An equation is a statement of equality which involves one or more unknown quantities, called the variables.
An equation involving only linear polynomials is called a Linear Equations.
The following are some examples of linear equations.

• 2x + 3 = 1
• $\frac{3}{2}$x + 5 = 2x + 7
• $\frac{5}{3}$y – 16 = 3y + 4

Solving a linear equation

Solving a linear means finding a value of the variable which satisfies the equation.
While solving a linear equation, following properties of equality are to be remembered.

• Same quantity can be added to both sides of an equality.
• Same quantity can be subtracted from both sides of an equality.
• Both sides of an equality can be multiplied by the same quantity (≠0).
• Both sides of an equality can be divided by the same Non-zero quantity.
• While transposing a term from one side to the other side of equality, the sign of the term must be changed.
• The terms involving the variable are kept on one side (LHS) of equality and terms not involving the variable are on other side.

Applications of linear equations

In Algebra, linear equations are used to solve some practical problems.
Following are the basic steps needed to solve such problems.
Step 1: Denote the unknown quantity by x. If there are more than one unknown quantities, then denote any one of these by ‘x’ and write the other in terms of ‘x’.

Step 2: From the information given in the problem, formulate a linear equation in x.

Step 3: Solve the linear equation to find ‘x’ (unknown quantity).

Step 4: Put the value of x is in the linear equation to find out the unknown quantities if there are more than one in number.

Systems of Linear Equations

• linear equation in one variable : A linear equation in one variable (x) of the form either ax+b>0 or ax+b$\geq$ 0 or ax + b note : step-1 collect all term containing the variable on the left side and the constants on the right side
• Divide this equation by the coefficient of the variable .This gives the solution of the given equation .

Example: solve the linear equation 2x + 7b $\geq$ 13 - x
Solution: 2x + 7b $\geq$ 13 – x
$\Rightarrow (2x+7)+x\: \geq (13-x)+x\Rightarrow 3x+7\geq 13$
$(3x+7)-7\geq 13-7\: 3x\geq 6\: \Rightarrow x\geq 2$
$\therefore x\in [2,\infty ]$ this solution can also be on the number line .

1. Two or more linear equation in one variable constitutes a system of llinear equation in one variable:

Step 1: solve each equation and write their solution sets in terms of interval s

Step 2: find the intersection of all the solution sets of inequation given in the system.

Step 3: the intersection represent the solution of the given system of linear equations.
You will recall that while solving linear equations, we followed the following rules

Rule 1: Equal numbers may be added to (or subtracted from) both sides of an equation.

Rule 2: Both sides of an equation may be multiplied (or divided) by the same non-zero number.

In the case of solving inequalities, we again follow the same rules except with a difference that in rule 2 the sign of inequality is reversed (that is < become >,$\leq$becomes $\geq$ and so on ) whenever we multiply (or divide) both sides of an inequality by a negative number. It is evident from the facts that.
3 > 2 while – 3 < – 2,
– 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14.

Thus, we state the following rules for solving an inequality:

Rule 1: Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality.

Rule 2: Both sides of an inequality can be multiplied (or divided) by the same positive number. But when both sides are multiplied or divided by a negative number, then thesign of inequality is reversed.

Now, let us consider some examples.

Example 1: Solve 30 x < 200 when

Step 1. x is a natural number, 2.x is an integer.
Solution: we have given 30x < 200
or$\frac{30x}{30}< \frac{200}{30}$(Rule 2), i.e., x < 20 / 3.

Step 2.When x is a natural number, in this case the following values of x make the
statement true. 1, 2, 3, 4, 5, 6.
The solution set of the inequality is {1,2,3,4,5,6}

Step 3.When x is an integer, the solutions of the given inequality are ..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6
The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}

Example 2: Solve 5x – 3 < 3x +1 when
(i) x is an integer, (ii) x is a real number.
Solution We have, 5x –3 < 3x + 1
or 5x –3 + 3 < 3x +1 +3 (Rule 1)
or 5x < 3x +4

or 5x – 3x < 3x + 4 – 3x (Rule 1)
or 2x < 4

or x < 2 (Rule 2)

(i) When x is an integer, the solutions of the given inequality are
..., – 4, – 3, – 2, – 1, 0, 1

(ii) When x is a real number, the solutions of the inequality are given by x < 2,
i.e., all real numbers x which are less than 2. Therefore, the solution set of the inequality is x ∈ (– ∞, 2).

We have considered solutions of inequalities in the set of natural numbers, set of integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall solve the inequalities in this Chapter in the set of real numbers.

Example 3: Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.
Solution: We have 7x + 3 < 5x + 9 or 2x < 6 or x < 3

The graphical representation of the solutions are given in

Graphical Solution of Linear Inequalities in Two Variables

In earlier section, we have seen that a graph of an inequality in one variable is a visual representation and is a convenient way to represent the solutions of the inequality. Now, we will discuss graph of a linear inequality in two variables. We know that a line divides the Cartesian plane into two parts. Each part is known as a half plane. A vertical line will divide the plane in left and right half planes and a non-vertical line will divide the plane into lower and upper half planes.
A point in the Cartesian plane will either lie on a line or will lie in either of the half planes I or II. We shall now examine the relationship, if any, of the points in the plane and the inequalities ax + by < c or ax + by > c. Let us consider the line ax + by = c, a ≠ 0, b ≠ 0

There are three possibilities namely:
(i) ax + by = c (ii) ax + by > c, (iii) ax + by < c.

In Case (i), clearly, all points (x, y) satisfying (i) lie on the line it represents and conversely. Consider

Case (ii), let us first assume that b > 0. Consider a point P (α,β) on the line ax + by = c, b > 0, so that aα + bβ = c. Take an arbitrary point Q (α , γ) in the half plane II Now, we interpret, γ > β (Why?) or bγ> bβ or aα + b γ > aα + bβ (Why?) or aα + b γ > c i.e., Q(α, γ ) satisfies the inequality ax + by > c.
Thus, all the points lying in the half plane II above the line ax + by = c satisfies
the inequality ax + by > c. Conversely, let (α, β) be a point on line ax + by = c and an arbitrary point Q(α, γ ) satisfying ax + by > c so that aα + bγ> c ⇒ aα + b γ > aα + bβ (Why?) ⇒ γ > β (as b > 0)
This means that the point (α, γ ) lies in the half plane II.
Thus, any point in the half plane II satisfies ax + by > c, and conversely any point satisfying the inequality ax + by > c lies in half plane II. In case b < 0, we can similarly prove that any point satisfying ax + by > c lies in the half plane I, and conversely. Hence, we deduce that all points satisfying ax + by > c lies in one of the half planes II or I according as b > 0 or b < 0, and conversely. Thus, graph of the inequality ax + by > c will be one of the half plane (called solution region) and represented by shading in the corresponding half plane.

Note:

• The region containing all the solutions of an inequality is called the solution region.
• In order to identify the half plane represented by an inequality, it is just sufficient to take any point (a, b) (not online) and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half plane and shade the region. which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point (0, 0) is preferred.
• If an inequality is of the type ax + by ≥ c or ax + by ≤ c, then the points on the line ax + by = c are also included in the solution region. So draw a dark line in the solution region.
• If an inequality is of the form ax + by > c or ax + by < c, then the points on the line ax + by = c are not to be included in the solution region. So draw a broken or dotted line in the solution region

Example: We obtained the following linear inequalities in two variables x and y: 40x + 20y ≤ 120

Solution: Let us now solve this inequality keeping in mind that x and y can be only whole numbers, since the number of articles cannot be a fraction or a negative number. In this case, we find the pairs of values of x and y, which make the statement (1) true. In fact, the set of such pairs will be the solution set of the inequality (1). To start with, let x = 0. Then L.H.S. of (1) is
40x + 20y = 40 (0) + 20y = 20y.
Thus, we have
20y ≤ 120 or y ≤ 6 ... (2)

For x = 0, the corresponding values of y can be 0, 1, 2, 3, 4, 5, 6 only. In this case, the
solutions of (1) are (0, 0), (0, 1), (0,2), (0,3), (0,4), (0, 5) and (0, 6).
Similarly, other solutions of (1), when
x = 1, 2 and 3 are: (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (3, 0)
This is shown in Fig 6.6

Let us now extend the domain of x and y from whole numbers to real numbers, and see what will be the solutions of
(1) In this case. You will see that the graphical method of solution will be very convenient in this case. For this purpose, let us consider the (corresponding) equation and draw its graph.
40x + 20y = 120 ... (3)
In order to draw the graph of the inequality

(1) We take one point say (0, 0), in half plane I
and check whether values of x and y satisfy the inequality or not We observe that x = 0, y = 0 satisfy the inequality. Thus, we say that the half plane I is the graph (Fig 6.7) of the inequality. Since the points on the line also satisfy the inequality (1) above, the line is also a part of the graph. Thus, the graph of the given inequality is half plane I including the line itself. Clearly half plane II is not the part of the graph. Hence, solutions of inequality (1) will consist of all the points of its graph (half plane I including the line).
We shall now consider some examples to explain the above procedure for solving a linear inequality involving two variables.

Point to be Remember

• Two real numbers or two algebraic expressions related by the symbols <, >, ≤or ≥ form an equality.
• Equal numbers may be added to (or subtracted from ) both sides of an inequality. Both sides of an inequality can be multiplied (or divided ) by the same positive number. But when both sides are multiplied (or divided) by a negative number,then the inequality is reversed.
• The values of x, which make an inequality a true statement, are called solutions of the inequality.
• To represent x < a (or x > a) on a number line, put a circle on the number a and dark line to the left (or right) of the number a.
• To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the numbers and dark the line to the left (or right) of the number x.
• If an inequality is having ≤ or ≥ symbol, then the points on the line are also included in the solutions of the inequality and the graph of the inequality lies left(below) or right (above) of the graph of the equality represented by dark line that satisfies an arbitrary point in that part.
• If an inequality is having < or > symbol, then the points on the line are not included in the solutions of the inequality and the graph of the inequality lies to the left (below) or right (above) of the graph of the corresponding equality represented by dotted line that satisfies an arbitrary point in that part.
• The solution region of a system of inequalities is the region which satisfies all the given inequalities in the system simultaneously.

Linear Equations Solutions and Questions

Linear Equations Shortcut Tricks and Study Material - Video