In earlier classes, you have encountered linear equations in one variable. For example, consider the following equation:

\[\ 2x + 1 = - 7\]

This is a linear equation in the variable *x*. There is no other variable. It can be solved to obtain \(x = - 4\) as the solution.

Now, we will consider *linear equations in two variables*. Consider the following equation:

\[x + 2y = 4\]

This is a linear equation in the variables *x* and *y*. It has two variable terms (the terms containing *x* and *y* and a constant term). We can write this equation as follows:

\[x + 2y - 4 = 0\]

When written this way (all the terms on one side and 0 on the other side), a two-variable linear equation is said to be in *standard form*. Thus, an arbitrary two-variable linear equation can be written in standard form as

\[ax + by + c = 0\]

Note that there are two variables (*x* and *y*) and three constants (*a*, *b* and *c*). Using *x* and *y* for the variables is the general convention, but is not necessary. For example, the equation \(3u + 4v + 5 = 0\) is a linear equation in the variables *u* and *v*. However, we will generally use *x* and *y* for the variables.

**Example 1:** Which of the following equations are two-variable linear equations? \(x,\,\,y,\,\,z\) are variables and all the other literals are constants.

(A) \(x - \sqrt 3 y + 1 = 0\)

(B) \({a^2}x + {b^2}y + {c^2} = {d^2}\)

(C) \(x - 3y + 4 = - 2z\)

(D) \(2y - 3x + xy = 1\)

(E) \(\) \(5 - 2x = \frac{y}{{\sqrt 2 }}\)

**Solution:**. The equations in (A), (B) and (E) are two-variable linear equations. Let us write each of them in standard form:

\[x - \sqrt 3 y + 1 = 0\,\,\, \to \,\,\,\left( 1 \right)x + \left( { - \sqrt 3 } \right)y + \left( 1 \right) = 0\]

\[{a^2}x + {b^2}y + {c^2} = {d^2}\,\,\, \to \,\,\,\left( {{a^2}} \right)x + \left( {{b^2}} \right)y + \left( {{c^2} - {d^2}} \right) = 0\]

\[5 - 2x = \frac{y}{{\sqrt 2 }}\,\,\, \to \,\,\,\left( { - 2} \right)x + \left( { - \frac{1}{{\sqrt 2 }}} \right)y + 5 = 0\]

The equation in (C) is a linear equation in three variables. The equation in (D) is not a linear equation, as it contains the *cross* term \(xy\).

To summarize, if you are able to write an equation in the form \(ax + by + c = 0\), this means that the equation is a two-variable linear equation.

**Example-2:** Write each of the following equations in standard form and identify the coefficients *a*, *b* and *c*.

I.\( - \sqrt 3 y = \pi - \sqrt 5 x\)

II.\(\left( {x - 1} \right)\left( {y - 3} \right) = xy + 1\)

III.\(\frac{{x - 1}}{{y - 3}} = \frac{2}{5}\)

**Solution:** I. We can write the given equation as

\[\sqrt 5 x - \sqrt 3 y - \pi = 0\]

Thus,

\[a = \sqrt 5 ,\,\,\,b = - \sqrt 3 ,\,\,\,c = \pi \]

We could also have written the same equation with all the signs reversed:

\[ - \sqrt 5 x + \sqrt 3 y + \pi = 0\]

In this case, the coefficients would have been

\[a = - \sqrt 5 ,\,\,\,b = \sqrt 3 ,\,\,\,c = - \pi \]

Both are correct.

II. Expanding the expression on the left, we have:

\[\begin{array}{l}xy - 3x - y + 3 = xy + 1\\ \Rightarrow \,\,\, - 3x - y + 2 = 0\\ \Rightarrow \,\,\,a = - 3,\,\,\,b = - 1,\,\,\,c = 2\end{array}\]

III. Cross-multiplying and rearranging to standard form, we have:

\[\begin{array}{l}5\left( {x - 1} \right) = 2\left( {y - 3} \right)\\ \Rightarrow \,\,\,5x - 5 = 2y - 6\\ \Rightarrow \,\,\,5x - 2y + 1 = 0\\ \Rightarrow \,\,\,a = 5,\,\,\,b = - 2,\,\,\,c = 1\end{array}\]