How do you solve: \[-14+6b+7-2b=1+5b\]?
Answer
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Hint: Rearrange the terms of the given equation by taking the terms containing the variable ‘b’ to the L.H.S. while taking the constant terms to the R.H.S. Now, use simple arithmetic operations like: - addition, subtraction, multiplication or division, whichever needed to simplify the equation. Make the coefficient of ‘b’ equal to 1 and accordingly change the R.H.S. to get the answer.
Complete answer:
Here, we have been provided with the linear equation: - \[-14+6b+7-2b=1+5b\] and we are asked to solve this equation. That means we have to find the value of ‘b’.
As we can see that the given equation is a linear equation in one variable which is ‘b’, so taking the terms containing the variable ‘b’ to the L.H.S. and taking all the constant terms to the R.H.S., we get,
\[\begin{align}
& \Rightarrow -14+6b+7-2b=1+5b \\
& \Rightarrow 6b-2b-5b=1-7+14 \\
& \Rightarrow 6b-7b=8 \\
& \Rightarrow -b=8 \\
\end{align}\]
Multiplying both the sides with (-1), we get,
\[\begin{align}
& \Rightarrow \left( -1 \right)\times \left( -b \right)=\left( -1 \right)\times 8 \\
& \Rightarrow b=-8 \\
\end{align}\]
Hence, the value of ‘b’ is -8.
Note: One may note that it is necessary to separate the variable terms and the constant terms. You cannot directly add a variable and a constant. You may see that here we were given a single equation only, this is because we have to find the value of only one variable, that is ‘b’. In general, if we have to solve an equation having ‘n’ number of variables then we must have ‘n’ number of equations given. Now, one can check the answer by substituting the obtained value of ‘b’ in the equation provided in the question. We have to determine the value of L.H.S. and R.H.S. separately and if they are equal then our answer is correct.
Complete answer:
Here, we have been provided with the linear equation: - \[-14+6b+7-2b=1+5b\] and we are asked to solve this equation. That means we have to find the value of ‘b’.
As we can see that the given equation is a linear equation in one variable which is ‘b’, so taking the terms containing the variable ‘b’ to the L.H.S. and taking all the constant terms to the R.H.S., we get,
\[\begin{align}
& \Rightarrow -14+6b+7-2b=1+5b \\
& \Rightarrow 6b-2b-5b=1-7+14 \\
& \Rightarrow 6b-7b=8 \\
& \Rightarrow -b=8 \\
\end{align}\]
Multiplying both the sides with (-1), we get,
\[\begin{align}
& \Rightarrow \left( -1 \right)\times \left( -b \right)=\left( -1 \right)\times 8 \\
& \Rightarrow b=-8 \\
\end{align}\]
Hence, the value of ‘b’ is -8.
Note: One may note that it is necessary to separate the variable terms and the constant terms. You cannot directly add a variable and a constant. You may see that here we were given a single equation only, this is because we have to find the value of only one variable, that is ‘b’. In general, if we have to solve an equation having ‘n’ number of variables then we must have ‘n’ number of equations given. Now, one can check the answer by substituting the obtained value of ‘b’ in the equation provided in the question. We have to determine the value of L.H.S. and R.H.S. separately and if they are equal then our answer is correct.
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