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Question Number 217291 by ArshadS last updated on 08/Mar/25
Solve: ((x+2)/(x−3))−((x−1)/(x+4))=((10)/(x^2 +x−12))
$$\mathrm{Solve}: \\ $$$$\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}−\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}}=\frac{\mathrm{10}}{{x}^{\mathrm{2}} +{x}−\mathrm{12}} \\ $$
Answered by Hanuda354 last updated on 08/Mar/25
(((x^2 +6x+8)−(x^2 −4x+3))/(x^2 +x−12)) = ((10)/(x^2 +x−12)) ((10x + 5)/(x^2 +x−12)) = ((10)/(x^2 +x−12)) 10x + 5 = 10 10x = 5 x = (1/2)
$$\frac{\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right)−\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right)}{{x}^{\mathrm{2}} +{x}−\mathrm{12}}\:=\:\frac{\mathrm{10}}{{x}^{\mathrm{2}} +{x}−\mathrm{12}} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{10}{x}\:+\:\mathrm{5}}{{x}^{\mathrm{2}} +{x}−\mathrm{12}}\:=\:\frac{\mathrm{10}}{{x}^{\mathrm{2}} +{x}−\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10}{x}\:+\:\mathrm{5}\:=\:\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10}{x}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by ArshadS last updated on 10/Mar/25
Thanks!
$$\mathrm{Thanks}! \\ $$
Answered by Rasheed.Sindhi last updated on 09/Mar/25
((x+2)/(x−3))−((x−1)/(x+4))=((10)/((x−3)(x+4))) ((x−3+5)/(x−3))−((x+4−5)/(x+4))=((10)/((x−3)(x+4))) 1+(5/(x−3))−1+(5/(x+4))=((10)/((x−3)(x+4))) (5/(x−3))+(5/(x+4))=((10)/((x−3)(x+4))) (1/(x−3))+(1/(x+4))=(2/((x−3)(x+4))) ((x+4+x−3)/((x−3)(x+4)))=(2/((x−3)(x+4))) 2x+1=2 x=1/2
$$\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}−\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}}=\frac{\mathrm{10}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\frac{{x}−\mathrm{3}+\mathrm{5}}{{x}−\mathrm{3}}−\frac{{x}+\mathrm{4}−\mathrm{5}}{{x}+\mathrm{4}}=\frac{\mathrm{10}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\mathrm{1}+\frac{\mathrm{5}}{{x}−\mathrm{3}}−\mathrm{1}+\frac{\mathrm{5}}{{x}+\mathrm{4}}=\frac{\mathrm{10}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\frac{\mathrm{5}}{{x}−\mathrm{3}}+\frac{\mathrm{5}}{{x}+\mathrm{4}}=\frac{\mathrm{10}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\frac{\mathrm{1}}{{x}−\mathrm{3}}+\frac{\mathrm{1}}{{x}+\mathrm{4}}=\frac{\mathrm{2}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\frac{{x}+\mathrm{4}+{x}−\mathrm{3}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)}=\frac{\mathrm{2}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\mathrm{2}{x}+\mathrm{1}=\mathrm{2} \\ $$$${x}=\mathrm{1}/\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Mar/25
((x+2)/(x−3))−((x−1)/(x+4))=((−10)/(7(x+4)))+((10)/(7(x−3))) ((10)/(7(x−3)))−((x+2)/(x−3))=((10)/(7(x+4)))−((x−1)/(x+4)) ((10−7(x+2))/(7(x−3)))=((10−7(x−1))/(7(x+4))) ((10−7x−14)/((x−3)))=((10−7x+7)/((x+4))) ((−7x−4)/(x−3))+7=((−7x+17)/(x+4))+7 ((−7x−4+7x−21)/(x−3))=((−7x+17+7x+28)/(x+4)) ((−4−21)/(x−3))=((+17+28)/(x+4)) −25(x+4)=45(x−3) 45x−135=−25x−100 70x=35 x=(1/2)
$$\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}−\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}}=\frac{−\mathrm{10}}{\mathrm{7}\left({x}+\mathrm{4}\right)}+\frac{\mathrm{10}}{\mathrm{7}\left({x}−\mathrm{3}\right)} \\ $$$$\frac{\mathrm{10}}{\mathrm{7}\left({x}−\mathrm{3}\right)}−\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}=\frac{\mathrm{10}}{\mathrm{7}\left({x}+\mathrm{4}\right)}−\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}} \\ $$$$\frac{\mathrm{10}−\mathrm{7}\left({x}+\mathrm{2}\right)}{\mathrm{7}\left({x}−\mathrm{3}\right)}=\frac{\mathrm{10}−\mathrm{7}\left({x}−\mathrm{1}\right)}{\mathrm{7}\left({x}+\mathrm{4}\right)} \\ $$$$\frac{\mathrm{10}−\mathrm{7}{x}−\mathrm{14}}{\left({x}−\mathrm{3}\right)}=\frac{\mathrm{10}−\mathrm{7}{x}+\mathrm{7}}{\left({x}+\mathrm{4}\right)} \\ $$$$\frac{−\mathrm{7}{x}−\mathrm{4}}{{x}−\mathrm{3}}+\mathrm{7}=\frac{−\mathrm{7}{x}+\mathrm{17}}{{x}+\mathrm{4}}+\mathrm{7} \\ $$$$\frac{−\mathrm{7}{x}−\mathrm{4}+\mathrm{7}{x}−\mathrm{21}}{{x}−\mathrm{3}}=\frac{−\mathrm{7}{x}+\mathrm{17}+\mathrm{7}{x}+\mathrm{28}}{{x}+\mathrm{4}} \\ $$$$\frac{−\mathrm{4}−\mathrm{21}}{{x}−\mathrm{3}}=\frac{+\mathrm{17}+\mathrm{28}}{{x}+\mathrm{4}} \\ $$$$−\mathrm{25}\left({x}+\mathrm{4}\right)=\mathrm{45}\left({x}−\mathrm{3}\right) \\ $$$$\mathrm{45}{x}−\mathrm{135}=−\mathrm{25}{x}−\mathrm{100} \\ $$$$\mathrm{70}{x}=\mathrm{35} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Mar/25
((x+2)/(x−3))−((x−1)/(x+4))=((10)/((x−3)(x+4))) Our goal is to achieve: (x−3)(x+4)=(y−p)(y+p) ((2(x+2))/(2(x−3)))−((2(x−1))/(2(x+4)))=((2∙2(10))/(2(x−3)∙2(x+4))) ((2x+4)/(2x−6))−((2x−2)/(2x+8))=((40)/((2x−6)(2x+8))) ((2x+1+3)/(2x+1−7))−((2x+1−3)/(2x+1+7))=((40)/((2x+1−7)(2x+1+7))) Let 2x+1=y ((y+3)/(y−7))−((y−3)/(y+7))=((40)/(y^2 −49)) ((y−7+3+7)/(y−7))−((y+7−3−7)/(y+7))=((40)/((y−7)(y+7))) ((y−7+3+7)/(y−7))−((y+7−3−7)/(y+7))=((40)/((y−7)(y+7))) 1+((10)/(y−7))−1+((10)/(y+7))=((40)/((y−7)(y+7))) (1/(y−7))+(1/(y+7))=(4/((y−7)(y+7))) (y+7)+(y−7)=4 2y=4 y=2 2x+1=2 x=(1/2)
$$\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}−\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}}=\frac{\mathrm{10}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$${Our}\:{goal}\:{is}\:{to}\:{achieve}: \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)=\left({y}−{p}\right)\left({y}+{p}\right) \\ $$$$\frac{\mathrm{2}\left({x}+\mathrm{2}\right)}{\mathrm{2}\left({x}−\mathrm{3}\right)}−\frac{\mathrm{2}\left({x}−\mathrm{1}\right)}{\mathrm{2}\left({x}+\mathrm{4}\right)}=\frac{\mathrm{2}\centerdot\mathrm{2}\left(\mathrm{10}\right)}{\mathrm{2}\left({x}−\mathrm{3}\right)\centerdot\mathrm{2}\left({x}+\mathrm{4}\right)} \\ $$$$\frac{\mathrm{2}{x}+\mathrm{4}}{\mathrm{2}{x}−\mathrm{6}}−\frac{\mathrm{2}{x}−\mathrm{2}}{\mathrm{2}{x}+\mathrm{8}}=\frac{\mathrm{40}}{\left(\mathrm{2}{x}−\mathrm{6}\right)\left(\mathrm{2}{x}+\mathrm{8}\right)} \\ $$$$\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{3}}{\mathrm{2}{x}+\mathrm{1}−\mathrm{7}}−\frac{\mathrm{2}{x}+\mathrm{1}−\mathrm{3}}{\mathrm{2}{x}+\mathrm{1}+\mathrm{7}}=\frac{\mathrm{40}}{\left(\mathrm{2}{x}+\mathrm{1}−\mathrm{7}\right)\left(\mathrm{2}{x}+\mathrm{1}+\mathrm{7}\right)} \\ $$$${Let}\:\mathrm{2}{x}+\mathrm{1}={y} \\ $$$$\frac{{y}+\mathrm{3}}{{y}−\mathrm{7}}−\frac{{y}−\mathrm{3}}{{y}+\mathrm{7}}=\frac{\mathrm{40}}{{y}^{\mathrm{2}} −\mathrm{49}} \\ $$$$\frac{{y}−\mathrm{7}+\mathrm{3}+\mathrm{7}}{{y}−\mathrm{7}}−\frac{{y}+\mathrm{7}−\mathrm{3}−\mathrm{7}}{{y}+\mathrm{7}}=\frac{\mathrm{40}}{\left({y}−\mathrm{7}\right)\left({y}+\mathrm{7}\right)} \\ $$$$\frac{{y}−\mathrm{7}+\mathrm{3}+\mathrm{7}}{{y}−\mathrm{7}}−\frac{{y}+\mathrm{7}−\mathrm{3}−\mathrm{7}}{{y}+\mathrm{7}}=\frac{\mathrm{40}}{\left({y}−\mathrm{7}\right)\left({y}+\mathrm{7}\right)} \\ $$$$\mathrm{1}+\frac{\mathrm{10}}{{y}−\mathrm{7}}−\mathrm{1}+\frac{\mathrm{10}}{{y}+\mathrm{7}}=\frac{\mathrm{40}}{\left({y}−\mathrm{7}\right)\left({y}+\mathrm{7}\right)} \\ $$$$\frac{\mathrm{1}}{{y}−\mathrm{7}}+\frac{\mathrm{1}}{{y}+\mathrm{7}}=\frac{\mathrm{4}}{\left({y}−\mathrm{7}\right)\left({y}+\mathrm{7}\right)} \\ $$$$\left({y}+\mathrm{7}\right)+\left({y}−\mathrm{7}\right)=\mathrm{4} \\ $$$$\mathrm{2}{y}=\mathrm{4} \\ $$$${y}=\mathrm{2} \\ $$$$\mathrm{2}{x}+\mathrm{1}=\mathrm{2} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by ArshadS last updated on 10/Mar/25
Thanks for multiple ways!
$${Thanks}\:{for}\:{multiple}\:{ways}! \\ $$
Answered by Nkosinathi last updated on 09/Mar/25
((x+2)/(x−3))−((x−1)/(x+4))=((10)/((x−3)(x+4))) (((x+4)(x+2)−(x−3)(x−1))/((x−3)(x+4)))=((10)/((x−3)(x+4))) x^2 +6x+8−(x^2 −4x+3)=10 x^2 +6x+8−x^2 +4x−3−10=0 10x−5=0 10x=5 x=(5/(10)) x=(1/2)
$$\frac{{x}+\mathrm{2}}{{x}−\mathrm{3}}−\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}}=\frac{\mathrm{10}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\frac{\left({x}+\mathrm{4}\right)\left({x}+\mathrm{2}\right)−\left({x}−\mathrm{3}\right)\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)}=\frac{\mathrm{10}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right)} \\ $$$${x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}−\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right)=\mathrm{10} \\ $$$${x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}−{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}−\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{10}{x}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{10}{x}=\mathrm{5} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{10}}\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by ArshadS last updated on 10/Mar/25
Thanks!
$${Thanks}! \\ $$

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