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Solving Rational Equations..PLEASE HELP
- Thread starter swynn
- Start date
Hello, swynn
Please check the exact form of the problem . . .
Since very few have typed mathematical expressions clearly,
. . I can guess what the form is supposed to be.
\(\displaystyle \text{It probably looks like this: }\;\frac{2}{y+3} + \frac{2y+3}{y-1} \:=\:\frac{6y-5}{x^2+2x-3}\)
Then there may be two typos . . . at least.
\(\displaystyle \text{If those }x\text{'s are replaced with }y\text{'s, we have: }\;\frac{2}{y+3} + \frac{2y+3}{y-1} \:=\:\frac{6y-5}{y^2+2y-3}\)
. . \(\displaystyle \text{which is a traditional problem: }\;\frac{2}{y+3} + \frac{2y+3}{y-1} \:=\:\frac{6y-5}{(y+3)(y-1)}\)
\(\displaystyle \text{The equation simplifies to: }\;2y^2 + 5y + 12\:=\:0\)
. . But this quadratic has no real roots.
Please check the exact form of the problem . . .
Solve: (2/y+3)+(2y+3/y-1)=(6y-5/x2+2x-3)
Since very few have typed mathematical expressions clearly,
. . I can guess what the form is supposed to be.
\(\displaystyle \text{It probably looks like this: }\;\frac{2}{y+3} + \frac{2y+3}{y-1} \:=\:\frac{6y-5}{x^2+2x-3}\)
Then there may be two typos . . . at least.
\(\displaystyle \text{If those }x\text{'s are replaced with }y\text{'s, we have: }\;\frac{2}{y+3} + \frac{2y+3}{y-1} \:=\:\frac{6y-5}{y^2+2y-3}\)
. . \(\displaystyle \text{which is a traditional problem: }\;\frac{2}{y+3} + \frac{2y+3}{y-1} \:=\:\frac{6y-5}{(y+3)(y-1)}\)
\(\displaystyle \text{The equation simplifies to: }\;2y^2 + 5y + 12\:=\:0\)
. . But this quadratic has no real roots.
(2/y+3)+(2y+3/y-1)=(6y-5/x2+2x-3) means \(\displaystyle \frac{2}{y}+3+2y+\frac{3}{y}-1=6y-\frac{5}{x}\cdot2 +2x-3\). I doubt that is what you want. You might use grouping symbols in conjunction with the order of operations convention to clarify your intent. Also, if you want to indicate an exponent, it is so designated with a "^" sign. For instance, a^2 means \(\displaystyle a^2\).
swynn said:I understand how to solve rational equations with 1 variable, but this one is more difficult and I can't figure it out
Solve: (2/y+3)+(2y+3/y-1)=(6y-5/x2+2x-3). Anyone who knows, I would appreciate some help
You've got ONE equation with TWO variables.
You can't solve this equation to get a single value for x and y....
Are you sure you typed the problem correctly?
A system of equations in two variables requires two equations involving these variables to get a unique solution (provided that one solution exists.)