Question

Solve the rational equation:
3x/x+1 + 2x/x-2 = 4x^-4/x^-x-2

Answer 1

no solution

Explanation:

`(3x)/(x+1)+(2x)/(x-2)=(4x^2-4)/(x^2-x-2)\\\\Domain:\\\\x+1\\eq0\ \wedge\ x-2\\eq0\ \wedge\ x^2-x-2\\eq0\\\\(1)\ x+1\\eq0\qquad\text{subtract 1 from both sides}\\x\\eq-1\\\\(2)\ x-2\\eq0\qquad\text{add 2 to both sides}\\x\\eq2\\\\(3)\ x^2-x-2\\eq0\\x^2+x-2x-2\\eq0\\x(x+1)-2(x+1)\\eq0\\(x+1)(x-2)\\eq0\to\ x+1\\eq0\ \wedge\ x-2\\eq0\\\\\bold{DOMAIN:}\ x\in\mathbb{R}-\{-1,\ 2\}`

`(3x)/(x+1)+(2x)/(x-2)=(4x^2-4)/(x^2-x-2)\\\\((3x)(x-2))/((x+1)(x-2))+((2x)(x+1))/((x+1)(x-2))=(4x^2-4)/((x+1)(x-2))\\\\((3x)(x-2)+(2x)(x+1))/((x+1)(x-2))=(4x^2-4)/((x+1)(x-2))\\\\\text{therefore}\\\\(3x)(x-2)+(2x)(x+1)=4x^2-4\qquad\text{use the distributive property}\\\\(3x)(x)+(3x)(-2)+(2x)(x)+(2x)(1)=4x^2-4\\\\3x^2-6x+2x^2+2x=4x^2-4\qquad\text{combine like terms}\\\\(3x^2+2x^2)+(-6x+2x)=4x^2-4`

`5x^2-4x=4x^2-4\qquad\text{subtract}\ 4x^2\ \text{from both sides}\\\\x^2-4x=-4\qquad\text{add 4 to both sides}\\\\x^2-4x+4=0\\\\x^2-2x-2x+4=0\\\\x(x-2)-2(x-2)=0\\\\(x-2)(x-2)=0\\\\(x-2)^2=0\iff x-2=0\qquad\text{add 2 to both sides}\\\\x=2\\otin\bold{D}`