Question

mates help to solve this by quadratic formula ​...

condition x is not equal to +1 and 3

Answer 1

$\large \bold \green{ \underline{answer}} \\ \\ \implies \bold{x = - 1 \: \: \: and \: \: \: \: x = 4}$

$\implies \bold \green{question} \\ \\ \implies \bold{ \frac{2}{x - 3} - \frac{3}{x - 1} = 1 } \\ \\ \implies \bold{ to \: \: find \: \: value \: \: of \: \: x} \\ \\ \implies \bold \orange{explanation} \\ \\ \implies \bold{ \frac{2(x - 1) - 3(x - 3)}{(x - 3)(x - 1)} = 1 } \\ \\ \implies \bold{ \frac{2x -2 - 3x + 9}{ {x}^{2} - 4x + 3 } = 1 } \\ \\ \implies \bold{ { - x + 7} = {x}^{2} - 4x + 3 } \\ \\ \implies \bold{ {x}^{2} - 3x - 4 = 0 } \\ \\ \implies \bold{ {x}^{2} - 4x + x -4 = 0 } \\ \\ \implies \bold{x(x - 4) + 1(x - 4) = 0} \\ \\ \implies \bold{(x + 1)(x - 4) = 0} \\ \\ \implies \bold{x = \: - 1} \: \: \: and \: \: \: x = 4$

Answer 2

QUESTION:

$\frac{2}{x - 3} - \frac{3}{x - 1} = 1$

solve for x.

ANSWER:

$\frac{2}{x - 3} - \frac{3}{x - 1} = 1$

$(taking \: lcm \: in \: lhs \: side)$

$\frac{2(x - 1) - 3(x - 3)}{(x - 3)(x - 1)} = 1$

$(solving \: numerator)$

$\frac{2x - 2 - 3x + 9}{(x - 3)(x - 1)} = 1$

$\frac{ - x + 7}{(x - 3)(x - 1)} = 1$

$(cross \: multiplication)$

$-x + 7 = (x - 3)(x - 1)$

$- x + 7 = {x}^{2} - 3x - x +3 \\ - x + 7 = {x}^{2} - 4x + 3 \\ {x}^{2} - 4x + x + 3 - 7 = 0 \\ {x}^{2} - 3x</p><p>[tex] - 4 = 0$

[/tex]

now,

we have to factorised it;

Splitting the middle term in such a way that it's sum equal to -3 and product equal to -4.

${x}^{2} - 4x + x - 4 \\ x(x - 4) + 1(x - 4)$

$(x - 4)(x + 1) = 0$

so,

$x - 4 = 0 \\ x = 4$

or

$x + 1 = 0 \\ x = - 1$