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

sa3b4or9i9

1) $\frac{2x-5}{x+1}-\frac{3}{{x}^{2}+x}=0$
$⇒\frac{2x-5}{x+1}-\frac{3}{x\left(x+1\right)}=0$
$⇒\frac{x\left(2x-5\right)-3}{x\left(x+1\right)}=0$
$⇒x\left(2x-5\right)-3=0$
$⇒2{x}^{2}-5x-3=0$
$⇒2{x}^{2}+x-6x-3=0$
$⇒x\left(2x+1\right)-3\left(2x+1\right)=0$
$⇒\left(x-3\right)\left(2x+1\right)=0$
Thus,
$x-3=0$ or $2x+1=0$
$x=3$ or $2x=-1$
$x=3$ or $x=-0.5$
2) $\frac{1}{y}-\frac{1}{6}=\frac{2}{3}$
$⇒\frac{1}{y}=\frac{1}{6}+\frac{2}{3}=\frac{1+4}{6}=\frac{5}{6}$
$⇒\frac{1}{y}=\frac{5}{6}$
$⇒y=\frac{6}{5}$

Marquis Ibarra

3. $\frac{3}{x+1}=\frac{9}{{x}^{2}-3x-4}$
$⇒\frac{3}{x+1}=\frac{9}{{x}^{2}-4x+x-4}$
$⇒\frac{1}{x+1}=\frac{3}{x\left(x-4\right)+1\left(x-4\right)}$
$⇒\frac{1}{x+1}=\frac{3}{\left(x+1\right)\left(x-4\right)}$
$⇒\frac{1}{1}=\frac{3}{x-4}$
$⇒x-4=3$
$⇒x=4+3$
$x=7$

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