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**Hint:**We are asked to find the solution of $x-9=-12$

Firstly, we learn what the solution of the equation means is, and then we will learn what linear equation is in 1 variable term. 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 a 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 answer:**

We are given that we have $x-9=-12$ We are asked to find the value of ‘x’, or we are asked how we will be able to solve this expression.

Solution of any problem is that value which when put into the given problem then equation is satisfied

Now we 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 $43x-43=-42x-42$ 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 let $x=0$

Putting $x=0$ in $x-9=-12$

We get $0-9=-12$

$-9=-12$

This is not true

Hence this is not the solution of the problem

We let $x=1$

Putting $x=1$ in $x-9=-12$

We get $1-9=-12$

$-8=-12$

This is not true

Hence this is not the solution of the problem

Now We let $x=-1$

Putting \[x=-1\] in $x-9=-12$

We get $-1-9=-12$

$-10=12$

This is not true

Hence this is not the solution of the problem

We are moving close to the term on the left so we moving along this side Now We let $x=-2$

Putting $x=-2$ in $x-9=-12$

We get $-2-9=-12$

$-11=-12$

This is not true

Hence this is not the solution of the problem

Now We let $x=-3$

Putting $x=-3$ in $x-9=-12$

We get $-3-9=-12$

$-12=-12$

This is true

Hence this is the solution of the problem

Hence $x-3$ is the solution of $x-9=-12$

We can see that above method is bit lengthy and of we move along wrong side we may not get our solution easily

We use algebraic tool to solve our problems

We have $x-9=-12$

We add 9 on both sides we will get

$x-9+9=-12+9$

So

$x=-3$ $\left[ \text{as }-9+9=0\text{ and }-12+9=-3 \right]$

Hence $x=-3$ 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 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 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 positive term by negative value the solution we get is a negative term this may happen that we skip - sign

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