Question

Solve the following pair of linear equations by the substitution method.
$0.2x+0.3y=1.3,\text{ }0.4x+0.5y=2.3$
(a) x =1, y =-2
(b) x = 6, y =-7
(c) x =5, y =1
(d) x=2, y=3

Answer 1

Hint: Here, we will first obtain the value of x in terms of y from one of the equations. Then, we will substitute that value of x in the other equation and obtain the value of y. After that we can get the value of x.

Complete step-by-step answer:

While using a substitution method to solve a pair of linear equations, we use the first equation to find the value of one variable in terms of another and substitute in the 2nd equation. Thereafter, on simplification we get the value of one variable and substituting it in any one of the given equations gives us the value of another variable.
Here, the equations given to us are:
$\begin{align}
  & 0.2x+0.3y=1.3..............\left( 1 \right) \\
 & 0.4x+0.5y=2.3..............\left( 2 \right) \\
\end{align}$
From equation we can write the value of x in terms of y as:
$x=\dfrac{1.3-0.3y}{0.2}$
On substituting this value of x in equation (2), we get:
$\begin{align}
  & 0.4\times \left( \dfrac{1.3-0.3y}{0.2} \right)+0.5y=2.3 \\
 & \Rightarrow 2\times \left( 1.3-0.3y \right)+0.5y=2.3 \\
 & \Rightarrow 2.6-0.6y+0.5y=2.3 \\
 & \Rightarrow 2.6-2.3=0.6y-0.5y \\
 & \Rightarrow 0.3=0.1y \\
 & \Rightarrow y=\dfrac{0.3}{0.1}=3 \\
\end{align}$
So, the value of y comes out to be y = 3.
Substituting this value of y in equation (1), we get:
$\begin{align}
  & 0.2x+0.3\times 3=1.3 \\
 & \Rightarrow 0.2x+0.9=1.3 \\
 & \Rightarrow 0.2x=1.3-0.9 \\
 & \Rightarrow 0.2x=0.4 \\
 & \Rightarrow x=\dfrac{0.4}{0.2}=\dfrac{4}{2}=2 \\
\end{align}$
So, the value of x comes out to be x = 2.
Therefore, the solution of the given pair of linear equations is x=2 and y =3.
Hence, option (d) is the correct answer.

Note: Students should note here that it is not necessary to obtain x in terms of y. We can also find the value of y in terms of x and then proceed. The calculations must be done properly to avoid mistakes.