Question

(x-2)/(x-3) + (3x-11)/(x-4) = (4x+13)/(x+1)

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,

Start by finding least common multiplier,,, that is,,,

(x-3), (x-4), (x+1).

LCM or least common multiplier is the LCM of a, b is the smallest multiplier that is divisible by both a and b.

Multiply by LCM,,,,,,

$\implies$

$\frac{x - 2}{x - 3} (x - 3)(x - 4)( x + 1) \\ \\ + \frac{3x - 11}{x - 4} (x - 3)(x - 4)(x + 1) \\ \\ = \frac{4x + 13}{x + 1} (x - 3)(x - 4)(x + 1)$

Multiply Fractions

$a \frac{b}{c} = \frac{a \times b}{c}$

So, by this we get,,,

$\frac{(x - 2)(x - 3)(x - 4)(x + 1)}{x - 3} \\ \\ + \frac{(3x - 11)(x - 3)(x - 4)(x + 1)}{x - 4} \\ \\ = \frac{(4x + 13)(x - 3)(x - 4)(x + 1)}{x + 1}$

By Cancelling the common factors of the respective denominator and nominator terms we get,,,

$((x - 2)(x - 4)(x + 1)) \\ \\ + ((3x - 11)(x - 3)(x + 1)) \\ \\ = ((4x + 13)(x - 3)(x - 4))$

By Applying FOIL method or Firsts, Outsides, Insides and Lasts.

(a + b) (c + d) =) a = x. b = - 2. c = x. d = - 4.

same goes for others, respectively.

$((xx + x( - 4) + ( - 2)x + ( - 2)( - 4) \times (x + 1)) \\ \\ ((3xx + 3x( - 3) + ( - 11)x + ( - 11)( - 3) \times (x + 1)) \\ \\ ((4xx + 4x( - 3) + 13x + 13( - 3) \times (x + 4))$

By adding similar elements,, - 4x - 2x = - 6x

$((xx - 6x + 2 \times 4)(x + 1)) \\ \\ + ((3xx - 3 \times 3x - 11x + 11 \times 3)(x + 1)) \\ \\ = ((4xx - 4 \times 3x + 13 \times - 13 \times 3)(x - 4))$

Apply the rule of exponents.

${a}^{b} \times {a}^{c} = {a}^{b + c} \\ \\ xx = {x}^{1 + 1} = {x}^{2}$

Therefore,,,,,,

$(( {x}^{2} - 6x + 2 \times 4)(x + 1)) \\ \\ ((3 {x}^{2} - 3 \times 3x - 11x + 11 \times 3)(x + 1)) \\ \\ = ((4 {x}^{2} + 4x \times 3 + 13x - 13 \times 3)(x - 4))$

$(( {x}^{2} - 6x + 8)(x + 1)) \\ \\ + ((3 {x}^{2} - 9x - 11 + 33)(x + 1)) \\ \\ = ((4 {x}^{2} - 12x + 13x - 39)(x - 4))$

$\implies$

$(( {x}^{2} - 6x + 8)(x + 1)) \\ \\ + ((3 {x}^{2} - 20x + 33)(x + 1)) \\ \\ = ((4 {x}^{2} + x - 39)(x + 4))$

Distributing parentheses,,,,

$( {x}^{2} x + {x}^{2} \times 1 + ( - 6x)x + ( - 6x) \times 1 + 8x + 8 \times 1) \\ \\ + (3 {x}^{2} + 3 \times 1 \times {x}^{2} - 20xx - 20 \times 1 \times x + 33x + 33 \times 1) \\ \\ = (4 {x}^{2} x + 4 {x}^{2} ( - 4) + xx - 4 + \times - 39 \times - 39 \times (+ 39) \times (4)$

$\implies$

$( {x}^{3} + {x}^{2} - 6 {x}^{2} - 6x + 8x + 8) \\ \\ + (3 {x}^{3} + 3 {x}^{2} - 20 {x}^{2} - 20x + 33x + 33) \\ \\ (4 {x}^{3} - 16 {x}^{2} + {x}^{2} - 43x + 156)$

$\implies$

$( {x}^{3} - 5 {x}^{2} + 2x + 8) \\ \\ + (3 {x}^{3} - 17 {x}^{2} + 13x + 133) \\ \\ = (4 {x}^{3} - 15 {x}^{2} - 43x + 156)$

Now,,, solving the two equations to equal the value on Right hand side,,,,

$(( {x}^{3} + 3 {x}^{3}) - (5 {x}^{2} + 17 {x}^{2} ) + (2 + 13)x + (8 + 33)) \\ \\ = (4 {x}^{3} - 22 {x}^{2} - 15x + 41)$

Subtraction of

$(4 {x}^{3} - 15 {x}^{2} - 43x + 156)$

From both the sides,, that is,, left hand side and right hand side.

$((4 {x}^{3} - 4 {x}^{3}) - (22 {x}^{2} + 15 {x}^{2}) + (15x + 43x) + (41 - 156)) \\ \\ \\ = (4 {x}^{3} - 15 {x}^{2} - 43x + 156) - (4 {x}^{3} - 15 {x}^{2} - 43x + 156) \\ \\ \\ 4 {x}^{3} - 4 {x}^{3} - 7 {x}^{2} + 58x + 41 - 156 = 0 \\ \\ \\ = - 7 {x}^{2} + 58x - 115 = 0$

NOW SOLVING WITH QUADRATIC EQUATION,,,,,,,,,,

For a Quadratic equation of the form,,,

${ax}^{2} + bx + c = 0 \\ \\ \\ x1 \\ x2 \\ = \\ \\ \frac{ - b + - \sqrt{ {b}^{2} - 2ac } }{2a}$

So by that formula or equation we get, ,,,,,,

FOR A = - 7 , B = 58 , C = - 115.

FIRST VALUE,,,,,

$\frac{ - 58 + \sqrt{ {58}^{2} - 4 \times 7 \times 115 } }{ - 2 \times 7} \\ \\ = \frac{ - 58 + \sqrt{144} }{14} \\ \\ = \frac{ 58 - \sqrt{144} }{14 } \\ \\ = \frac{58 - 12}{14} \\ \\ = \frac{46}{14} \\ \\ \\ 1st \: \: answer = \frac{23}{7}$

AND THE SECOND VALUE IS,,,,,,,

$\frac{ - 58 - \sqrt{ {58}^{2} - 4 \times 7 \times 115 } }{ - 2 \times 7} \\ \\ = \frac{ - 58 - \sqrt{144} }{ - 2 \times 7} \\ \\ = \frac{ - (58 + \sqrt{144}) }{ - 14} \\ \\ = \frac{58 + 12}{14} \\ \\ = \frac{70}{14} \\ \\ \\ 2nd \: \: answer = 5$

Final Solutions for this answer are,,,,,,,,

$x = \frac{23}{7} \: \: \: \: \: \: \: \: \: \\ \\ and\\ \\ \\ x = 5$

HOPE THIS HELPS YOU AND GIVES YOU TWO UNIQUE SOLUTIONS!!!!

Answer 2

Hello,

Solution of your problem attached.
Steps
1) Take LCM on LHS.
2) solve brackets
3) cross multiply the terms.
4) Simplify the expression
5) You will get Quadratic equation,solve it by Quadratic formula.
Hope it helps you.