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Solving Rational Equations: 3y/3y+9 + 8y+16/2y+6 = y-4/y+3
- Thread starter Rujaxso
- Start date
This is basic algebra.
The kinds of rules that we have been discussing are generalizations about numbers. If you don't understand a mechanical rule, try it with numbers (generally not 0 or 1 because they have some special properties.
\(\displaystyle \dfrac{6 + 2}{2} = \dfrac{8}{2} = 4.\) Agree?
\(\displaystyle \dfrac{6 + 2}{2} = \dfrac{6 + 1}{1} = 7.\) WRONG, WRONG, WRONG.
\(\displaystyle \dfrac{6 + 2}{2} = \dfrac{2(3 + 1)}{2} = \dfrac{3 + 1}{1} = 4.\) Perfectly OK.
A rule of algebra is
\(\displaystyle \dfrac{ab + ac}{ad + ae} \equiv \dfrac{a(b + c)}{a(d + e)} \equiv \dfrac{b + c}{d + e}.\)
If you study foundations of mathematics, you can prove that rule. But if you want to understand why it works, do some examples.
The kinds of rules that we have been discussing are generalizations about numbers. If you don't understand a mechanical rule, try it with numbers (generally not 0 or 1 because they have some special properties.
\(\displaystyle \dfrac{6 + 2}{2} = \dfrac{8}{2} = 4.\) Agree?
\(\displaystyle \dfrac{6 + 2}{2} = \dfrac{6 + 1}{1} = 7.\) WRONG, WRONG, WRONG.
\(\displaystyle \dfrac{6 + 2}{2} = \dfrac{2(3 + 1)}{2} = \dfrac{3 + 1}{1} = 4.\) Perfectly OK.
A rule of algebra is
\(\displaystyle \dfrac{ab + ac}{ad + ae} \equiv \dfrac{a(b + c)}{a(d + e)} \equiv \dfrac{b + c}{d + e}.\)
If you study foundations of mathematics, you can prove that rule. But if you want to understand why it works, do some examples.
I have been saying that there is something off about this problem FOR EXACTLY THIS REASON.BTW i get y = -3 which creates 0 for one denominator making it NO SOLUTION
Interesting, Thanks Denis. That is an imagination for my figment.
No solution was the actual answer that MyMathLab wanted. (we use MyMathLab for the course I'm taking)
I already knew the destination I just wanted to know how to get there.
I like how everybody pitches in on this forum. You get a lot of different but valuable ways of approaching things.
No solution was the actual answer that MyMathLab wanted. (we use MyMathLab for the course I'm taking)
I already knew the destination I just wanted to know how to get there.
I like how everybody pitches in on this forum. You get a lot of different but valuable ways of approaching things.
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