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.

- anonymous

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- anonymous

unpack
"L= 3 yds less than triple the W"

- anonymous

Length = 3 yards less than triple the Width

- anonymous

"triple with width W" would be written as \(3\times W\) or \(3W\)

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## More answers

- anonymous

3 yards less than triple the width: \(3W-3\)

- anonymous

so far so good?

- 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

still lost

- anonymous

on how to figure out the Length

- anonymous

ok lets try this:
we are forgetting everything but the line "L= 3 yds less than triple the W"

- anonymous

only focusing on that one
suppose i say the width is 50
what is the length?

- anonymous

if it is not clear, let me know and we can walk through it

- anonymous

L = 147??

- anonymous

exactly

- anonymous

and suppose the width is 25
what is the length?

- anonymous

L = 72

- anonymous

k good
now we know what we are doing
how did you get your answers?

- anonymous

oopss forgetting ft

- anonymous

i don't see the mention of feet in the problem
so lets assume the units are yards, and ignore feet

- anonymous

3(25) - 3 = 3*25 = 75 - 3 = 72

- 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

right

- anonymous

ok so that part is done
now how about this part
" P= 450"

- anonymous

kk

- 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

actually . no

- anonymous

|dw:1377785738464:dw|

- anonymous

wouldn't it be p =l * w

- anonymous

so since we have the Perimeter wouldn't we divide the L and W

- anonymous

to get the Dimensions

- 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

oooppps wouldn't we divide to get them?

- anonymous

so no, it is not \(P=LW\) but rather \(A=LW\) we need an expression for the perimeter

- anonymous

oh ok

- anonymous

|dw:1377785947293:dw|

- 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

now with this picture |dw:1377786057826:dw|
what is the perimeter?

- anonymous

you have \(P=W+W+L+L\) which we would write as
\[P=2L+2W\]

- anonymous

good so far?

- anonymous

yeah

- anonymous

ok now we know that \(P=450\) i.e. we know
\[2L+2W+450\]

- anonymous

right

- 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

typo there sorry
equation is
\[2(3W-3)+2W=450\]

- anonymous

final job, which will take a few steps, is to solve this for \(W\)

- anonymous

should we work through the steps?

- anonymous

w = 8

- anonymous

is that correct??

- anonymous

that is not what i get
lets work through the steps

- anonymous

k

- 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

ok, that is what I got

- anonymous

second step is to combine like terms on the left, i.e. add \(6w+2W\) to get
\[8W-6=450\]

- anonymous

ok, that is what I got

- anonymous

add \(6\) to both sides and get
\[8W=456\]

- anonymous

then I added 6 to 450

- anonymous

k good
last step?

- anonymous

ooppss ans. should be 57

- anonymous

ok good, that is the width, right
now we also need the length

- anonymous

I miss read my calculations. I switched the numbers around

- anonymous

ok did you get the length?

- anonymous

so, we would put 3(57) -3 = L

- anonymous

right

- anonymous

L = 168

- anonymous

that is what i get as well

- anonymous

ok, so the actual set up for the problem would be: 3w-3 = L; 2(3w -3) + 2w = 450;
6w - 6

- 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

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

my daughter has to show her work, and how the equation should be set up.

- anonymous

show her work to whom?

- anonymous

her teacher

- anonymous

i guess so
i have no way of knowing what her teacher wants

- 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

i suppose she could ask her teacher what kind of work she wants to see

- anonymous

on line class?

- anonymous

yes

- 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

thanks. so far she is doing good. just this one question through her

- anonymous

Thank you so much for you help.

- anonymous

yw

- anonymous

bye

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