Question

X-1/x-2+x-3/x-4=3 1/3,x not equal to 2,4 solve by using quadratic formula

Answer 1

$\begin{lgathered}\red{\bf{Given}}\begin{cases}\bf\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3(\dfrac{1}{3}) = \dfrac{10}{3}\\ \\ \bf \:x \ne \: 2 \: and \: 4. \end{cases}\end{lgathered}$

$\huge{\mathfrak{\overbrace{\underbrace{\pink{\fbox{\green{\blue{\bf\:S}\pink{o}\red{l}\orange{u}\purple{t}\blue{i}\green{o}\red{n}}}}}}}}$

$\red\leadsto\footnotesize\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3} \\ \\ \red\leadsto\footnotesize\frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\frac{10}{3} \\ \\ \red\leadsto\footnotesize\frac{x^{2}-4x-x+4+x^{2}-2x-3x+6}{x^{2}-4x-2x+8}=\frac{10}{3} \\ \\ \red\leadsto\footnotesize\frac{2x^{2}-10x+10}{x^{2}-6x+8}=\frac{10}{3} \\ \\ \red\leadsto\footnotesize\:3(2x^{2}-10x+10)=10(x^{2}-6x+8) \\ \\ \red\leadsto\footnotesize\:6x^{2}-30x+30=10x^{2}-60x+80 \\ \\\red\leadsto\footnotesize \: 10x^{2}-60x+80-6x^{2}+30x-30 = 0 \\ \\\red\leadsto\footnotesize \:4x^{2}-30x+50=0 \\ \\\red\leadsto\footnotesize \: 2(2x^{2}-15x+25)=0 \\ \\\red\leadsto\footnotesize \:2x^{2}-15x+25=0 \\ \\ \underline{\textbf{\green{Splitting the Middle Term Now,}}} \\ \\\red\leadsto\footnotesize 2x^{2}-10x-5x+25=0 \\ \\ \red\leadsto\footnotesize 2x(x-5)-5(x-5)=0 \\ \\ \red\leadsto\footnotesize(x-5)(2x-5)=0 \\ \\\red\leadsto\footnotesize\:x-5 = 0 \\ \\ \blue{\bf{Or}} \\ \\\red\leadsto\footnotesize \: 2x-5 = 0 \\ \\\bf \: Hence \\ \\ \red\leadsto\large\red{\boxed{\tt\blue{x}\purple{=} \green {5}\orange,\pink{ \frac{5}{2}}}}\: \: \: \: \bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\huge\bold{\red{\ddot{\smile}}}}}}}}}}}}}$

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$\boxed{\begin{minipage}{7 cm}\boxed{\underline{\underline{\bigstar\:\bf\:Algebric\:Identity\:\bigstar}}}\\\\1] (A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2] (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3] A^{2} - B^{2} = (A+B)(A-B)\\\\4] (A+B)^{2} = (A-B)^{2} + 4AB\\\\5] (A-B)^{2} = (A+B)^{2} - 4AB\\\\6] (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7] (A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8] A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9] A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\10] (A+B+C)^{2} = A^{2} + B^{2} + C^{2} + 2(AB + BC + CA)\\\\\end{minipage}}$

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Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$: \implies{\sf{ \frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = \frac{10}{3} }}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ \frac{(x - 1)(x - 3) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{10}{3} }}$

$: \implies{\sf{ \frac{ {x}^{2} - 4x - x + 4 - {x}^{2} - 3x - 2x + 6 }{(x - 2)(x - 4)} = \frac{10}{3} }}$

$: \implies{\sf{ 3({x}^{2} - 4x - x + 4 - {x}^{2} - 3x - 2x + 6 ) = 10(x - 2)(x - 4)}}$

$: \implies{\sf{ 3(2 {x}^{2} - 10x + 10 ) = 10( {x}^{2} - 4x - 2x + 8 )}}$

$: \implies{\sf{ 6 {x}^{2} - 30x + 30 = 10{x}^{2} - 40x - 20x + 8 0}}$

$: \implies{\sf{ 6 {x}^{2} - 30x + 30 - 10{x}^{2} + 60x - 8 0 = 0}}$

$: \implies{\sf{ - 4 {x}^{2} + 30x - 5 0 = 0}}$

$: \implies{\sf{ 4 {x}^{2} - 30x + 5 0 = 0}}$

$: \implies{\sf{ 2 {x}^{2} - 15x + 25 = 0}}$

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By quadratic formula,

$\longrightarrow{\sf{ x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$

$\longrightarrow{\sf{ x = \frac{ 15 \pm \sqrt{ { (- 15)}^{2} - 4(2)(25) } }{2a} }}$

$\longrightarrow{\sf{ x = \frac{ 15 \pm \sqrt{ 225 - 200} }{4} }}$

$\longrightarrow{\sf{ x = \frac{ 15 \pm \sqrt{ 25} }{4} }}$

$\longrightarrow{\sf{ x = \frac{ 15 \pm 5}{4} }}$

$\longrightarrow{\sf{ x = \frac{ 15 + 5}{4}, \frac{ 15 - 5}{4} }}$

$\longrightarrow{\sf{ x = \frac{ 20}{4}, \frac{ 10}{4} }}$

$\longrightarrow{\sf{ x = 5, \frac{ 5}{2} }}$