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How do you solve $\dfrac{{x + 12}}{{x - 4}} = \dfrac{{x - 3}}{{x - 7}}$?

Answer
VerifiedVerified
462.3k+ views
Hint: As, the coefficients of $x$ is 1. So, the given equation is the linear equation in one variable. We have to find the value of the variable. For that, we have to divide both sides of the equation by the coefficient of the variable and do simplification. The value of the variable will be the desired result.

Complete step-by-step solution:
As, the coefficients of $x$ is 1. So, the given equation is the linear equation in one variable.
Let us know what is the linear equation in one variable.
When you have a variable of a maximum of one order in an equation, then it is known as a linear equation in one variable. The linear equation is usually expressed in the form of \[ax + b = 0\]. Here a and b are two integers and the solution of $x$ can be only one. The value of a and b can never be equal to 0.
For Example, \[5x + 6 = 10\] is an equation that is linear and has only a single variable in it. The only solution to this equation would be $x = \dfrac{4}{5}$.
Now the given equation is $\dfrac{{x + 12}}{{x - 4}} = \dfrac{{x - 3}}{{x - 7}}$.
Cross multiply the terms,
$ \Rightarrow \left( {x + 12} \right)\left( {x - 7} \right) = \left( {x - 3} \right)\left( {x - 4} \right)$
Open the brackets and multiply the terms,
$ \Rightarrow {x^2} + 12x - 7x - 84 = {x^2} - 3x - 4x + 12$
Simplify the terms,
$ \Rightarrow 5x - 84 = - 7x + 12$
Move variable part on left side and constant part on right side,
$ \Rightarrow 5x + 7x = 84 + 12$
Simplify the terms,
$ \Rightarrow 12x = 96$
Divide both sides by 12,
$\therefore x = 8$

Hence, the value of $x$ is 8.

Note: When you have to solve an equation that has always only one solution, then the steps given below are followed.
Step 1: Find the LCM. In case any fractions exist, clear them.
Step 2: In this step simplification of both sides of the equation happens.
Step 3: Here, you will be isolating the variable on one side.
Step 4: You will verify the obtained result.
To verify our result we must substitute the answer back in the original equation. If the substituted value satisfies the equation then our answer is correct.
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