Question

How do you solve 7r+7-6r=13?

Answer 1

Hint: We are asked to find the solution of 7r+7-6r=13. Firstly, we will learn what linear equations are in 1 variable term. We will use a hit and trial method to find the value of ‘r’. In this method, we put the value of ‘r’ one by one by hitting arbitrary values and looking for the 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 to do the questions using algebraic operations which makes them easy.

Complete answer:
We are given that we have 7r+7-6r=13. We are asked to find the value of ‘r’, or we are asked how we will be able to solve this expression.
We will first 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, 5-x=2, 2x, 2y etc.
Our equation 7r+7-6r=13 also has just one variable ‘r’.
We have to find the value of ‘r’ which will satisfy our given equation.
Firstly we try by the method of hit and trial, in which we will put different values of ‘r’ and take which
one fits the solution correctly.
Let r=0.
Putting r=0 in 7r+7-6r=13, we get,
\[7\times 0+7-6\times 0=13\]
So, 7=13.
Which is not the solution.
Let r=2.
Putting r=2 in 7r+7-6r=13, we get,
\[\begin{align}
  & 7\times 2+7-6\times 2=13 \\
 & \Rightarrow 14+7-12=13 \\
\end{align}\]
So, we get 9=13.
Which is not the solution.
Now let r=6.
Putting r=6 in 7r+7-6r=13, we get,
\[\begin{align}
  & 7\times 6+7-6\times 6=13 \\
 & \Rightarrow 42+7-36=13 \\
\end{align}\]
So, 13=13.
So, we get that r=6 is the solution of this problem. As on putting r=6, we get RHS=LHS.
Now, as we can see, this method is very tricky as we could have moved towards the wrong path once we started solving it.
So, now let us use the algebraic operators like $+,-\times ,\div $ to solve this problem.
We have 7r+7-6r=13.
As 7r-6r=1r or r. So, we can write,
r+7=13
Now, on subtracting 7 on both the sides, we get,
r+7-7=13-7
So, we get,
r=6
Hence, r=6 is our required 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: 6x+2=8x, here if one adds ‘2’ with 6x and gets it as 8x, then it is wrong, since we cannot add constants and variables at once. Only the same variables can be added to each other. When we add the variable the only constant part is added or subtracted, the variable remains the same. That is we know that 3x+2x=5x but doing it as $3x+2x=5{{x}^{2}}$is wrong so we have to be careful. Remember, when we divide positive term by negative value the solution we get is a negative term but it can occur that we forget to put the negative sign in that term.