Question

How do you solve $2x+7y=5$ and $6x+y=5$?

Answer 1

Hint: From the first equation try to find the value of ‘x’ in terms of ‘y’. Then put the value of ‘x’ obtained earlier to get the value of ‘y’. After getting the value of ‘y’ put that value in any of the equations to get the value of ‘x’.

Complete step by step solution:
Solving the simultaneous equations: First we have to take any one of the two equations. Then we have to find the value of ‘x’ in terms of ‘y’. Putting that value of ‘x’ in the second equation we can get the value of ‘y’. Again putting that value of ‘y’ in any of the equations we can get the value of ‘x’.
Considering the first equation
$2x+7y=5$
From here ‘x’ can be obtained in terms of ‘y’ as
$\begin{align}
  & 2x=5-7y \\
 & \Rightarrow x=\dfrac{5-7y}{2} \\
\end{align}$
Now considering the second equation $6x+y=5$
It can be written as
$y=5-6x$
Putting the value of ‘x’ we got earlier in the above equation, we get
$\begin{align}
  & \Rightarrow y=5-6\times \left( \dfrac{5-7y}{2} \right) \\
 & \Rightarrow y=\dfrac{10-6\times \left( 5-7y \right)}{2} \\
 & \Rightarrow y\times 2=10-30+42y \\
 & \Rightarrow 2y-42y=-20 \\
 & \Rightarrow -40y=-20 \\
 & \Rightarrow y=\dfrac{-20}{-40} \\
 & \Rightarrow y=\dfrac{1}{2} \\
\end{align}$
Putting the value of ‘y’ in the second equation, we get
\[\begin{align}
  & 6x+y=5 \\
 & \Rightarrow 6x+\dfrac{1}{2}=5 \\
 & \Rightarrow 6x=5-\dfrac{1}{2} \\
 & \Rightarrow 6x=\dfrac{10-1}{2} \\
 & \Rightarrow 6x\times 2=9 \\
 & \Rightarrow 12x=9 \\
 & \Rightarrow x=\dfrac{9}{12} \\
 & \Rightarrow x=\dfrac{3}{4} \\
\end{align}\]
Hence the solution of the system of linear equations $2x+7y=10$ and $x-2y=15$ is $\left( x,y \right)=\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
This is the required solution.

Note: The two given equations can also be solved by taking simultaneously. We have to make the coefficient of one variable equal by multiplying suitable constants with both the equations. Then either by subtracting or adding the modified equations, one variable can be obtained. Putting that value in one of the given equations the other variable can also be obtained. This is the alternative method.