Introduction Linear Equations and Inequations

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:

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

Download Linear Equations Worksheets
Linear Equations
grade 9 | Answers Set 1
Linear Equations
grade 9 | Questions Set 2
Linear Equations
grade 9 | Questions Set 1
Linear Equations
grade 9 | Answers Set 2
  
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