How do you solve: $\dfrac{5}{2}t - t = 3 + \dfrac{3}{2}t$?
Answer
558k+ views
Hint: The value of t in $\dfrac{5}{2}t - t = 3 + \dfrac{3}{2}t$ 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 solution:
We would use the method of transposition to find the value of t in $\dfrac{5}{2}t - t = 3 + \dfrac{3}{2}t$. 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 t, we need to isolate t from the rest of the parameters.
So, $\dfrac{5}{2}t - t = 3 + \dfrac{3}{2}t$
Taking all the terms consisting t to left side of the equation and constant terms to the right side of equation, we get,
$ \Rightarrow $$\dfrac{5}{2}t - t - \dfrac{3}{2}t = 3$
We must remember to reverse the signs of the terms while shifting the terms from one side of the equation to the other side.
Now, taking t common from all the terms in left side of the equation, we get,
$ \Rightarrow $$t\left( {\dfrac{5}{2} - 1 - \dfrac{3}{2}} \right) = 3$
Adding up the like terms,
$ \Rightarrow $$t\left( {\dfrac{5}{2} - \dfrac{5}{2}} \right) = 3$
$ \Rightarrow $$0t = 3$
Now, we can see that $0t = 3$ is the equation obtained by solving the original equation.
But, there exists no value of t for which $0t = 3$ can hold true.
Hence, there is no solution to the given linear equation in one variable.
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 solution:
We would use the method of transposition to find the value of t in $\dfrac{5}{2}t - t = 3 + \dfrac{3}{2}t$. 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 t, we need to isolate t from the rest of the parameters.
So, $\dfrac{5}{2}t - t = 3 + \dfrac{3}{2}t$
Taking all the terms consisting t to left side of the equation and constant terms to the right side of equation, we get,
$ \Rightarrow $$\dfrac{5}{2}t - t - \dfrac{3}{2}t = 3$
We must remember to reverse the signs of the terms while shifting the terms from one side of the equation to the other side.
Now, taking t common from all the terms in left side of the equation, we get,
$ \Rightarrow $$t\left( {\dfrac{5}{2} - 1 - \dfrac{3}{2}} \right) = 3$
Adding up the like terms,
$ \Rightarrow $$t\left( {\dfrac{5}{2} - \dfrac{5}{2}} \right) = 3$
$ \Rightarrow $$0t = 3$
Now, we can see that $0t = 3$ is the equation obtained by solving the original equation.
But, there exists no value of t for which $0t = 3$ can hold true.
Hence, there is no solution to the given linear equation in one variable.
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

