Question

How do you solve 5x – 4 = 21?

Answer 1

Hint: We are asked to find the solution of 5x – 4 = 21. Firstly, we learn what the solution of the equation means is then we will learn what a linear equation in 1 variable term is. We use a hit and trial method to find the value of ‘x’. In this method, we put the value of ‘x’ one by one by hitting arbitrary values and looking for needed values. Once we work with the hit and trial method we will try another method where we apply algebra. We subtract, add or multiply terms to get to our final term and get our required solution. We will also learn that doing the questions using algebraic tools makes them easy.

Complete step-by-step solution:
We are given that we have 5x – 4 = 21 and we are asked to find the value of ‘x’ or we are asked how we will be able to solve this expression. A solution of any problem is that value which when put into the given problem then the equation is satisfied. Now we will learn about the equation on one variable. One variable simply represents the equation that has one variable (say x, y, or z) and another one constant. For example x + 2 = 4, 2 – x = 2, 2x, 2y, etc. Our equation 5x – 4 = 21 also has just one variable ‘x’. We have to find the value of ‘x’ which will satisfy our given equation. Firstly we try by the method of hit and trial. In which we will put a different value of ‘x’ and take which one fits the solution correctly.
We have 5x – 4 = 21. We will put x = 0 in 5x – 4 = 21. We get,
\[5\left( 0 \right)-4=21\]
\[\Rightarrow -4=21\]
This is not true. So, x = 0 is not the solution.
We will put x = 1 in 5x – 4 = 21. We get,
\[5\left( 1 \right)-4=21\]
\[\Rightarrow 5-4=21\]
On simplifying, we get,
\[\Rightarrow 1=21\]
This is not true. So, x = 1 is not the solution.
As we can see that this is increasing on the left side. So, we are moving in the right direction.
We will put x = 3 in 5x – 4 = 21. So, we get,
\[5\left( 3 \right)-4=21\]
\[\Rightarrow 15-4=21\]
On simplifying, we get,
\[\Rightarrow 11=21\]
This is not true. So, x = 3 is not the solution.
We will put x = 5 in 5x – 4 = 21, so we get,
\[5\left( 5 \right)-4=21\]
\[\Rightarrow 25-4=21\]
\[\Rightarrow 21=21\]
This is true. So, x = 5 is the solution.
In this method, we need to check whether we are moving on the right path or not every time by checking the distance between the right side and the left side so this makes it a little time consuming. We have another way in which we use algebraic tools to solve the problem.
We have 5x – 4 = 21
We add 4 on both the sides, so we get,
\[\Rightarrow 5x-4+4=21+4\]
On simplifying, we get,
\[\Rightarrow 5x=25\]
Now, we divide both the side by 5, so we get,
\[\Rightarrow \dfrac{5x}{5}=\dfrac{25}{5}\]
On simplifying, we get,
\[\Rightarrow x=5\]
Hence the solution is x = 5.

Note: Remember that, we cannot add the variable to the constant. Usual mistakes like this where one adds constants with variables usually happen. For example, 3x + 6 = 9x, here one added ‘6’ with 3 of x made it 9x, this is wrong, we cannot add constant and variable at once. Only the same variables are added to each other. When we add the variable the only constant part is added or subtracted variable remains the same. That is 2x + 2x = 4x error like doing it \[2x+2x=4{{x}^{2}}\] may happen. So, be careful there. Remember when we divide a positive term by a negative value the solution we get is a negative term this may happen that we skip – sign. We need to choose the best techniques as per the question if the question is defined using variable only time so hit and trials would be ok but if the equation has more time variable then we go for the arithmetic tool.