- #1

chwala

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- see attached.

My interest is on the highlighted part only...my understanding is that one should use simultaneous equation... unless there is another way hence my post query.

In my working i have;

##y=\dfrac{2ξ+η}{10}## and ##x=\dfrac{2η-ξ}{10}## giving us;

##x+3y=\dfrac{2η-ξ+6ξ+3η}{10}=\dfrac{5ξ+5η}{10}=\dfrac{ξ+η}{2}## cheers guys.

In my working i have;

##y=\dfrac{2ξ+η}{10}## and ##x=\dfrac{2η-ξ}{10}## giving us;

##x+3y=\dfrac{2η-ξ+6ξ+3η}{10}=\dfrac{5ξ+5η}{10}=\dfrac{ξ+η}{2}## cheers guys.