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unpack "L= 3 yds less than triple the W"
Length = 3 yards less than triple the Width
"triple with width W" would be written as \(3\times W\) or \(3W\)
3 yards less than triple the width: \(3W-3\)
so far so good?
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 \(300-3=297\)
on how to figure out the Length
ok lets try this: we are forgetting everything but the line "L= 3 yds less than triple the W"
only focusing on that one suppose i say the width is 50 what is the length?
if it is not clear, let me know and we can walk through it
L = 147??
and suppose the width is 25 what is the length?
L = 72
k good now we know what we are doing how did you get your answers?
oopss forgetting ft
i don't see the mention of feet in the problem so lets assume the units are yards, and ignore feet
3(25) - 3 = 3*25 = 75 - 3 = 72
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
ok so that part is done now how about this part " P= 450"
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)
actually . no
wouldn't it be p =l * w
so since we have the Perimeter wouldn't we divide the L and W
to get the Dimensions
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
oooppps wouldn't we divide to get them?
so no, it is not \(P=LW\) but rather \(A=LW\) we need an expression for the perimeter
measure of the perimeter is the total length in this case it would be \(10+10+15+15=50\) or i could write (suggestively) \[2\times 10+2\times 15=50\]
now with this picture |dw:1377786057826:dw| what is the perimeter?
you have \(P=W+W+L+L\) which we would write as \[P=2L+2W\]
good so far?
ok now we know that \(P=450\) i.e. we know \[2L+2W+450\]
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 \(L\)
typo there sorry equation is \[2(3W-3)+2W=450\]
final job, which will take a few steps, is to solve this for \(W\)
should we work through the steps?
w = 8
is that correct??
that is not what i get lets work through the steps
\[2(3W-3)+2W=450\] first step is to multiply on the left using the distributive property \[6W-6+2W=450\] so far so good?
ok, that is what I got
second step is to combine like terms on the left, i.e. add \(6w+2W\) to get \[8W-6=450\]
ok, that is what I got
add \(6\) to both sides and get \[8W=456\]
then I added 6 to 450
k good last step?
ooppss ans. should be 57
ok good, that is the width, right now we also need the length
I miss read my calculations. I switched the numbers around
ok did you get the length?
so, we would put 3(57) -3 = L
L = 168
that is what i get as well
ok, so the actual set up for the problem would be: 3w-3 = L; 2(3w -3) + 2w = 450; 6w - 6
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??
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
my daughter has to show her work, and how the equation should be set up.
show her work to whom?
i guess so i have no way of knowing what her teacher wants
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
i suppose she could ask her teacher what kind of work she wants to see
on line class?
algebra is hard to learn on line you have to do it all yourself, and it requires a lot of maturity good luck
thanks. so far she is doing good. just this one question through her
Thank you so much for you help.