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Mathematics 7 Online
OpenStudy (anonymous):

Brad can make 4 key chains in an hour. Velma can make only 3 key chains in an hour, but she already has 6 completed key chains. Explain to Brad how he can use a system of equations to determine when he will have the same number of key chains as Velma. Please explain each step too?

OpenStudy (freckles):

Let t represent the amount of hours.

OpenStudy (freckles):

Say molly can make 3 keys per hour than the number of keys per hour she can make is given by the expression 3t so can you make an expression for Brad?

OpenStudy (anonymous):

so it's the number of key chains multiplied by the amount of hours?

OpenStudy (freckles):

no. M(t)=3t because Molly can make 3 keys per hour for example when t=1 hour then M=3 this means she has made 3 keys in 1 hour if t=2 then M=3+3=6 this mean she has made 6 keys in 2 hours because she can make 3 keys in 1 hour

OpenStudy (freckles):

Say Velma has already made 2 keys and can make 5 keys an hour. So her function would be V(t)=2+5t because at 0 hours V=2 which this means before she began she had 2 completed keys at t=1 she would have V=7 this means she has had 7 completed keys after 1 hour

OpenStudy (freckles):

You can find when they have the same number of keys by setting M(t)=V(t) by solving the following for t: 3t=2+5t

OpenStudy (freckles):

So try making an expression for Brad and Velma.

OpenStudy (anonymous):

okay, one second while i do that.

OpenStudy (freckles):

Oops didn't realize you had a velma someone said something about velma so i thought she wasn't being used

OpenStudy (freckles):

my velma is different from your velma

OpenStudy (freckles):

Brad can make 4 key chains in an hour. This means in 1 hour he can make 4. In 2 hours he can make 4+4=2(4)=8. In 3 hours he can make 4+4+4=3(4)=12. So in t hours he can make t*4 or 4t is how it is normally written. B(t)=4t. Do you see how I came up with that?

OpenStudy (anonymous):

do i have it right so far? Let (t) represent the amount of hours. Let (V) represent Velma and (B) for Brad. Since Velma is able to make 3 key chains per hour, vt = 3t. Brad will be able find when he will have the same number of key chains by setting B(t)=V(t) Solve for (t). 4t = 6 + 3t

OpenStudy (freckles):

you did it !: )

OpenStudy (freckles):

B(t)=4t and V(t)=6+3t

OpenStudy (freckles):

4t=3t+6 try subtract 3t on both sides

OpenStudy (anonymous):

Solve for (t). 4t = 6 + 3t Subtract 3t on both sides. 4t - 3t = 6 + 3t - 3t Dive 1t by both sides. 1t = 6 t=6

OpenStudy (anonymous):

is that correct?

OpenStudy (freckles):

yeah! :) so in 6 hours they will have same amount of key chains made

OpenStudy (freckles):

np sorry for using you velma in my example

OpenStudy (freckles):

:p

OpenStudy (anonymous):

no worries

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