Question

How do you solve \[3(4x - 5) = 5(2x - 5)\] ?

Answer 1

Hint: The first thing that needs to be done for solving this question is to simplify the expressions on each side of the equation by opening the brackets. Then, we need to take the variable term on one side of the equation, and the other terms on the other side of the equation. The obtained equation will be solvable for the final answer.

Complete step by step answer:
We have,
 \[3(4x - 5) = 5(2x - 5)\]
First, we will open the brackets to simplify the expressions on both sides of the equation.
We will get,
 \[3 \times 4x - 15 = 5 \times 2x - 25\]
Simplifying further, we get,
 \[12x - 15 = 10x - 25\]
Now, we need to isolate the terms having an \[x\] on one side, and the remaining terms on the other side. For this we can either transpose terms, or we can add and subtract the terms.
Here, first, we will add 15 on both sides of the equation.
Adding 15, we get
 \[
  12x - 15 + 15 = 10x - 25 + 15 \\
 \Rightarrow 12x = 10x - 10 \\
 \]
Now, subtracting \[10x\] from both sides, we get
 \[
  12x - 10x = 10x - 10 - 10x \\
 \Rightarrow 2x = - 10 \\
 \]
Now simplifying, we get
 \[x = \dfrac{{ - 10}}{2} = - 5\]

Thus, the final value of x comes out to be \[x = - 5\] .

Note: There are two methods to solve an equation. In the first method, we can perform operations on both sides of an equation. The condition is that an operation must be performed on both sides of an equation. We can’t perform operations on only one side of the equation. We can also not perform different operations on both sides of the equation.
In the second method, we can directly take a term from one side of the equation to another, by changing the operation being performed on the term. For e.g. If you want to shift a term from one side of the equation to another, and that term is having addition as the operation, then on the other side, its operation will be subtraction. Similarly, if a term is being multiplied on one side, then for taking on the other side, its operation will change to division.