Please help me with this equations:
1.) 13/x^2-9x+14 = 5/x-3 + 3x-7/x^2+2x-15
2.) 3x/x+1 - 1/3 = 5/x+1
3.) 11/x+5 - 5/x-3 = 3x-7/x^2+2x-15
4.) 17/y + 13/y+2 = 14/y^2-12y
5.) y/y-8 - 48/y^2-64 = y/y+8
thank you for your kind consideraton :-)
Please help me with this equations:
1.) 13/x^2-9x+14 = 5/x-3 + 3x-7/x^2+2x-15
2.) 3x/x+1 - 1/3 = 5/x+1
3.) 11/x+5 - 5/x-3 = 3x-7/x^2+2x-15
4.) 17/y + 13/y+2 = 14/y^2-12y
5.) y/y-8 - 48/y^2-64 = y/y+8
thank you for your kind consideraton :-)
Hello, whoa!
Is there a typo in #1?
We end up with an ugly cubic equation.
I'll do #2 . . .
$\displaystyle 2)\;\frac{3x}{x+1} - \frac{1}{3} \:= \:\frac{5}{x+1}$
Find the LCD and multiply through by it . . .
The LCD is: $\displaystyle 3(x+1)$
We have: .$\displaystyle 3(x+1)\cdot\frac{3x}{x+1} -\, 3(x+1)\cdot\frac{1}{3} \;= \;3(x+1)\cdot\frac{5}{x+1} $
Cancel and get: .$\displaystyle 3\cdot3x - (x + 1) \;=\;3\cdot5$
. . .Solve for $\displaystyle x\!:$ . . . .$\displaystyle 9x - x - 1 \;=\;15$
. . . . . . . . . . . . . . . . . . . . $\displaystyle 8x \;= \;16$
. . . . . . . . . . . . . . . . . . . . .$\displaystyle \boxed{x\;=\;2}$
Hello, whoa!
$\displaystyle 1)\;\;\frac{13}{x^2-9x+14} \:= \:\frac{5}{x-3} + \frac{3x-7}{x^2+2x-15}$
We have: .$\displaystyle \frac{13}{(x-2)(x-7)} \;=\;\frac{5}{x-3} + \frac{3x-7}{(x-3)(x+5)}$
The LCD is:$\displaystyle (x-2)(x-3)(x-7)(x+5)$
Multiply through by the LCD:
. . $\displaystyle 13(x-3)(x+5) \:=\:5(x-2)(x-7)(x+5) + (3x-7)(x-2)(x-7)$
which simplifies to that ugly cubic I promised you:
. . $\displaystyle 8x^3 - 80x^2 + 94x + 447 \;= \;0$ . . . ack!
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I conclude that you copied the first fraction incorrectly.
Problem #1 is expected to be simple and straight-forward.
. . I suspect that its denominator is $\displaystyle x + 5$.
Can you please get back to us?