
How do you solve: $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$ ?
Answer
481.2k+ views
Hint: The value of x in $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$ can be found by using the method of transposition. Method of transposition involves doing the exact same mathematical thing on both sides of an equation with the aim of simplification in mind. This method can be used to solve various algebraic equations like the one given in question with ease.
Complete step-by-step answer:
We would use the method of transposition to find the value of x in $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$. Method of transposition involves doing the exact same thing on both sides of an equation with the aim of bringing like terms together and isolating the variable or the unknown term in order to simplify the equation and finding the value of the required parameter.
Now, In order to find the value of x, we need to isolate x from the rest of the parameters.
So, $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$
First opening all the brackets first, we get,
$ \Rightarrow \dfrac{x}{3} - 5 = 3x - 3 \times 2$
Simplifying the calculations, we get,
$ \Rightarrow \dfrac{x}{3} - 5 = 3x - 6$
Taking all the terms consisting x to right side of the equation and constant terms to the left side of equation, we get,
$ \Rightarrow - 5 + 6 = 3x - \dfrac{x}{3}$
We must remember to reverse the signs of the terms while shifting the terms from one side of the equation to the other side.
Now, adding up the like terms, we get,
$ \Rightarrow 1 = \dfrac{{8x}}{3}$
Multiplying both sides of equation by $\left( {\dfrac{3}{8}} \right)$, we get,
$ \Rightarrow \dfrac{{8x}}{3} \times \dfrac{3}{8} = \dfrac{3}{8}$
Cancelling common factors in numerator and denominator, we get,
$ \Rightarrow x = \dfrac{3}{8}$
Hence, the value of x in $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$ is $\dfrac{3}{8}$.
Note: If we add, subtract, multiply or divide by the same number on both sides of a given algebraic equation, then both sides will remain equal. The given problem deals with algebraic equations. There is no fixed way of solving a given algebraic equation. Algebraic equations can be solved in various ways. Linear equations in one variable can be solved by a transposition method with ease. We must take care while doing the calculations.
Complete step-by-step answer:
We would use the method of transposition to find the value of x in $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$. Method of transposition involves doing the exact same thing on both sides of an equation with the aim of bringing like terms together and isolating the variable or the unknown term in order to simplify the equation and finding the value of the required parameter.
Now, In order to find the value of x, we need to isolate x from the rest of the parameters.
So, $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$
First opening all the brackets first, we get,
$ \Rightarrow \dfrac{x}{3} - 5 = 3x - 3 \times 2$
Simplifying the calculations, we get,
$ \Rightarrow \dfrac{x}{3} - 5 = 3x - 6$
Taking all the terms consisting x to right side of the equation and constant terms to the left side of equation, we get,
$ \Rightarrow - 5 + 6 = 3x - \dfrac{x}{3}$
We must remember to reverse the signs of the terms while shifting the terms from one side of the equation to the other side.
Now, adding up the like terms, we get,
$ \Rightarrow 1 = \dfrac{{8x}}{3}$
Multiplying both sides of equation by $\left( {\dfrac{3}{8}} \right)$, we get,
$ \Rightarrow \dfrac{{8x}}{3} \times \dfrac{3}{8} = \dfrac{3}{8}$
Cancelling common factors in numerator and denominator, we get,
$ \Rightarrow x = \dfrac{3}{8}$
Hence, the value of x in $\dfrac{x}{3} - 5 = 3\left( {x - 2} \right)$ is $\dfrac{3}{8}$.
Note: If we add, subtract, multiply or divide by the same number on both sides of a given algebraic equation, then both sides will remain equal. The given problem deals with algebraic equations. There is no fixed way of solving a given algebraic equation. Algebraic equations can be solved in various ways. Linear equations in one variable can be solved by a transposition method with ease. We must take care while doing the calculations.
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