Question

How do you solve \[x+2y=5\] and \[2x-3y=-4\]?

Answer 1

Hint: Assume the given equations as equation (1) and (2) respectively. Now, write the variable x in terms of variable ‘y’ considering the first equation. Substitute this value of ‘x’, founded in terms of ‘y’, in equation (2) and solve the equation for the value of y. Once the value of y is found, substitute it in equation (1) to find the value of x.

Complete step by step solution:
Here, we have been provided with two equations: \[x+2y=5\] and \[2x-3y=-4\] and we are asked to solve this system, that means we have to find the values of the variables x and y.
Now, let us assume the two given equations as equation (1) and equation (2) respectively, so we have,
\[\Rightarrow x+2y=5\] - (1)
\[\Rightarrow 2x-3y=-4\] - (2)
The method of substitution states that we have to select one of the two equations and write one of the variables in terms of the other. Now, we have to substitute this obtained value in the non – selected equation to convert it into an equation containing only one variable that can be solved easily.
Here, let us select equation (1), i.e., \[x+2y=5\]. So, writing the variable x in terms of the variable y, we get,
\[\Rightarrow x=5-2y\]
So, substituting the value of x in equation (2), we get,
\[\Rightarrow 2\left( 5-2y \right)-3y=-4\]
Removing the bracket and simplifying, we get,
\[\begin{align}
  & \Rightarrow 10-4y-3y=-4 \\
 & \Rightarrow -7y=-4-10 \\
 & \Rightarrow -7y=-14 \\
\end{align}\]
Dividing both the sides with -7 we get,
\[\Rightarrow y=2\]
So, we have obtained the value of variable y, therefore, substituting it in equation (1), we get,
\[\begin{align}
  & \Rightarrow x+2\times 2=5 \\
 & \Rightarrow x+4=5 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow x=5-4 \\
 & \Rightarrow x=1 \\
\end{align}\]
Hence, the solution of the given system of equations can be given as (x, y) = (1, 4).

Note: One may note that we can also select equation (2) in place of equation (1) and find the value of x in terms of y and proceed. This will also give the same answer. You may check the answer by substituting the values of x and y obtained, in the given equations. If L.H.S. and R.H.S. turns out to be the same for both the cases then our answer is correct. Remember that we can also solve the question by the elimination method or by the cross – multiplication method. But here you will be advised to use the substitution or elimination method because the chances of making mistakes in the cross – multiplication method is more due to its long expression.