3-x-1-1-x-3-4-x-2- Find x-

We have,3/x 1+1/x 3=4/x 2Taking L.C.M of (x 1) and (x 3).[ 3(x 3)+1(x 1)/(x 1)(x 3) = 4/x 2; 3x 9+x 1/x^2 3x x+3 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Properties of Determinants

3/x-1+1/x-3=4...

Question

$\frac{3}{x - 1} + \frac{1}{x - 3} = \frac{4}{x - 2}$ . Find $x$ .

Open in App

Solution

We have, $\frac{3}{x - 1} + \frac{1}{x - 3} = \frac{4}{x - 2}$
Taking L.C.M of $(x - 1)$ and $(x - 3)$ .
$\begin{array}{rcl} \frac{3 (x - 3) + 1 (x - 1)}{(x - 1) (x - 3)} & = & \frac{4}{x - 2} \\ \frac{3 x - 9 + x - 1}{x^{2} - 3 x - x + 3} & = & \frac{4}{x - 2} \\ \frac{4 x - 10}{x^{2} - 4 x + 3} & = & \frac{4}{x - 2} \\ (4 x - 10) (x - 2) & = & 4 (x^{2} - 4 x + 3) \\ 4 x^{2} - 8 x - 10 x + 20 & = & 4 x^{2} - 16 x + 12 \\ - 18 x + 16 x & = & 12 - 20 \\ - 2 x & = & - 8 \\ x & = & 4 \end{array}$
Hence, the value of $x$ is $4$ .

Suggest Corrections

Similar questions

Let f(x) be a function defined by $f (x) = {\begin{matrix} \frac{3}{| x | + 2 x}, & x \neq 0 0, & x = 0 \end{matrix}$ Show that ${lim}_{x \to 0}$ f(x)does not exist.

Q. If

x^{4} + \frac{1}{x^{4}} = 194,

find

x^{3} + \frac{1}{x^{3}}, x^{2} + \frac{1}{x^{2}}

and

x + \frac{1}{x}

If $f = \frac{x}{1 + x^{2}} + \frac{1}{3} {(\frac{x}{1 + x^{2}})}^{3} + \frac{1}{5} {(\frac{x}{1 + x^{2}})}^{5} + . . .$ to $\infty$ and $g = x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} + \frac{1}{7} x^{7} - \frac{2}{9} x^{9} + . . .$ , then $f = d \times g$ . Find 4d.

Q. If

f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}, and h (x) = \frac{f (x)}{g (x)}

then the local minimum value of

h (x)

Q. Find the value of the following expression:

81^{\frac{1}{{log}_{5} 3}} + 27^{{log}_{9} 36} + 3^{\frac{4}{{log}_{7} 9}}

Join BYJU'S Learning Program

3x-1+1x-3=4x-2. Find x.

We have,3x-1+1x-3=4x-2Taking L.C.M of (x-1) and (x-3).3(x-3)+1(x-1)(x-1)(x-3)=4x-23x-9+x-1x2-3x-x+3=4x-24x-10x2-4x+3=4x-2(4x-10)(x-2)=4(x2-4x+3)4x2-8x-10x+20=4x2-16x+12-18x+16x=12-20-2x=-8x=4Hence, the value of x is 4.

$\frac{3}{x - 1} + \frac{1}{x - 3} = \frac{4}{x - 2}$ . Find $x$ .