# Introduction Linear Equations and Inequations

Go back to  'Two Variable Linear Equations'

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:

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

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