
How do you solve: \[-14+6b+7-2b=1+5b\]?
Answer
557.1k+ views
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

