Question

Solve for x:

(i) $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3};x\ne 2,4$
(ii) $\frac{1}{x}-\frac{1}{x-2}=3,x\ne 0,2$
(iii) $x+\frac{1}{x}=3,x\ne 0$
(iv) $\frac{16}{x}-1=\frac{15}{x+1},x\ne 0,-1$
(v) $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6},x\ne 3,-5$

Answer 1

(i) We have been given,

,

Now we solve the above equation as follows,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the value of

(ii) We have been given,

,

Now we solve the above equation as follows,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the value of

(iii) We have been given,

,

Now, we solve the equation as follows:

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the value of

(iv) We have been given,

$\frac{16}{x}-1=\frac{15}{x+1},x\ne 0,-1$

Now we solve the above equation as follows,

$$\begin{gathered} \frac{16-x}{x}=\frac{15}{x+1} \\ \Rightarrow (16-x)(x+1)=15x \\ \Rightarrow 16x+16-x^2-x=15x \\ \Rightarrow 15x+16-x^2-15x=0 \\ \Rightarrow 16-x^2=0 \\ \Rightarrow x^2-16=0 \end{gathered}$$

Now we also know that for an equation $a{x}^2+bx+c=0$, the discriminant is given by the following equation:

$D=b^2-4ac$

Now, according to the equation given to us, we have, $a=1$, $b=0$ and $c=-16$.

Therefore, the discriminant is given as,

$$\begin{gathered} D=(0{)}^2-4\left(1\right)\left(-16\right) \\ =64 \end{gathered}$$

Now, the roots of an equation is given by the following equation,

$x=\frac{-b\pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$$\begin{gathered} x=\frac{-0\pm \sqrt{64}}{2\left(1\right)} \\ =\frac{\pm 8}{2} \\ =\pm 4 \end{gathered}$$

Therefore, the value of $x=\pm 4.$

(v) $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6},x\ne 3,-5$
$$\begin{gathered} \frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6} \\ \Rightarrow \frac{x+5-x+3}{\left(x-3\right)\left(x+5\right)}=\frac{1}{6} \\ \Rightarrow \frac{8}{\left(x-3\right)\left(x+5\right)}=\frac{1}{6} \end{gathered}$$
$$\begin{gathered} \Rightarrow 48=x^2+2x-15 \\ \Rightarrow x^2+2x-15-48=0 \\ \Rightarrow x^2+2x-63=0 \\ \Rightarrow x^2+9x-7x-63=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x\left(x+9\right)-7\left(x+9\right)=0 \\ \Rightarrow \left(x-7\right)\left(x+9\right)=0 \\ \Rightarrow x=7,-9 \end{gathered}$$