Answer
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Hint: Rearrange the terms of the given equation by leaving the terms containing the variable ‘r’ in the L.H.S. and taking all other terms to the R.H.S. Now, make the coefficient of ‘r’ in the L.H.S. equal to 1 and accordingly change the R.H.S. to find the value of ‘r’ in terms of ‘s’ and get the answer.
Complete step-by-step solution:
Here, we have been provided with the equation: \[\dfrac{1}{2}r+3s=1\] and we are asked to solve this equation for the value of ‘r’. That means we have to find the value of ‘r’ in terms of ‘s’.
As we can see that the given equation is a linear equation in two variables ‘r’ and ‘s’, so we can only find the value of one of the variables in terms of the other. So, now leaving the terms containing the variable ‘r’ in the L.H.S. and taking all other terms to the R.H.S., we get,
\[\Rightarrow \dfrac{1}{2}r=1-3s\]
Multiplying both the sides with 2 to make the coefficient of ‘r’ equal to 1, we get,
\[\begin{align}
& \Rightarrow r=2\left( 1-3s \right) \\
& \Rightarrow r=2-6s \\
\end{align}\]
Note: One may note that we haven’t found any numerical value of r. This is because there were two variables and only one equation. To find the numerical value we will need one more relation between ‘r’ and ‘s’. In general, remember that to solve the system of equations containing ‘n’ variables we need ‘n’ number of relations between these variables. If it is not present then we can only find the value of one variable in terms of the other.
Complete step-by-step solution:
Here, we have been provided with the equation: \[\dfrac{1}{2}r+3s=1\] and we are asked to solve this equation for the value of ‘r’. That means we have to find the value of ‘r’ in terms of ‘s’.
As we can see that the given equation is a linear equation in two variables ‘r’ and ‘s’, so we can only find the value of one of the variables in terms of the other. So, now leaving the terms containing the variable ‘r’ in the L.H.S. and taking all other terms to the R.H.S., we get,
\[\Rightarrow \dfrac{1}{2}r=1-3s\]
Multiplying both the sides with 2 to make the coefficient of ‘r’ equal to 1, we get,
\[\begin{align}
& \Rightarrow r=2\left( 1-3s \right) \\
& \Rightarrow r=2-6s \\
\end{align}\]
Note: One may note that we haven’t found any numerical value of r. This is because there were two variables and only one equation. To find the numerical value we will need one more relation between ‘r’ and ‘s’. In general, remember that to solve the system of equations containing ‘n’ variables we need ‘n’ number of relations between these variables. If it is not present then we can only find the value of one variable in terms of the other.
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