Question

Solve \[x+3y=16, 2x-y=4\] by using the substitution method.

Answer 1

Hint: The value of one of the variables from one equation is substituted in the other equation. Then simplify the new equation formed by using elementary arithmetic operations and find values of x and y.

Complete step-by-step answer:
The substitution method is the algebraic method to solve simultaneous linear equations. In this method the value of one variable from one equation is substituted in the other equation. In this way a pair of the linear equation gets transformed into one linear equation with only one variable, which can then be easily solved.
Now we have been given two linear equations.
\[\begin{align}
  & x+3y=16......(1) \\
 & 2x-y=4.......(2) \\
\end{align}\]
Now let us first simplify the equation and solve equation (1) for x.
\[\begin{align}
  & x+3y=16 \\
 & \Rightarrow x=16-3y \\
\end{align}\]
Now let us first simplify the equation and solve equation (2) for x. We get,
\[\begin{align}
  & 2x-y=4 \\
 & 2\left( 16-3y \right)-y=4 \\
\end{align}\]
Now let us solve the new equation obtained using elementary arithmetic operations.
\[\begin{align}
  & 32-6y-y=4 \\
 & \Rightarrow 32-7y=4 \\
 & 7y=32-4 \\
 & 7y=28\Rightarrow y={}^{28}/{}_{7}=4 \\
\end{align}\]
Now we get y = 4, now solve the equation to find the value of the second variable, put y = 4 in equation (1).
\[x=16-3y=16-\left( 3\times 4 \right)=16-12=4\]
Thus we got x = 4 and y = 4.
Thus we solved both the equations using the substitution method and we get x = 4 and y = 4.

Note: We can check the above answers using other algebraic methods.
Subtracting both the equations,
\[\begin{align}
  & x+3y=16......(1) \\
 & 2x-y=4.......(2) \\
\end{align}\]
\[\begin{align}
  & 2x+6y=32 \\
 & 2x-1y=04 \\
 & \overline{0x+7y=28} \\
\end{align}\]
Thus y = 4 and putting it in equation (2), we get,
\[\begin{align}
  & 2x=4+4 \\
 & \therefore x=4 \\
\end{align}\]