l + w) (2) l = 2 + 2w
Take what l equals in equation (2) and substitute it into equation (1). This will make an equation for the perimeter in terms of ONLY the width, so we can then solve for the width.
P = 2l + 2w P = 2(2 + 2w) + 2w I just substituted in 2 + 2w for l in equation (1)
when a number is like this (7) it means multiply that number, right?
It sure does!
So (7)(7) = 7 * 7 = 49
ahhhhhhhhhhh, much clearer now
So you can simplify and get P = 4 + 4w + 2w P = 4 + 6w We are given a perimeter of 40. 40 = 4 + 6w You can now solve for w. Once you have w you can substitute it into equation (2) and get the length.
so could i just do 2+2(1) 2+2(2) 2+2(3) 2+2(4) until i reach 40??
Well, no. What you would have to do is this. width length perimeter 1 2+2(1) 2(1 + 4)
But yes, that would work; however, it is not the most efficient method. Using algebra and a system of two equations and two unknowns makes it more efficient.
ok
If you do it that way, you will see the pattern. There's really no one correct way to solve a problem. As long as you get the correct solution, you are correct. :)
you have been a big help dude, thanks.
You're welcome.
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