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Mathematics 9 Online
OpenStudy (bbb911):

The numbers on Sam's three raffle tickets are consecutive integers. The sum of the numbers is 126. What are the numbers?

OpenStudy (anonymous):

x + (x-1) + (x-2) = 126 x, plus the number before x, plus the number behind that is what i have written

OpenStudy (anonymous):

3x - 3 = 126 simplified down can you solve from here?

OpenStudy (bbb911):

im confused

OpenStudy (anonymous):

well, you want three consecutive numbers that add up to 126, yes?

OpenStudy (bbb911):

yes

OpenStudy (anonymous):

so we have 3 consecutive numbers = 126

OpenStudy (anonymous):

how do we write "3 consecutive numbers" with math? well, there are a couple different ways, but i chose: x + (x-1) + (x-2)

OpenStudy (anonymous):

x, some random number, plus the number before x, plus the number before that

OpenStudy (anonymous):

for instance, if x = 5, then we have x + (x-1) + (x-2) 5 + 4 + 3 3 consecutive numbers does it make sense why i did that now?

OpenStudy (bbb911):

and do you have to guess any number tht will sum up to 126?

OpenStudy (anonymous):

no, here's where the clever bit comes in: re-write "3 consecutive numbers = 126" as x + (x-1) + (x-2) = 126

OpenStudy (jdoe0001):

consecutive numbers, meaning one is bigger than another by 1 3, 4, 5 17 , 18 , 19 and so on so, if a number is "whatever" whatever, whatever + 1, (whatever + 1)+1

OpenStudy (anonymous):

now you can solve for x, and find your 3 consecutive numbers

OpenStudy (bbb911):

so would the number that i will plug in be 3 5 and 4?

OpenStudy (anonymous):

no, okay let's look at it step-by-step: x + (x - 1) + (x-2) how can we simplify this? what is the first step to collect all the terms?

OpenStudy (anonymous):

we want to add all the x's together, and add everything that isn't an x together

OpenStudy (jdoe0001):

you'd get 3 numbers that add up to 126, we dunno which they're, we do know that whatever, whatever + 1, (whatever + 1)+1 = 126

OpenStudy (bbb911):

but how if there aren't any other numbers, or do we start with the brackets first?

OpenStudy (anonymous):

well, it can be re-written like this x + x - 1 + x - 2 = 126 i just wrote it with () to make it clearer. do you know how to simplify that down now?

OpenStudy (bbb911):

ok one sec

OpenStudy (jdoe0001):

there would be only 1 set of 3 that'd add up to that

OpenStudy (bbb911):

i got 3x-3

OpenStudy (anonymous):

right, remember it's equal to 126 though 3x - 3 = 126 can you solve for x now?

OpenStudy (bbb911):

yes one sec

OpenStudy (bbb911):

is it x=41

OpenStudy (anonymous):

almost, you subtracted 3 instead of adding it though 3x - 3 = 126 3x - 3 + 3 = 126 + 3 3x = 129 x = 43

OpenStudy (anonymous):

okay, if x = 43, do you know how you would find the other 2 side lengths now?

OpenStudy (bbb911):

ooo my mistake lemme do it again

OpenStudy (bbb911):

kk got 43 now you said how would i find the other 2 side lengths right?

OpenStudy (anonymous):

yep, do you think you can do it? remember, we know now that x = 43. 43 is one of the three numbers you need remember we wrote the other two as (x - 1) and (x-2)

OpenStudy (anonymous):

or, look at it another way: either 43 + 44 + 45 is equal to 126, or 41 + 42 + 43 is cause that's the only trio of numbers you can have

OpenStudy (bbb911):

and do u guess the other 2 numbers?

OpenStudy (anonymous):

noo, remember we had x + (x - 1) + (x - 2) = 126? well, we know x is 43 now, right? so plug x back in and see what the other 2 numbers are

OpenStudy (anonymous):

aka write it as 43 + (43 - 1) + (43 - 2) = 126 the 3 numbers should equal 126

OpenStudy (bbb911):

oh we plug in 43 in back right ?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

basically: 1. create an equation with a variable 2. solve for the variable 3. use the now-solved variable to find what you need

OpenStudy (bbb911):

kk one sec

OpenStudy (bbb911):

kk got it thanks soo much :)

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