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Mathematics 15 Online
OpenStudy (anonymous):

x+1=y/4 and 4x=y/5

OpenStudy (kc_kennylau):

what's \(s\)? :)

OpenStudy (kittiwitti1):

He's probably asking us to find s...

OpenStudy (kc_kennylau):

But \(n\) equations \(n\) unknowns...

OpenStudy (kittiwitti1):

He just changed it to 5...

OpenStudy (kc_kennylau):

LOL wait how do you edit questions

OpenStudy (kittiwitti1):

Click edit? c:

OpenStudy (anonymous):

x=1/4 y=5

OpenStudy (kc_kennylau):

lol ok @gioson5 There are 4 approaches: 1. Make \(x\) the subject of \((1)\) and substitute it into \((2)\). 2. Make \(y\) the subject of \((1)\) and substitute it into \((2)\). 3. Make \(x\) the subject of \((2)\) and substitute it into \((1)\). 4. Make \(y\) the subject of \((2)\) and substitute it into \((1)\).

OpenStudy (kittiwitti1):

How the heck does x=1/4? O_O WOAH. @kc_kennylau

OpenStudy (kc_kennylau):

lol @kittiwitti1 @Takuno don't just give out answers please :)

OpenStudy (kittiwitti1):

\[x=\frac{y}{4}-1,~y=5(4x)\]you decide which one to use in substitution

OpenStudy (anonymous):

ok, because equation one equals y/4 and equation two equals y/5 so the ratio of (equation one)/(equation two) = 5/4. therefore look at the left hand side of both of the equations, (x+1)/(4x) = 5/4 and then you can use cross multiplication.

OpenStudy (kc_kennylau):

As I said there are 4 approaches lol @kittiwitti1

OpenStudy (kittiwitti1):

Well I don't want to consider all of 'em lol @kc_kennylau

OpenStudy (kittiwitti1):

gotta go bye D:

OpenStudy (kc_kennylau):

bye :/

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