anonymous
  • anonymous
My daughter is in Algebra 2 for home schooling, and I am her learning coach. She has a word problem that is this: A playing field: P= 450, L= 3 yds less than triple the W. What are the dimensions of the playing field. Question is How do I explain to her how to set this up. It has been 5 years since I have done this, and I am brain dead.
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
unpack "L= 3 yds less than triple the W"
anonymous
  • anonymous
Length = 3 yards less than triple the Width
anonymous
  • anonymous
"triple with width W" would be written as \(3\times W\) or \(3W\)

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anonymous
  • anonymous
3 yards less than triple the width: \(3W-3\)
anonymous
  • anonymous
so far so good?
anonymous
  • anonymous
for example, if the width was say \(100\) (which it is not, just an example) then the length would be 3 less than triple 100 or \(300-3=297\)
anonymous
  • anonymous
still lost
anonymous
  • anonymous
on how to figure out the Length
anonymous
  • anonymous
ok lets try this: we are forgetting everything but the line "L= 3 yds less than triple the W"
anonymous
  • anonymous
only focusing on that one suppose i say the width is 50 what is the length?
anonymous
  • anonymous
if it is not clear, let me know and we can walk through it
anonymous
  • anonymous
L = 147??
anonymous
  • anonymous
exactly
anonymous
  • anonymous
and suppose the width is 25 what is the length?
anonymous
  • anonymous
L = 72
anonymous
  • anonymous
k good now we know what we are doing how did you get your answers?
anonymous
  • anonymous
oopss forgetting ft
anonymous
  • anonymous
i don't see the mention of feet in the problem so lets assume the units are yards, and ignore feet
anonymous
  • anonymous
3(25) - 3 = 3*25 = 75 - 3 = 72
anonymous
  • anonymous
k good so if you know the width, you know the length how do you find it? multiply by 3, then subtract 3 if we replace \(25\) by the variable \(W\) then the equation tells you \[L=3W-3\] which is what "L= 3 yds less than triple the W" says in math
anonymous
  • anonymous
right
anonymous
  • anonymous
ok so that part is done now how about this part " P= 450"
anonymous
  • anonymous
kk
anonymous
  • anonymous
we need an expression for the perimeter in terms of the width \(W\) and length \(L\) do you know it? ("no" is a fine answer, just asking)
anonymous
  • anonymous
actually . no
anonymous
  • anonymous
|dw:1377785738464:dw|
anonymous
  • anonymous
wouldn't it be p =l * w
anonymous
  • anonymous
so since we have the Perimeter wouldn't we divide the L and W
anonymous
  • anonymous
to get the Dimensions
anonymous
  • anonymous
that is an expression for the Area and as you see by your multiplication it comes in square units (2 dimension) if the width is 4 and the length is 3 yards, the area is \(3\times 4=12\) square yards perimeter is a length, one dimension, a number of yards in this case
anonymous
  • anonymous
oooppps wouldn't we divide to get them?
anonymous
  • anonymous
so no, it is not \(P=LW\) but rather \(A=LW\) we need an expression for the perimeter
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
|dw:1377785947293:dw|
anonymous
  • anonymous
measure of the perimeter is the total length in this case it would be \(10+10+15+15=50\) or i could write (suggestively) \[2\times 10+2\times 15=50\]
anonymous
  • anonymous
now with this picture |dw:1377786057826:dw| what is the perimeter?
anonymous
  • anonymous
you have \(P=W+W+L+L\) which we would write as \[P=2L+2W\]
anonymous
  • anonymous
good so far?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
ok now we know that \(P=450\) i.e. we know \[2L+2W+450\]
anonymous
  • anonymous
right
anonymous
  • anonymous
and we also know that \(L=3W-3\) so we can replace the \(L\) in \(2L+2W=450\) by \(2W-3\) with judicious use of parenthese and get the equation \[2(2W-3)+2W=450\] which we can then solve for \(L\) because now there is only one variable used \(L\)
anonymous
  • anonymous
typo there sorry equation is \[2(3W-3)+2W=450\]
anonymous
  • anonymous
final job, which will take a few steps, is to solve this for \(W\)
anonymous
  • anonymous
should we work through the steps?
anonymous
  • anonymous
w = 8
anonymous
  • anonymous
is that correct??
anonymous
  • anonymous
that is not what i get lets work through the steps
anonymous
  • anonymous
k
anonymous
  • anonymous
\[2(3W-3)+2W=450\] first step is to multiply on the left using the distributive property \[6W-6+2W=450\] so far so good?
anonymous
  • anonymous
ok, that is what I got
anonymous
  • anonymous
second step is to combine like terms on the left, i.e. add \(6w+2W\) to get \[8W-6=450\]
anonymous
  • anonymous
ok, that is what I got
anonymous
  • anonymous
add \(6\) to both sides and get \[8W=456\]
anonymous
  • anonymous
then I added 6 to 450
anonymous
  • anonymous
k good last step?
anonymous
  • anonymous
ooppss ans. should be 57
anonymous
  • anonymous
ok good, that is the width, right now we also need the length
anonymous
  • anonymous
I miss read my calculations. I switched the numbers around
anonymous
  • anonymous
ok did you get the length?
anonymous
  • anonymous
so, we would put 3(57) -3 = L
anonymous
  • anonymous
right
anonymous
  • anonymous
L = 168
anonymous
  • anonymous
that is what i get as well
anonymous
  • anonymous
ok, so the actual set up for the problem would be: 3w-3 = L; 2(3w -3) + 2w = 450; 6w - 6
anonymous
  • anonymous
6w - 6 + 2w = 450; 8w = 456; 456/8 = 57 = W = 57; 3(57) -3 = L --- L= 168. Is that all my daughter needs to do to show her work for this type of problem??
anonymous
  • anonymous
wow ok i am not sure what your sentence "Is that all my daughter needs to do to show her work" means lets go back and see what all we need first we need to know how to write "L is 3 yards less than triple the width" in math then we need to know the formula for a perimeter also we need to know how to substitute \(3W-3\) for \(L\) in \[2L+2W=450\] and finally we need to know how to solve \(2(3W-3)+2W=450\) for \(W\) the last part, finding \(3\times 57-3\) is not much
anonymous
  • anonymous
my daughter has to show her work, and how the equation should be set up.
anonymous
  • anonymous
show her work to whom?
anonymous
  • anonymous
her teacher
anonymous
  • anonymous
i guess so i have no way of knowing what her teacher wants
anonymous
  • anonymous
is sitting her with me and she did all the calculations except for the last one where I came up with 8 instead of 57
anonymous
  • anonymous
i suppose she could ask her teacher what kind of work she wants to see
anonymous
  • anonymous
on line class?
anonymous
  • anonymous
yes
anonymous
  • anonymous
algebra is hard to learn on line you have to do it all yourself, and it requires a lot of maturity good luck
anonymous
  • anonymous
thanks. so far she is doing good. just this one question through her
anonymous
  • anonymous
Thank you so much for you help.
anonymous
  • anonymous
yw
anonymous
  • anonymous
bye

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