Question

How do you solve $2a + 9 = 4a - 19$?

Answer 1

Hint: Here we are given a simple equation and asked to solve for the variable `$a$’. We need to rearrange or alter the given equation. Firstly, we have to separate the variables from the constant terms (numbers) by transferring it to the other side. We need to make sure that terms containing $a$ as variables must be on one side of the equation and the constant terms must be on the other side. Then we solve the given problem by using the basic mathematical operations and transferring methods for solving the equation.

Complete step-by-step solution:
Here we have given the equation $2a + 9 = 4a - 19$ ……(1)
Now we reorder the terms on L.H.S. we get,
$ \Rightarrow 9 + 2a = 4a – 19$
Now we reorder the terms on R.H.S. we get,
$ \Rightarrow 9 + 2a = - 19 + 4a$
Now we solve the equation $9 + 2a = - 19 + 4a$ ……(2)
Solving for the variable ‘$a$’.
Here first we have to transfer all the terms containing $a$ to the R.H.S. of the equation.
When transferring any variable or number to the other side, the sign of the same will be changed to its opposite sign.
So transferring the terms containing $a$ to the other side, we get,
$ \Rightarrow 9 = - 19 + 4a - 2a$
Combine the like terms, $4a - 2a = 2a$
$\therefore $ We get, $9 = - 19 + 2a$
Now transfer the constant terms to the L.H.S. of the equation, so we obtain,
$ \Rightarrow 9 + 19 = 2a$
$ \Rightarrow 28 = 2a$
Now transferring the number 2 to the L.H.S. we get,
$\dfrac{{28}}{2} = a$
$ \Rightarrow a = 14$.
Here we are dividing by 2 on the L.H.S. because in the R.H.S. 2 was multiplied to the variable $a$ and so in order to find $a$ we need to divide the L.H.S. by 2.
Hence we get $a = 14$.
Now to verify whether the obtained value of $a$ satisfies the given equation $2a + 9 = 4a - 19$, we substitute $a = 14$ in the equation and we must obtain L.H.S. is equal to R.H.S.
Substituting $a = 14$ in $2a + 9 = 4a - 19$ we get,
$ \Rightarrow 2(14) + 9 = 4(14) - 19$
$ \Rightarrow 28 + 9 = 56 - 19$
$ \Rightarrow 37 = 37$
Hence L.H.S. is equal to R.H.S.
Therefore, the required value of $a$ is, $a = 14$.

Note: If the question is given in the form of MCQ , we can directly apply the values of $a$ mentioned. If the equation satisfies the value of $a$, then it is the required solution for the given problem.
We need to be careful while taking the terms to the other side. When transferring any variable or number to the other side, the sign of the same will be changed to its opposite sign.
It is important to know the following basic facts.
An equation remains unchanged or undisturbed if it satisfies the following conditions.
(1) If L.H.S. and R.H.S. are interchanged.
(2) If the same number is added on both sides of the equation.
(3) If the same number is subtracted on both sides of the equation.
(4) When both L.H.S. and R.H.S. are multiplied by the same number.
(5) When both L.H.S. and R.H.S. are divided by the same number.