Actually for me this is very simple, I already had the actual answer in a minute, it's just that I want him to understand but I was too tired to work-out the problem quickly last night.
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Engineering mathematics is much more complex algebraically.
---------- Post added at 07:53 AM ---------- Previous post was at 07:47 AM ----------
It made me laugh at first, even in middle-school I could have solved this easily.
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I'm curious to know why when I transform the equations into:
2(x- 30) = y + 30; 3(y-50) = x +50
to
2x-y=90; 3y-x=100
The fucking result is not RIGHT???!!!
Originally Posted by user590177
I'm curious to know why when I transform the equations into:
2(x- 30) = y + 30; 3(y-50) = x +50
to
2x-y=90; 3y-x=100
The fucking result is not RIGHT???!!!
The whole point is to get one variable alone by itself on one side, and then you have what that variable is equal to, in-terms of the other variable. Don't put two variables on the same side, or you won't be able to solve it.
when you transcribe to
2x-y=90; 3y-x=100: 2x-y=90; neither the x nor y value is known, that makes two unknowns on one side of the equation, we can't solve the problem like this.
We can however find out what y is equal to in terms of 'x', when we get 'y' by itself on the right side, 2(x- 30) = y + 30 -> 2x-60-30 = y: now we have a way to express what 'y' is equal to in-terms of 'x', 'y' is 2x - 90, our second expression is 3(y-50) = x +50, because we now know y = 2x - 90: we can substitute "2x - 90" wherever 'y' appears and by doing so we'll make it so we're only dealing with one variable, 'x', with their being only one unknown value the expression will be solvable
y
3[(2x-90) - 50] = x +50: solve for x, and then use the 'x' value to solve for y
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There are of course different approaches to take.
Originally Posted by Arhk
