Sorry for being so late... You probably already finished it but.
The sum of the length and width is 19 inches. The length is one less than twice the width. Find the length and width of the rectangle. Sum is add so the first step is to write the equation. L + W = 19 It says the length is equivalent to something. You can use the substitution method. You should change the equation to standard form and solve the equation using the elimination method. Sorry I didn't come to the Math counts meeting... I forgot between the time the intercom went on and the time I started eating lunch... Nick H. Period 5
So #33 says the sum of the length and width of a rectangle is 19 in. so x+y=19 and the length is one less than twice the width so x=2y-1 which is also -x+2y=1 x+y=19 + -x+2y=1 3y=120 y=20/3 then solve for x
For 33: + l w The sum of the length and width of a = 19 l = rectangle is 19. The length is one - 2* w less than twice the width. (I would do something for than, but I can't find the right arrow.)
The first equation would be: l + w = 19
Your second equation would be: l = 2w - 1 You probably want w and l to equal something so: l = 2w - 1 -2w -2w l - 2w = -1 Now multiply -1 to be able able to get eliminate one of the variables when combining. -1(l - 2w)=-1(-1) -l + 2w = 1 Now combine the two equations. l + w = 19 -l +2w = 1 3w = 20 1/3(3w)= 1/3(20) w = 20/3 Now plug this in for w in the other equation. -l + 2w = 1 -l + 2(20/3)= 1 -l + 40/3 = 1 -40/3 -40/3 -l = -37/3 -(-l) = -(-37/3) l = 37/3 So you should have your length as 37/3 inches and width as 20/3 inches. Mr. Gerson, will 1st period get their tests back soon? Saloni B. Period 1
5 comments:
can somebody help me with #33?
HAnnah P
per.5
I don't get #33, can someone please help me!? Thanks!
Linda L
Per.2
Sorry for being so late... You probably already finished it but.
The sum of the length and width is 19 inches. The length is one less than twice the width. Find the length and width of the rectangle.
Sum is add so the first step is to write the equation.
L + W = 19
It says the length is equivalent to something. You can use the substitution method. You should change the equation to standard form and solve the equation using the elimination method.
Sorry I didn't come to the Math counts meeting... I forgot between the time the intercom went on and the time I started eating lunch...
Nick H.
Period 5
So #33 says the sum of the length and width of a rectangle is 19 in.
so
x+y=19
and the length is one less than twice the width
so
x=2y-1
which is also -x+2y=1
x+y=19
+ -x+2y=1
3y=120
y=20/3
then solve for x
Jay M Per4
For 33:
+ l w
The sum of the length and width of a
= 19 l =
rectangle is 19. The length is one
- 2* w
less than twice the width.
(I would do something for than, but I can't find the right arrow.)
The first equation would be:
l + w = 19
Your second equation would be:
l = 2w - 1
You probably want w and l to equal something so:
l = 2w - 1
-2w -2w
l - 2w = -1
Now multiply -1 to be able able to get eliminate one of the variables when combining.
-1(l - 2w)=-1(-1)
-l + 2w = 1
Now combine the two equations.
l + w = 19
-l +2w = 1
3w = 20
1/3(3w)= 1/3(20)
w = 20/3
Now plug this in for w in the other equation.
-l + 2w = 1
-l + 2(20/3)= 1
-l + 40/3 = 1
-40/3 -40/3
-l = -37/3
-(-l) = -(-37/3)
l = 37/3
So you should have your length as 37/3 inches and width as 20/3 inches.
Mr. Gerson, will 1st period get their tests back soon?
Saloni B.
Period 1
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