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**Hint:**We are asked to find the solution of 8j = 96. 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 will use the hit and trial method to find the value of ‘j’. In this method, we put the value of ‘j’ 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 question using an algebraic tool makes them easy.

**Complete step-by-step solution:**

We are given that we have 8j = 96 and we are asked to find the value of ‘j’, or we are asked how we will be able to solve this expression. The 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 8j = 96 also has just one variable ‘j’. We have to find the value of ‘j’ 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 ‘j’ and take which one fits the solution correctly.

We let j = 0. Putting j = 0 in 8j = 96, we get,

\[\Rightarrow 8\times 0=96\]

\[\Rightarrow 0=96\]

This is not true. Hence this is not the solution to the problem.

We let j = 1. Putting j = 1 in 8j = 96, we get,

\[\Rightarrow 8\times 1=96\]

\[\Rightarrow 8=96\]

This is not true. Hence this is not the solution to the problem.

We let j = 8. Putting j = 8 in 8j = 96, we get,

\[\Rightarrow 8\times 8=96\]

\[\Rightarrow 64=96\]

This is not true. Hence this is not the solution to the problem.

As we see the gap between terms is decreasing so it means we are moving along the right way.

We let j = 10. Putting j = 10 in 8j = 96, we get,

\[\Rightarrow 8\times 10=96\]

\[\Rightarrow 80=96\]

This is not true. Hence this is not the solution to the problem. So, we move a little more ahead.

We let j = 12. Putting j = 12 in 8j = 96, we get,

\[\Rightarrow 8\times 12=96\]

\[\Rightarrow 96=96\]

This is true. Hence j = 12 is the solution to the problem.

We can see that the above method is a bit lengthy and if we move along the wrong side we may not get our solution easily. We use algebraic tools to solve our problems.

We have 8j = 96.

Dividing 8 on both the sides, we get,

\[\Rightarrow \dfrac{8j}{8}=\dfrac{96}{8}\]

On solving, we get,

\[\Rightarrow j=12\]

**Hence, x = 12 is the solution.**

**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 constants and variables 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, the error can occur like \[2x+2x=4{{x}^{2}}\] 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 if we tend to skip the negative sign.

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