Question

How do you solve the simultaneous equations $3x+7y=26$ and $4x+5y=13$?

Answer 1

Hint: For solving simultaneous equations first get the value of ‘x’ in terms of ‘y’ from the given first equation. Then put that value in the given second equation to get the value of ‘y’. Again put the value of ‘y’ 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 equation we can get the value of ‘x’.
Considering the first equation
$3x+7y=26$
From here ‘x’ can be obtained in terms of ‘y’ as
$\begin{align}
  & 3x=26-7y \\
 & \Rightarrow x=\dfrac{26-7y}{3} \\
\end{align}$
Now considering the second equation $4x+5y=13$
It can be written as
$\Rightarrow 5y=13-4x$
Putting the value of ‘x’ we got earlier in the above equation, we get
$\begin{align}
  & \Rightarrow 5y=13-4\times \dfrac{26-7y}{3} \\
 & \Rightarrow 5y=\dfrac{39-4\left( 26-7y \right)}{3} \\
 & \Rightarrow 5y\times 3=39-104+28y \\
 & \Rightarrow 15y-28y=-65 \\
 & \Rightarrow -13y=-65 \\
 & \Rightarrow y=\dfrac{-65}{-13} \\
 & \Rightarrow y=5 \\
\end{align}$
Putting the value of ‘y’ in the second equation, we get
$\begin{align}
  & 4x+5y=13 \\
 & \Rightarrow 4x+5\times 5=13 \\
 & \Rightarrow 4x+25=13 \\
 & \Rightarrow 4x=13-25 \\
 & \Rightarrow 4x=-12 \\
 & \Rightarrow x=\dfrac{-12}{4} \\
 & \Rightarrow x=-3 \\
\end{align}$

Hence the solution of the system of linear equations $3x+7y=26$ and $4x+5y=13$ is $\left( x,y \right)=\left( -3,5 \right)$.

Note: For solving simultaneous equations we can also make the coefficient of one variable same in both the equations by multiplying constants. Then adding or subtracting the modified equations one variable can be obtained. Putting that value in one of the equations and doing necessary calculations another variable can also be obtained.