# Ex.4.1 Q2 Linear Equations in Two Variables Solution - NCERT Maths Class 9

## Question

** **Express the following linear equations in the form \(ax + by + c = 0\) and indicate the values of \(a, b, c\) in each case:

(i) \(\begin{align}2x + 3y = 9.3\overline 5\end{align}\)

(ii) \(\begin{align}x - \frac{y}{5} - 10 = 0\end{align}\)

(iii) \(\begin{align}{-2 x+3 y}={6}\end{align}\)

(iv) \(\begin{align}x={3 y}\end{align}\)

(v) \(\begin{align}{2 x}={-\,5 y}\end{align}\)

(vi) \(\begin{align}3 x+2={0}\end{align}\)

(vii) \(\begin{align}y-2={0}\end{align}\)

(viii) \(\begin{align}5={2 x}\end{align}\)

## Text Solution

**Steps:**

(i) Consider

\(\begin{align}&2x + 3y = 9.3\overline 5 \quad \dots {\rm{ Equation }}\left( {\rm{1}} \right)\\&\Rightarrow 2x + 3y - 9.3\overline 5 = 0\end{align}\)

Comparing this equation with the standard form of the linear equation in two variables, \(ax + by + c = 0\) we have,

- \(a=2{,}\)
- \(b=3{,}\)
- \(c=-9.3\overline {5}\)

(ii) Consider \(\begin{align}x-\frac{y}{5}-10=0 \quad \dots \text{Equation (1)}\end{align}\)

Comparing this equation with the standard form of the linear equation in two variables, \(ax + by + c = 0\) we have,

- \(a=1{,}\)
- \(\begin{align}b=-\frac{1}{5} \end{align}\)
- \(\begin{align}c=-10\end{align}\)

(iii) Consider

\(\begin{align}&-2 x+3 y=6 \quad \dots {\rm{ Equation }}\left( {\rm{1}} \right) \\ &{ \Rightarrow -2 x+3 y-6=0}\end{align}\)

Comparing this equation with the standard form of the linear equation in two variables, \(ax + by + c = 0\) we have,

- \(a=-2{,}\)
- \(a=-2{,}\)
- \(c=-6\)

(iv) Consider *\(x = 3y \quad \dots \text{Equation (1)}\)*

\(=> 1x - 3y + 0 = 0 \)

- \(a=1{,}\)
- \(b=-3\)
- \(c=0\)

(v) Consider \(2x = -5y \qquad \dots \text{Equation (1)}\)

\(\begin{align}{ \Rightarrow 2x + 5y + 0 = 0}\end{align}\)

- \(a=2{,}\)
- \(b=5{,}\)
- \(c=0\)

(vi) Consider \(3x + 2 = 0 \quad \dots \text{Equation (1)}\)

We can re-write Equation (\(1\)) as shown below,

\(\begin{align}{ 3x + 0y + 2 = 0}\end{align}\)

- \(a=3{,}\)
- \(b=0{,}\)
- \(c=2\)

(vii) Consider \(\rm y - 2 = 0\quad \dots\text{Equation (1)}\)

We can re-write Equation (\(1\)) as shown below,

\(\begin{align}{ 0 x+1 y-2=0}\end{align}\)

- \(a = 0,\)
- \(b = 1,\)
- \(c = -2\)

(vii) \(5 = 2x \quad \dots \text{Equation (1)}\)

\(\begin{align}{ 2x + 0y - 5 =\rm 0}\end{align}\)

- \(a = 2,\)
- \(b = 0,\)
- \(c = -5\)