Question

How do you solve the system $\dfrac{1}{3}x-y=3$ and $2x+y=25$ using substitution?

Answer 1

Hint: To solve the given system of the linear equations using the substitution, we must pick one equation out of the two and find the value of one variable in terms of the other. So we will pick the first equation $\dfrac{1}{3}x-y=3$ and separate $y$ in the form of $x$ as $y=\dfrac{1}{3}x-3$. Then we will substitute this value in the second equation $2x+y=25$. Substituting this we will get a linear equation only in the variable $x$ which can be easily solved by algebraic manipulations. Finally, on substituting this value of $x$ in either of the two equations, we will obtain the value of $y$ and hence the solution of the given system of equations.

Complete step-by-step solution:
Let us write the given system of linear equations as
$\begin{align}
  & \dfrac{1}{3}x-y=3........(i) \\
 & 2x+y=25........(ii) \\
\end{align}$
Let us pick the first equation (i) to write one variable in terms of the other.
$\Rightarrow \dfrac{1}{3}x-y=3$
Subtracting $\dfrac{1}{3}x$ from both the sides, we get
$\begin{align}
  & \Rightarrow \dfrac{1}{3}x-y-\dfrac{1}{3}x=3-\dfrac{1}{3}x \\
 & \Rightarrow -y=3-\dfrac{1}{3}x \\
\end{align}$
Multiplying $-1$ on both sides of the above equation, we get
\[\Rightarrow y=\dfrac{1}{3}x-3.......(iii)\]
Substituting this value of $y$, that is the equation (iii) in the second equation (ii) we get
\[\begin{align}
  & \Rightarrow 2x+\dfrac{1}{3}x-3=25 \\
 & \Rightarrow \dfrac{7}{3}x-3=25 \\
\end{align}\]
Adding $3$ both the sides, we get
$\begin{align}
  & \Rightarrow \dfrac{7}{3}x-3+3=25+3 \\
 & \Rightarrow \dfrac{7}{3}x=28 \\
\end{align}$
Multiplying both sides by $3$ we get
$\begin{align}
  & \Rightarrow \dfrac{7}{3}x\times 3=28\times 3 \\
 & \Rightarrow 7x=84 \\
\end{align}$
Dividing both the sides by $7$ we get
$\begin{align}
  & \Rightarrow \dfrac{7x}{7}=\dfrac{84}{7} \\
 & \Rightarrow x=12 \\
\end{align}$
Finally, substituting this value of $x$ in equation (iii) we get
\[\begin{align}
  & \Rightarrow y=\dfrac{1}{3}\times 12-3 \\
 & \Rightarrow y=4-3 \\
 & \Rightarrow y=1 \\
\end{align}\]
Hence, we have solved the given system of equations using the substitution method and obtained the solution as $x=12,y=1$.

Note: Do not substitute the value \[y=\dfrac{1}{3}x-3\] in the first equation $\dfrac{1}{3}x-y=3$. This is because this value is derived from the first equation only and on back substituting it in the first equation will give you nothing. So make sure that you substitute it in the second equation $2x+y=25$ as we have done in the above solution.