Question

Solve for x

$\sf\ \dfrac{x-4}{x-5}+\dfrac{x-6}{x-5}=\dfrac{10}{3}\ \ \ \ \ x\neq 5\ , 7$

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Answer 1

$\huge\rm{Question} \\ \rm1) \: \frac{x-4}{x-5} + \frac{x-6}{x-7} = \frac{10}{3}$

$\rm\red{Solution:-}$

$\rm→ \frac{(x-4)(x-7)+(x-6)(x-5)}{(x-5)(x-7)} \\ \rm→ \frac{x²-7x-4x+28+x²-5x-6x+30}{x²-7x-5x+35} = \frac{10}{3} \\ \rm→ \frac{2x²-22x+58}{x²-12x+35} \neq \frac{10}{3} \\ \rm→6x²-66x+174=10x²-120x+350 \\ \rm→0=10x²-120x+350-6x²+66x-174 \\ \rm→0=4x²-54x+176 \\ \rm→0=2(2x²-27x+88) \\ \rm→2x²-27x+88=0 \\ \rm→2x²-16x-11x+88=0 \\ \rm→2x(x-8)-11(x-8)=0 \\ \rm→(x-8)(2x-11)=0 \\ \rm→x=8 → x = 8 \: \: \: \: \: 2x-11=0 \\ \rm→2x=11 \\ \rm→x= \frac{11}{2}$

$\huge\rm{Hence \: verified!}$

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Linear Equations has been used . We see that we are given a equation . If we solve the given equation, then we will get value which is not correct according to question . So correct question is mentioned below . After simplifying the question, we will get a Quadratic equation and we need to solve that equation to get answer .

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★ Correct Question :-

Solve for x ::

$\\\;\large{\sf{\mapsto\;\;\dfrac{x\:-\:4}{x\:-\:5}\;+\;\dfrac{x\:-\:6}{x\:-\:7}\;=\;\dfrac{10}{3}}}$

where x ≠ 5,7

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★ Solution :-

Given,

$\\\;\sf{:\rightarrow\;\;\dfrac{x\:-\:4}{x\:-\:5}\;+\;\dfrac{x\:-\:6}{x\:-\:7}\;=\;\dfrac{10}{3}}$

• On taking the LCM of the denominators, we get,

$\\\;\sf{:\rightarrow\;\;\dfrac{(x\:-\:4)\blue{(x\:-\:7)}\;+\;(x\:-\:6)\blue{(x\:-\:5)}}{\orange{(x\:-\:5)(x\:-\:7)}}\;=\;\bf{\dfrac{10}{3}}}$

• Here the highlighted part shows the multiples which are done to make denominators as their LCM .

$\\\;\sf{:\rightarrow\;\;\dfrac{(x^{2}\:-\:7x\:-\:4x\:+\:28)\;+\;(x^{2}\:-\:5x\:-\:6x\:+\:30)}{(x^{2}\:-\:7x\:-\:5x\:+\:35)}\;=\;\bf{\dfrac{10}{3}}}$

$\\\;\sf{:\rightarrow\;\;\dfrac{(x^{2}\:-\:11x\:+\:28)\;+\;(x^{2}\:-\:11x\:+\:30)}{(x^{2}\:-\:7x\:-\:5x\:+\:35)}\;=\;\bf{\dfrac{10}{3}}}$

$\\\;\sf{:\rightarrow\;\;\dfrac{x^{2}\:-\:11x\:+\:28\;+\;x^{2}\:-\:11x\:+\:30}{(x^{2}\:-\:12x\:+\:35)}\;=\;\bf{\dfrac{10}{3}}}$

$\\\;\sf{:\rightarrow\;\;\dfrac{(2x^{2}\:-\:22x\:+\:58)}{(x^{2}\:-\:12x\:+\:35)}\;=\;\bf{\dfrac{10}{3}}}$

• Now cross - multiplying LHS and RHS, we get,

$\\\;\sf{:\rightarrow\;\;\red{3}\:\times\:(2x^{2}\:-\:22x\:+\:58)\;=\;\red{10}\:\times\:(x^{2}\:-\:12x\:+\:35)}$

$\\\;\sf{:\rightarrow\;\;6x^{2}\:-\:66x\:+\:174\;=\;10x^{2}\:-\:120x\:+\:350}$

• Now taking here 2 as common, we get,

$\\\;\sf{:\rightarrow\;\;\gray{2}\:\times\:(3x^{2}\:-\:33x\:+\:87)\;=\;\gray{2}\:\times\:(5x^{2}\:-\:60x\:+\:175)}$

• Cancelling 2 from both sides, we get,

$\\\;\sf{:\rightarrow\;\;(3x^{2}\:-\:33x\:+\:87)\;=\;(5x^{2}\:-\:60x\:+\:175)}$

$\\\;\sf{:\rightarrow\;\;(3x^{2}\:-\:33x\:+\:87)\;-\;(5x^{2}\:-\:60x\:+\:175)\;=\;\bf{0}}$

$\\\;\sf{:\rightarrow\;\;3x^{2}\:-\:33x\:+\:87\;-\;5x^{2}\:+\:60x\:-\:175\;=\;\bf{0}}$

$\\\;\sf{:\rightarrow\;\;-\:2x^{2}\;+\;27x\;-\;88\;=\;\bf{0}}$

• Now since the x² is negative terms here, so we need to transpose this to other side. Then changing sign, we get,

$\\\;\bf{\pink{:\rightarrow\;\;2x^{2}\;-\;27x\;+\;88\;=\;\bf{0}}}$

• This is a quadratic equation. Let's now solve this. Using the method of Splitting the Middle Term, we get,

$\\\;\sf{:\Longrightarrow\;\;2x^{2}\;-\;16x\;-\;11x\;+\;88\;=\;\bf{0}}$

• Now taking the common terms, we get,

$\\\;\sf{:\Longrightarrow\;\;2x(x\;-\;8)\;-\;11(x\;-\;8)\;=\;\bf{0}}$

• Then, combining the Common terms, we get,

$\\\;\bf{:\Longrightarrow\;\;(2x\;-\;11)(x\;-\;8)\;=\;\bf{0}}$

• Here since both terms are being multiplied, then

$\\\;\sf{:\Longrightarrow\;\;\tt{either}\;\sf{(2x\;-\;11)}\;=\;\bf{0}\;\;\tt{or}\;\sf{(x\;-\;8)}\;=\;\bf{0}}$

• Then equating both separately, we get

$\\\;\sf{:\Longrightarrow\;\;\tt{either}\;\sf{2x}\;=\;\bf{11}\;\;\tt{or}\;\sf{x}\;=\;\bf{8}}$

$\\\;\sf{:\Longrightarrow\;\;\tt{either}\;\sf{x}\;=\;\bf{\dfrac{11}{2}}\;\;\tt{or}\;\sf{x}\;=\;\bf{8}}$

• This means that,

$\\\;\bf{:\Longrightarrow\;\;x\;=\;\purple{8\;,\;\dfrac{11}{2}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\:value\:\;of\:\;x\;=\;\bf{\purple{8,\;\;\dfrac{11}{2}}}}}}$

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★ Why the given Question is incorrect ?

Let's find the value as according to given question.

$\\\;\tt{\leadsto\;\;\dfrac{x\:-\:4}{x\:-\:5}\;+\;\dfrac{x\:-\:6}{x\:-\:5}\;=\;\dfrac{10}{3}}$

→ And x ≠ 5, 7

So,

$\\\;\tt{\leadsto\;\;\dfrac{(x\:-\:4)\;+\;(x\:-\:6)}{x\:-\:5}\;=\;\dfrac{10}{3}}$

$\\\;\tt{\leadsto\;\;\dfrac{2x\:-\:10}{x\:-\:5}\;=\;\dfrac{10}{3}}$

Now cross - multiplying, we get,

$\\\;\tt{\leadsto\;\;3\:\times\:(2x\:-\;10)\;=\;10\:\times\:(x\:-\:5)}$

$\\\;\tt{\leadsto\;\;6x\:-\;30\;=\;10x\:-\:50}$

$\\\;\tt{\leadsto\;\;10x\;-\;6x\;=\;50\;-\;30}$

$\\\;\tt{\leadsto\;\;4x\;=\;20}$

$\\\;\tt{\leadsto\;\;x\;=\;\dfrac{20}{4}\;=\;\bf{5}}$

But we see that in question its given that => x ≠ 5, 7 . So our answer here is wrong and we also know that Linear Equation in One Variable cannot have the condition of No Solution .

• Hence, given question is wrong .

Thus correct question and answer is given above ↑ .