Destinee Hensley

2022-03-20

Solve the following rational equations and inequalities.

1.$2x-\frac{5}{x}+1-\frac{3}{{x}^{2}}+x=0$

2.$\frac{1}{y}-\frac{1}{6}=\frac{2}{3}$

3.$\frac{3}{x}+1=\frac{9}{{x}^{2}}-3x-4$

1.

2.

3.

sa3b4or9i9

Beginner2022-03-21Added 14 answers

1) $\frac{2x-5}{x+1}-\frac{3}{{x}^{2}+x}=0$

$\Rightarrow \frac{2x-5}{x+1}-\frac{3}{x(x+1)}=0$

$\Rightarrow \frac{x(2x-5)-3}{x(x+1)}=0$

$\Rightarrow x(2x-5)-3=0$

$\Rightarrow 2{x}^{2}-5x-3=0$

$\Rightarrow 2{x}^{2}+x-6x-3=0$

$\Rightarrow x(2x+1)-3(2x+1)=0$

$\Rightarrow (x-3)(2x+1)=0$

Thus,

$x-3=0$ or $2x+1=0$

$x=3$ or $2x=-1$

$x=3$ or $x=-0.5$

2)$\frac{1}{y}-\frac{1}{6}=\frac{2}{3}$

$\Rightarrow \frac{1}{y}=\frac{1}{6}+\frac{2}{3}=\frac{1+4}{6}=\frac{5}{6}$

$\Rightarrow \frac{1}{y}=\frac{5}{6}$

$\Rightarrow y=\frac{6}{5}$

Thus,

2)

Marquis Ibarra

Beginner2022-03-22Added 9 answers

I made the third one

3.$\frac{3}{x+1}=\frac{9}{{x}^{2}-3x-4}$

$\Rightarrow \frac{3}{x+1}=\frac{9}{{x}^{2}-4x+x-4}$

$\Rightarrow \frac{1}{x+1}=\frac{3}{x(x-4)+1(x-4)}$

$\Rightarrow \frac{1}{x+1}=\frac{3}{(x+1)(x-4)}$

$\Rightarrow \frac{1}{1}=\frac{3}{x-4}$

$\Rightarrow x-4=3$

$\Rightarrow x=4+3$

$x=7$

3.

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?

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