Chapter: Quadratic and Linear Equations - Class 10

Linear equations are equations that involve only one variable of degree 1. A quadratic equation is an equation of degree 2 which means the highest power of the variable in the equation is 2. This chapter includes linear equations in two variables, important properties of inequalities, the nature of roots, the sum and product of roots, the formation of equations with given roots and concepts related to quadratic and linear equations.

Equation

Linear Equation in One Variable

Linear Equation in Two Variables

Pair of Linear Equations in Two Variables

Graphical Method of Solution of a Pair of Linear Equations

Algebraic Methods of Solving a Pair of Linear Equations

Important Properties of Linear Inequalities

Quadratic Equations

Solved Questions on Quadratic and Linear Equations

Equation

An equation is a statement that shows two expressions are equal. It can have one or more unknown variables.

Linear Equation in One Variable

A linear equation in one variable is an expression that involves only one variable of degree 1. The general form of a linear equation with one variable is:

ax + b = c

Where,

a and b are the constants.

x is the variable.

For example, 3x − 4 = 10 is a linear equation in one variable.

Solution of a Linear Equation

A solution to a linear equation is the value of the variable that makes the equation true. In other words, it is the value that satisfies the equation when substituted into the variable.

Example: What is the solution of equation 2x + 1.85 = 1.2?

a) x = 0.325
b) x = − 0.325
c) x = 0.625
d) x = − 0.625

Answer: b) x = − 0.325

Explanation: A solution of a linear equation is a value of the variable that makes the equation true. 

We are given 2x + 1.85 = 1.2

→ 2x = 1.2 − 1.85
→ 2x = − 0.65
→ x = − 0.65 / 2
→ x = − 0.325

Thus, x = − 0.325 is the solution of 2x + 1.85 = 1.2.

Note:

→ When a term is transposed from LHS to RHS, addition changes to subtraction and subtraction changes to addition.

→ When a term is transposed from LHS to RHS, multiplication changes to division and division changes to multiplication.

→ An equation remains unchanged if the same number is added or subtracted from both sides.

→ An equation remains unchanged if both sides are multiplied or divided by the same non-zero number.

→ While solving a linear equation in one variable, take all the terms involving the variable on one side and the constant terms on the other.

Linear Equation in Two Variables

→ A linear equation in two variables is an expression that involves two variables of degree 1.

The general form of a linear equation with two variables is:

ax + by = c

Where,

a, b and c are constants.

x and y are the variables.

For example, 3x + 5y = 12 is a linear equation in two variables.

→ A linear equation in two variables has infinitely many solutions.

→ The graph of every linear equation in two variables is a straight line.

→ x = 0 is the equation of Y axis and y = 0 is the equation of X axis.

→ The graph of x = a and y = b is a straight line parallel to the Y-axis and X axis, respectively.

→ An equation y = mx represents a line passing through the origin.

Pair of Linear Equations in Two Variables

A pair of linear equations in two variables is a system of two equations where each equation is of the form ax + by = c and there are two variables.
This system can be represented as:

a₁x + b₁y = c₁

a₂x + b₂y = c₂

Where,

a₁, a₂, b₁, b₂, c₁ and c₂ are constants.

x and y are variables.

a₁ and a₂ are coefficients of x.

b₁ and b₂ are coefficients of y.

Solution of Pair of Linear Equations in Two Variables

The solution of a pair of linear equations in two variables represents the values of the variables that make both equations true simultaneously. 

A pair of linear equations can be solved and represented by graphical and algebraic methods.

Graphical Method of Solution of a Pair of Linear Equations

The graphical method of solving a pair of linear equations involves plotting the lines represented by the equations on a coordinate plane and finding the point of intersection, if any.

The graph of a pair of linear equations in two variables is represented by two lines  a₁x + b₁y = c₁ and a₂x + b₂y = c₂.

→ If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.

→ If the lines coincide, then there are infinitely many solutions - each point on the line being a solution. In this case, the pair of equations is consistent.

→ If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent.

Example: On comparing the ratios a₁/a_2, b₁/b₂ and c₁/c₂, which pair of linear equations are consistent?

a) 2x + 3y = 4 and 4x + 6y = 12
b) ³⁄₂ x + ⁴⁄₅ y = 12 and 6x + ¹⁶⁄₅ y = 24
c) 1.5x + 4.2y = 2.2 and 4.5x + 12.6y = 6.6
d) √2 x + √5 y = 3 and √6 x + √15 y = 6

Answer: c) 1.5x + 4.2y = 2.2 and 4.5x + 12.6y = 6.6

Explanation: We know that a pair of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = _c2 is consistent if 

In option (a),

2x + 3y = 4 and 4x + 6y = 12
a₁ = 2, a₂ = 4, b₁ = 3, b₂ = 6, c₁ = 4 and c₂ = 12

So, the pair of linear equations in option (a) are inconsistent.

In option (b),

So, the pair of linear equations in option (b) are inconsistent.

In option (c),

1.5x + 4.2y = 2.2 and 4.5x + 12.6y = 6.6

a₁ = 1.5, a₂ = 4.5, b₁ = 4.2, b₂ = 12.6, c₁ = 2.2 and c₂ = 6.6

So, the pair of linear equations in option (c) are consistent.

In option (d),

√2 x + √5 y = 3 and √6 x + √15 y = 6

a₁ = √2, a₂ = √6, b₁ = √5, b₂ = √15, c₁ = 3 and c₂ = 6

So, the pair of linear equations in option (d) are inconsistent.

Algebraic Methods of Solving a Pair of Linear Equations

A pair of linear equations can be solved algebraically by any of the following methods:

1. Substitution Method

The substitution method is an algebraic technique used to solve a system of simultaneous linear equations.

In this method, the value of one variable from one equation is substituted in the other equation.

The steps involved are:

→ Find the value of one variable say x in terms of the other variable which is y.

→ Substitute this value of x in the other equation and reduce it to an equation in one variable.

→ Substitute the value of y obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

Example: If the cost of 3 apples and 4 oranges is $18 and the cost of 9 apples and 8 oranges is $42, then what is the cost of each apple and orange, respectively?

a) $3 and $2
b) $3 and $5
c) $2 and $3
d) $5 and $3

Answer: c) $2 and $3

Explanation: Let the cost of one apple be $x and the cost of one orange be $y.

Thus, the pair of linear equations formed are:

Cost of 3 apples and 4 oranges is $18.

3x + 4y = 18 … (1)

Cost of 9 apples and 8 oranges is $42.

9x + 8y = 42 … (2)

Now, express the value of x in terms of y from the equation (1),

x = ^18-4y/₃ … (3)

Now we substitute this value of x in Equation (2),

→ 9(^18-4y/₃) + 8y = 42

→ 3(18 − 4y) + 8y = 42

→ 54 − 12y + 8y = 42|

→ − 12y + 8y = 42 − 54

→ − 4y = − 12

→ y = ^-12/_-4

→ y = 3

Substitute this value of y = 3 in equation (3) to find the value of x.

→ x = ^18-4(3)⁄₃

→ x = ^18-12⁄₃

→ x = ⁶⁄₃

→ x = 2

Thus, the cost of one apple is $2 and the cost of one orange is $3.

2. Elimination Method

The elimination method is a technique used to solve systems of linear equations. It involves performing addition or subtraction operations on the equations to eventually obtain an equation in a single variable.

The steps involved are:

→ Arrange the equations so that the variables line up and multiply both equations by some non-zero numbers so that coefficients of either x or y are the same.

→ Add or subtract the equations to eliminate one variable.

→ Solve the resulting equation for the remaining variable.

→ Substitute this value in any of the given equations to find the other variable.

Example: What is the solution of the following pair of linear equations

5x + 4y = 10 and 15x + 24y = 40?

a) x = 4 / 5 and y = 5 / 6
b) x = 5 / 6 and y = 4 / 5
c) x = 5 / 6 and y = 4 / 3
d) x = 4 / 3 and y = 5 / 6

Answer: d) x = 4 / 3 and y = 5 / 6

Explanation: We are given

5x + 4y = 10 …(1)

15x + 24y = 40 …(2)

Multiply equation (1) by 3 to eliminate x,

⇒ 3 × (5x + 4y) = 3 × 10

⇒ 15x + 12y = 30 … (3)

Now, subtracting equation (3) from equation (2):

→ y = 10 / 12

→ y = 5 / 6

Put the value of y in equation (2) to find the value of x,

→ 15x + 24 (5 / 6) = 40

→ 15x + 4(5) = 40

→ 15x + 20 = 40

→ 15x = 40 − 20

→ 15x = 20

→ x = 20 / 15

→ x = 4 / 3

Thus, x = 4 / 3 and y = 5 / 6 are the solutions of the given pair of linear equations.

Important Properties of Linear Inequalities

→ A linear inequality is a mathematical statement that describes a relationship between two expressions using inequality symbols like "<" (less than), ">" (greater than), "≤" (less than or equal to) or "≥" (greater than or equal to).

→ The inequality will stay valid even if both sides are changed by the same positive value whether by addition, subtraction, multiplication or division.

→ You can switch any term from one side of the inequality to the other as long as you reverse its sign.

→ If both sides of an inequality are multiplied or divided by the same negative number, the direction of the inequality sign must be reversed.

For example: 3x + 5 < 8, 2x  − 7 > 11, x + 3 ≤ 7 and 4x − 9 6

Quadratic Equations

Quadratic equations are equations that involve a variable of degree 2.

The general form of a quadratic equation is:

ax² + bx + c = 0

Where,

a, b and c are constants and a 0.

x is the variable.

For example, 2x² + 3x + 4 = 0 is a quadratic equation.

Roots of a Quadratic Equation

The roots of a quadratic equation are the values of the variable that make the quadratic equation true. In other words, they are the values of x for which the quadratic equation equals zero.

Nature of Roots

→ The expression b² − 4ac is called the discriminant (D).

→ The nature of the roots depends entirely on the value of the discriminant (D).

Example: What is the discriminant of the quadratic equation 3x² + 5x − 12 = 0?

a) −144
b) 144
c) −169
d) 169

Answer: d) 169

Explanation: We are given 3x² + 5x − 12 = 0

Here a = 3, b = 5, c = − 12

We know that D = b2 − 4ac

→ D = (5)² − 4(3)(− 12)
→ D = 25 − (12)(− 12)
→ D = 25 + 144
→ D = 169

Thus, if a, b and c are rational, then

→ If D = b² − 4ac > 0 (i.e. positive), then the roots are real and unequal.

1. If D = b² − 4ac is a perfect square, then the roots are rational.
2. If D = b² − 4ac is not a perfect square, then the roots are irrational.

→ If D = b² − 4ac = 0 (i.e zero), then the roots are real and equal, each root is equal to − b / 2a. Thus, ax² + bx + c = 0 is a perfect square if D = 0. 

→ If D = b² − 4ac < 0 (i.e. negative), then the roots are imaginary (complex).

Sum and Product of Roots

If the two roots of a quadratic equation ax2 + bx + c = 0 are and , then

Formation of Equations with Given Roots

If we have to form the quadratic equation whose roots are α and β, then x = α and x = β.

So, (x − α) = 0 and (x − β) = 0

→ (x − α)(x − β) = 0

→ x² − (α + β)x + αβ = 0

→ x² − (Sum of roots)x + Product of roots = 0

Example: What is the quadratic equation whose roots are 3 and 5?

a) x² − 8x + 15 = 0
b) x² + 8x − 15 = 0
c) x² − 8x − 15 = 0
d) x² + 8x + 15 = 0

Answer: a) x² − 8x + 15 = 0

Explanation: We have 3 and 5 as two roots of the quadratic equation.

Sum of roots = 3 + 5 = 8

Product of roots = 3 × 5 = 15

We know that a quadratic equation is given by:

x² − (Sum of roots)x + Product of roots = 0

⇒ x² − 8x + 15 = 0

Special Roots

For a quadratic equation ax² + bx + c = 0, a ≠ 0 and a, b, c R whose roots are α and β.

→ Reciprocal Roots: If c = a, then the roots of a quadratic equation will be reciprocal to each other if the constant term is equal to the coefficient of x2.

→ Zero Roots:

Case 1: If c = 0, then one root is zero.
Case 2: If b = 0 and c = 0, then both the roots are zero.

→ Infinite Roots:

Case 1: If a = 0, then one root will be infinite.
Case 2: If a = 0 and b = 0, then both the roots will be infinite.