Solve and explain using a chart and algebra
Jack plants two varieties of Bean stalks. Variety A was 12 inches tall when planted and grow 6 inches per day. Variety B was planted the same day, was 30 inches tall and grows 3 inches each day. When will the bean stalks be the same height?
Please help me explain this and solve it by using a chart and algebra (no need for graph)
let's use x to represent number of days "Variety A was 12 inches tall when planted and grow 6 inches per day." growth rate can be expressed as 6 * x, then add on the 12 inches the plant already had to begin with, to get height of A = 6x + 12 repeat this process with plant B then set the equations equal to each other to get the time when the beanstalks are the same height
They both have the same height is the 6th day with the same height of 48 inches but im not sure how to put this onto a table
Y = 6x + 12 Y = 3x + 30
@Vocaloid
oh, a table you just need to make two columns, 1 for the day and 1 for the height |dw:1541027874282:dw|
*three columns, sorry
then to fill out the first row, you'd calculate the heights of the plants on day 1, fill them into the table, then keep going until both A and B have the same height, then you can stop
do i do a table for each variety?
you can combine them into 1 table so the first column is days, second column is height of plant A, third column is height of plant B
ohh
for some reason they both dont meet on day 6... did i do something wrong
@Vocaloid
two things, the table should technically start at 0 since on day 0 the plants were planted, your calculations are correct, you need to keep going, the heights should be equal at 48 not 42
OH
yeah, in order to get the right value the heights need to be the same on the same day (same row)
so for day 1, id put both to heigh 0?
"Variety A was 12 inches tall when planted and grow 6 inches per day. Variety B was planted the same day, was 30 inches tall and grows 3 inches each day" so the height on day 0 (planting day) is 12 for A and 30 for B
it still doesnt look right
|dw:1541030140786:dw|
Thank u... that makes much more sense.... Ive done the graph part and then we did the table part and now i have to do the algebra part.... how do u think i could start this?
algebraically? you have two equations Y = 6x + 12 Y = 3x + 30 we can set them equal to each other and solve for x 6x + 12 = 3x + 30 know where to go from here?
can u show me what to do next... im rly bad at math lol i get rly confused
cuz i ended up with this @Vocaloid
hm not quite you should have ended up with x = 6 as per our table
6x + 12 = 3x + 30 try subtracting 3x from both sides
wouldnt u divide 3 from both sides
sure, you can do that if you'd like
hows this
the x's don't just disappear also you have to apply the /3 to every term
( 6x + 12 ) / 3 = 2x + 4 ( 3x + 30) / 3 = x + 10
ohhh
there we go
good then you'd subtract x from both sides
uhh does subtracting x do anything
at this point you have 2x + 4 = x + 10 subtracting x from both sides will eliminate x from the right side, thus making it so that x is only on one side
oh ok
2x - x is x, not 2x so at this point you should have x + 4 = 10 then you just need to subtract 4 from both sides, that should end up w/ x = 6
thanks.... a lot.... B)
What about the height? (48), where would i put this into an algebraic equation?
solving for x gave us the number of days to reach equal height so plugging x into either Y = 6x + 12 or Y = 3x + 30 should give 48 as the result both ways
oh ok B)
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