Question

Solving equations, thank you!

Answer 1

Answer:

x = -4 or x = 1

Step-by-step explanation:

$\dfrac{x}{2x-3}-\dfrac{8-3x}{4x^2-9}=0\\\\\bold{DOMAIN:}\\\\2x-3\neq0\ \wedge\ 4x^2-9\neq0\\\\2x-3\neq0\qquad\text{add 3 to both sides}\\2x\neq3\qquad\text{divide both sides by 2}\\\boxed{x\neq1.5}\\\\4x^2-9\neq0\qquad\text{add 9 to both sides}\\4x^2\neq9\qquad\text{divide both sides by 4}\\x^2\neq2.25\\x\neq\pm\sqrt{2.25}\\\boxed{x\neq-1.5\ \wedge\ x\neq1.5}$

$\dfrac{x}{2x-3}-\dfrac{8-3x}{4x^2-9}=0\\\\\dfrac{x}{2x-3}-\dfrac{8-3x}{(2x)^2-3^2}=0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{x}{2x-3}-\dfrac{8-3x}{(2x-3)(2x+3)}=0\\\\\dfrac{x(2x+3)}{(2x-3)(2x+3)}-\dfrac{8-3x}{(2x-3)(2x+3)}=0\\\\\dfrac{x(2x+3)-(8-3x)}{4x^2-9}=0$

$\text{The fraction is equal to 0 if the numerator is equal to 0. Therefore}\\\\\dfrac{x(2x+3)-(8-3x)}{4x^2-9}=0\iff x(2x-3)-(8-3x)=0\\\\\text{use the distributive property}\\\\(x)(2x)+(x)(3)-8-(-3x)=0\\\\2x^2+3x-8+3x=0\qquad\text{combine like terms}\\\\2x^2+(3x+3x)-8=0\\\\2x^2+6x-8=0\qquad\text{divide both sides by 2}\\\\x^2+3x-4=0\\\\x^2+4x-1x-4=0\\\\x(x+4)-1(x+4)=0\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\x+4=0\qquad\text{subtract 4 from both sides}\\x=-4\in D\\\\x-1=0\qquad\text{add 1 to both sides}\\x=1$