How do you solve $\dfrac{3x-4}{{{x}^{2}}-10x+21}-\dfrac{x-8}{{{x}^{2}}-18x+77}=\dfrac{x-5}{{{x}^{2}}-14x+33}$ and check for extraneous solutions?

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Vedantu Content Team · Accepted Answer

Hint: Here in this question we have been asked to solve the given expression $\dfrac{3x-4}{{{x}^{2}}-10x+21}-\dfrac{x-8}{{{x}^{2}}-18x+77}=\dfrac{x-5}{{{x}^{2}}-14x+33}$ . For that we will simplify the expression using some arithmetic operations. First we will express the denominators as factors and then we will take LCM of terms on the LHS of the question. We will be able to cancel out common terms and obtain the solutions.Complete step by step solution:Now considering from the question we have been asked to solve the given expression $\dfrac{3x-4}{{{x}^{2}}-10x+21}-\dfrac{x-8}{{{x}^{2}}-18x+77}=\dfrac{x-5}{{{x}^{2}}-14x+33}$ .For that we will simplify the expression using some arithmetic operations.Firstly we will simplify ${{x}^{2}}-10x+21$ and write it as$ \Rightarrow {{x}^{2}}-3x-7x+3\left( 7 \right) $$  \Rightarrow x\left( x-3 \right)-7\left( x-3 \right) $$ \Rightarrow \left( x-7 \right)\left( x-3 \right) $Now we will simplify ${{x}^{2}}-18x+77$ and write it as$ \Rightarrow {{x}^{2}}-7x-11x+\left( 11 \right)7 $$  \Rightarrow x\left( x-7 \right)-11\left( x-7 \right) $$ \Rightarrow \left( x-7 \right)\left( x-11 \right) $Now we will simplify ${{x}^{2}}-14x+33$ and write it as$ \Rightarrow {{x}^{2}}-3x-11x+\left( 11 \right)3 $$ \Rightarrow x\left( x-3 \right)-11\left( x-3 \right) $$ \Rightarrow \left( x-3 \right)\left( x-11 \right) $Now by using these simplifications we will have $ \dfrac{3x-4}{\left( x-7 \right)\left( x-3 \right)}-\dfrac{x-8}{\left( x-7 \right)\left( x-11 \right)}=\dfrac{x-5}{\left( x-3 \right)\left( x-11 \right)}$Now we will make the denominator common for all the terms by performing simple arithmetic operations. After doing that we will have $\Rightarrow \dfrac{\left( 3x-4 \right)\left( x-11 \right)}{\left( x-11 \right)\left( x-7 \right)\left( x-3 \right)}-\dfrac{\left( x-3 \right)\left( x-8 \right)}{\left( x-3 \right)\left( x-7 \right)\left( x-11 \right)}=\dfrac{\left( x-5 \right)\left( x-7 \right)}{\left( x-3 \right)\left( x-11 \right)\left( x-7 \right)}$ .Now we can say that the expression here will be undefined for the following values of $x$ they are $3,7,11$ .Now by cancelling out the denominators on the both sides we will have $\Rightarrow \left( 3x-4 \right)\left( x-11 \right)-\left( x-3 \right)\left( x-8 \right)=\left( x-5 \right)\left( x-7 \right)$ .By further simplifying this expression using multiplication operation we will have $\Rightarrow 3{{x}^{2}}-33x-4x+44-\left( {{x}^{2}}-8x-3x+24 \right)={{x}^{2}}-7x-5x+35$ .Now we need to perform further arithmetic simplifications. After doing that we will have$ \Rightarrow 2{{x}^{2}}-26x+20={{x}^{2}}-12x+35 $$  \Rightarrow {{x}^{2}}-14x=15 $Now we will simplify ${{x}^{2}}-14x-15=0$ and write it as$  \Rightarrow {{x}^{2}}-15x+1x-\left( 15 \right)1=0 $$  \Rightarrow x\left( x-15 \right)+1\left( x-15 \right)=0 $$  \Rightarrow \left( x-15 \right)\left( x+1 \right) =0$$ \Rightarrow x=15, x=-1$Therefore we can conclude that the solution for the given expression $\dfrac{3x-4}{{{x}^{2}}-10x+21}-\dfrac{x-8}{{{x}^{2}}-18x+77}=\dfrac{x-5}{{{x}^{2}}-14x+33}$ will be given as 15 and $-1$ and it is not defined for $x=3,7,11$.Note: While answering questions of this type we should be sure with the simplifications that we are performing in between the steps. Someone make a calculation mistake and consider it as ${{x}^{2}}-14x-10=0$ $  \dfrac{14\pm \sqrt{{{\left( -14 \right)}^{2}}-4\left( -10 \right)}}{2}=\dfrac{14\pm \sqrt{196+60}}{2} $$  \Rightarrow \dfrac{14\pm \sqrt{256}}{2}=\dfrac{14\pm 16}{2} $$ \Rightarrow 7\pm 8=15,-1 $then we will have a wrong answer.