Question 1When $x$ is added to both the numerator and the denominator of the fraction $\frac{a}{b},$ $a\ne b,$ $b\ne0,$ the value of the fraction is changed to $\frac{c}{d}.$ Then $x$ equals:$\dfrac1{c-d}$$\dfrac{ad-bc}{c-d}$$\dfrac{ad-bc}{c+d}$$\dfrac{bc-ad}{c-d}$$\dfrac{bc-ad}{c+d}$