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Find R and P if
[tex] \frac{R}{3x-2}+ \frac{P}{x+3}= \frac{19-x}{3x^2+7x-6} [/tex]


Sagot :

[tex]\frac{R}{3x-2}+ \frac{P}{x+3}= \frac{19-x}{3x^2+7x-6} \\\\ \frac{R(x+3)+P(3x-2)}{(3x-2)(x+3)} = \frac{19-x}{ 3x^2+9x-2x-6 }\\\\\frac{R(x+3)+P(3x-2)}{ (3x-2) (x +3) } = \frac{19-x}{3x (x +3)-2(x+3) }\\\\\frac{R(x+3)+P(3x-2)}{ (3x-2) (x +3) } = \frac{19-x}{ (3x-2) (x +3) }[/tex]

[tex]R(x+3)+P(3x-2) =19-x\\\\R(x+3) =19-x-P(3x-2)\ \ :(x+3)\\\\R =\frac{19-x-P(3x-2)}{x+3}=\frac{19-x- 3Px+2P}{x+3}=\frac{ -x(1+ 3P+19 )+2P}{x+3} \\\\P(3x-2) =19-x-R(x+3)\ \ /:(3x-2)\\\\P =\frac{19-x-R(x+3)}{3x-2}=\frac{19-x-Rx-3R}{3x-2}= \frac{ -x(1+ R )-3R+19}{3x-2}[/tex]


[tex](3x-2)(x+3) \neq 0 \\\\3x-2 \neq 0 \ \ \ or \ \ \ x+3\neq 0\\\\3x \neq 2\ \ /:3 \ \ \ or \ \ \ x \neq -3\\\\x\neq \frac{2}{3}\ \ \ or \ \ \ x\neq -3 \\ \\ D=R\setminus \left \{ -3,\frac{2}{3} \right \}[/tex]