Question

Can someone show me how to do this problem

Answer 1

`\bf \cfrac{3x}{x^2+1}=\cfrac{30}{11x-3}\implies \stackrel{cross-multiplying}{3x(11x-3)=30(x^2+1)}\implies 33x^2-9x=30x^2+30 \\\\\\ 3x^2-9x-30=0\implies (3x-15)(x+2)=0\implies \begin{cases} 3x-15=0\\ \qquad 3x=15\\[0.8em] \qquad x=\cfrac{15}{3}\\[0.8em] \qquad x=5\\[-0.5em] \hrulefill\\ x+2=0\\ \qquad \boxed{x=-2} \end{cases}`