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# How do you solve $\dfrac{x-3}{x-4}=\dfrac{x-5}{x+4}$?

Last updated date: 13th Jun 2024
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Hint: From the question given, we have been asked to solve $\dfrac{x-3}{x-4}=\dfrac{x-5}{x+4}$.We can solve the given equation by transforming it and then simplifying the equation. By doing this, we can get the solution for the given question. We can clearly say that the values of $\left( x-4 \right)\text{ and }\left( x+4 \right)$ will not be equal to zero, because if the values of $\left( x-4 \right)\text{ and }\left( x+4 \right)$ will be zero, then the given equation will become undefined.

Complete step by step solution:
From the question, we have been given that $\dfrac{x-3}{x-4}=\dfrac{x-5}{x+4}$
Now, multiply both sides of the equation with $\left( x-4 \right)\left( x+4 \right)$.
By multiplying the both sides of the equation with $\left( x-4 \right)\left( x+4 \right)$, we get
$\Rightarrow \dfrac{x-3}{x-4}\left( x-4 \right)\left( x+4 \right)=\dfrac{x-5}{x+4}\left( x-4 \right)\left( x+4 \right)$
On further more simplifying the above equation we get $\Rightarrow \left( x-3 \right)\left( x+4 \right)=\left( x-5 \right)\left( x-4 \right)$
Now, simplify the equation by multiplying all the terms.
By multiplying the terms, we get $\Rightarrow {{x}^{2}}+x-12={{x}^{2}}-9x+20$
Now, simplify the equation further more.
Shift ${{x}^{2}}$ in the right hand side of the equation to the left hand side of the equation.
B y shifting ${{x}^{2}}$ from the right hand side of the equation to the left hand side of the equation, we get $\Rightarrow {{x}^{2}}-{{x}^{2}}+x-12=-9x+20$
By the cancellation of terms, we get $\Rightarrow x-12=-9x+20$
Shift whole right hand side terms to the left hand side of the equation.
By shifting, we get
$\Rightarrow x-12+9x-20=0$
$\Rightarrow 10x-32=0$
Now, shift $-32$ to the right hand side of the equation. By shifting we get,
$\Rightarrow 10x=32$
$\Rightarrow x=\dfrac{32}{10}$
$\Rightarrow x=\dfrac{16}{5}$
Hence, the given equation is solved.

Note: We should be very careful while doing the calculation in this problem. Also, we should be very careful while solving the equation. Also, we should be very careful while transforming the equation. Also, we should do exact transformations to the given equation to obtain the perfect answer. Similarly we can solve \begin{align} & \dfrac{x-9}{x-5}=\dfrac{x-3}{x+5}\Rightarrow \left( x-9 \right)\left( x+5 \right)=\left( x-3 \right)\left( x-5 \right)\Rightarrow {{x}^{2}}-4x-45={{x}^{2}}-8x+15\Rightarrow -45=-4x+15 \\ & \Rightarrow 4x=60\Rightarrow x=15 \\ \end{align}