Maths Tutorials

Algebra

Algebra is the branch of mathematics that studies algebraic structures and the operations they use. It is the study of solving for unknown quantities and values.

What is an algebraic expression? It is an expression built up from constants, variables, basic algebraic operators (+,-,x,/), whole number powers, and roots (fractional powers).

What is an algebraic equation? An equation can be defined as a mathematical statement in which two expressions are equal to each other.

algebra
algebra

Solving for unknown variables:

In an algebraic equation, there are variables whose value has to be determined. If there is one variable, one equation is sufficient to determine its value. If there are 2 variables, 2 equations will be needed to determine the values of both variables. Similarly for 3, and so on so forth.

i) 2x + 6 = 13x - 9
In this equation, in order to solve for ‘x’, first all terms containing the variable ‘x’ should be taken to one side and the remaining constants to the other. When the sides are switched, the operator changes either from + to -, - to +, x to / or / to x.
6 + 9 = 13x - 2x
15 = 11x
In 2 or more terms, if the variables in that term have the same degree (power), their coefficients can be summed up or subtracted depending on the operator.
Now x can be evaluated by dividing both sides by 11: x = 15/11

Example

Following the exact same rules as mentioned in example:

  • ii) 3x - 11x = 7x + 19
  • iii) -8x = 7x + 19
  • iv) -19 = 7x + 8x
  • v) -19 = 15x
  • vi) -19/15 = x

Solving Simultaneous Equations

When there are 2 variables to be solved for, 2 equations are needed in order to calculate the values of both variables. Similarly, for 3, 4, 5, and so on number of variables.

i) 2x + 3y = 17, 6x + 2y = 13

In order to solve for x and y, follow these steps:

Step 1: The coefficient of any one variable should be the same in both equations.
Step 2: Multiply the equations by a certain number so that after multiplication, the coefficients of one variable are the same in both equations.

Multiply equation 1 by 3:
6x + 9y = 51, 6x + 2y = 13

Step 3: Now, in both equations the coefficient of x is the same.
Step 4: Subtract the equations from each other so that one variable is completely removed.

6x + 9y - 6x - 2y = 51 - 13
7y = 38
y = 38 / 7

Step-by-Step Solution: Example i

  1. Substitute the value of y in one of the equations to solve for x:
    6x + 9 * (38/7) = 51
  2. Simplify the equation:
    6x = 51 - 342 / 7
    6x = (357 - 342) / 7
    6x = 15 / 7
    x = 15 / 42

Note

Verify Solutions: Always check if the derived values satisfy the original equations.
Fractions: Added a tip to clear fractions early for easier manipulation.

Step-by-Step Solution: Example ii

Equations: 2x = 81 - 9y - 3x, 5y - 3x = 21 - 7y

  1. Rearrange to:
    5x + 9y = 81, 12y - 3x = 21
  2. Multiply equation 1 by 3 and equation 2 by 5:
    15x + 27y = 243, 60y - 15x = 105
  3. Add the equations to eliminate x:
    27y + 60y = 348
    87y = 348
    y = 348 / 87
  4. Substitute y back into an equation to solve for x:
    5x + 9(348/87) = 81
    5x + 3132 / 87 = 81
    5x = 81 - 3132 / 87
    5x = (7047 - 3132) / 87
    5x = 3915 / 87
    x = 3915 / 435 = 783 / 87 = 261 / 29
Quadratic Equations

A quadratic equation is an equation of degree two. It is a second order polynomial equation.

A polynomial equation is an equation that has at least one algebraic term with at least one variable, and all exponents are integers that are equal to or greater than zero

Graph of a Quadratic Equation

Graph of a linear equation

An equation of degree 2 will always have 2 solutions. In linear equations, there exists only one solution as the degree is one, however, in quadratic equations, there exists two solutions to an equation. They are of the form Solving Quadratic Equations

Solving Quadratic Equations

i) x² + 6x + 8 = 0

Given: a = 1, b = 6, c = 8

To solve this, we need to find two numbers whose product is (a × c) and whose sum is b.

In this case, a × c = 8 and b = 6. The two numbers are 2 and 4:
2 × 4 = 8 and 2 + 4 = 6.

The equation then becomes: x² + 2x + 4x + 8 = 0

Factorizing the equation:

Now, we can factorize the equation by adding 2x and 4x:

x(x + 2) + 4(x + 2) = 0 → (x + 2)(x + 4) = 0

For the equation to be equal to 0, either (x + 2) should be 0 or (x + 4) should be 0.

Therefore, x + 2 = 0 or x + 4 = 0, which gives the solutions:

Solution: x = -2, x = -4

ii) 2x² − x = 6

Given: 2x² − x − 6 = 0

In this case, a × c = 2 × (-6) = -12 and b = -1.

We need to find two numbers whose product is -12 and whose sum is -1.

The two numbers are -4 and 3:
-4 × 3 = -12 and -4 + 3 = -1.

The equation becomes: 2x² − 4x + 3x − 6 = 0

Factorizing the equation:

Now, we can factorize the equation as follows:

2x(x − 2) + 3(x − 2) → (2x + 3)(x − 2) = 0

For the equation to be equal to 0, either (2x + 3) should be 0 or (x − 2) should be 0.

Therefore, 2x + 3 = 0 or x − 2 = 0, which gives the solutions:

Solution: x = 2 or x = -3/2

If the quadratic equation does not give 2 whole numbers whose product is a x c and whose sum is b we can use the quadratic formula to find the solutions which won’t be whole numbers

This formula gives 2 solutions because there is a + and a - in the numerator Since there are two solutions to a quadratic equation, sometime the sum and product of both the solutions need to be found

The product of both solutions to a quadratic equation is given by: c/a The sum of both solutions to a quadratic equation is given by: -b/a

Discriminant of a Quadratic Equation

A quadratic equation can have two solutions, but these solutions may not always be distinct. They can be the same, or they could be imaginary.

The discriminant of a quadratic equation is given by D.

If a quadratic equation ax² + bx + c = 0 has distinct solutions, then D > 0.

If the solutions are equal, then D = 0.

If the solutions do not exist and are imaginary, then D < 0.

The discriminant is calculated using the formula: D = b² − 4ac

This value is inside the square root function of the quadratic formula.

What is an Imaginary Number?

It is a number that does not exist in the real plane

Here, to find the value of i, we need to find the square root of a negative number which cannot be calculated. Hence i is the square root of negative one. This is called the imaginary unit.

Understanding Exponents

An exponent is the degree or power to which a variable or function is raised. For example, in 3², the number 3 is raised to the power of 2. When a number or function is raised to a power n, it means you multiply that number or function by itself n times.

Examples:

  • 3² = 3 × 3
  • 4⁶ = 4 × 4 × 4 × 4 × 4 × 4
  • 7⁹ = 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7

Laws of Exponents

Functions and Inverse Functions

Functions

A function is a relation between a set of inputs each producing an output. A function is usually defined starting with a letter and the variable in bracket, i.e., f(x).

Example:

f(x) = 2x³ + 6x² + 5x + 3

For an input x = 1: f(1) = 2(1)³ + 6(1)² + 6(1) + 3 = 17 is the output

x = 3: f(3) = 2(3)³ + 6(3)² + 5(3) + 3 = 54 + 54 + 15 + 3 = 126 is the output

Types of Function

Inverse Functions

Inverse functions do exactly the opposite of what functions do. A function takes an input to give an output. The inverse of that very function takes that output and produces the input.

Example:

Let y = 7x + 18

Here x is the input which gives y as the output. The inverse of this function will be in terms of x:

x = (y − 18) / 7 is the inverse of the function

The inverse of a function f(x) is given by f⁻¹(x)

Another Example:

f(x) = 9x − 17 = y

x = (y + 17) / 9 → f⁻¹(y) = (y + 17) / 9

If 1 is the input for the function f then f(1) = 9(1) − 17 = −8

-8 is the output and 1 was the input.

If we substitute -8 in our inverse function we get f⁻¹(−8) = (−8 + 17) / 9 = 1

Hence the output produced in the function f gives the input of f using its inverse

Note: Inverse functions only exist if the function is bijective i.e. one to one and onto. Refer to types of functions to understand more. Trigonometric functions (sin, cos, tan, cot, cosec and sec) all have inverse functions however this is only possible in certain intervals and not everywhere.

Trigonometric Example:

sinx = y → sin⁻¹(y) = x

sin 90 = 1 → sin⁻¹(1) = 90

Principal Intervals for Inverse Trigonometric Functions:

Function Principal Interval
Sin [-90°, 90°]
Cos [0°, 180°]
Tan (-90°, 90°)
Cot (0°, 180°)
Sec [0°, 90°) and (90°, 180°]
Cosec [-90°, 0°) and (0°, 90°]

Arithmetic and Geometric Progression

Arithmetic Progression (AP)

Arithmetic progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value.

1, 4, 7, 10, 13, 16, 19, 22… is an AP as the difference between successive terms is constant

1, 7, 10, 12, 19, 25, 34…. Is not an AP as difference between successive terms isn't constant

Formula for nth term of an AP:

t = a + (n − 1)d

where t is the nth term, a is the first term, d is the difference between successive terms

Formula for sum of terms up to the nth term of an AP:

S = n/2[2a + (n − 1)d]

where S is the sum up till n terms

Examples:

1. Find the 7th and 11th terms of the series 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31….

t = 1 + (7 − 1)(3) = 19

t = 1 + (11 − 1)(3) = 31

2. Find the sum up till 6 terms in the series 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 …..

S = 6/2[2(5) + (6 − 1)(2)] = 3(20) = 60

Geometric Progression (GP)

A geometric progression (GP) is a series with a constant ratio between successive numbers in the series (e.g. 1, 3, 9, 27, 81, 243…)

Formula for nth term in a GP:

t = aqm where m = n-1

Formula for sum of n terms in a GP:

S = a(1 - rn) / (1 - r), where r ≠ 1

S = na, where r = 1

Sum of n terms in an infinite geometric series:

Examples:

1. Find the 8th term in the series 3, 6, 12, 24, 48 …… n=8

m=8-1=7, a = 3, q = 2

t = 3(2)7 = 3(128) = 384

2. Find the 9th term in the series 2, 8, 32, 128, 512 …

m = 9-1 = 8 a=2, q = 4

t = 2(4)8 = 2(65536) = 131072

3. Find the sum up till the 5th term in the series 1, 3, 9 ….. a=1, q = 3, n=5

S = (1(35 − 1))/(3 − 1) = 1(242)/2 = 121

Linear Inequalities

A linear inequality shows the relation between two or more expressions.

Example: 2x + 3 > 3x - 9

To solve this inequality, the rules are followed for finding the solution for an equation:

12 > 3x - 2x —> x < 12

From this relation, the value of x is found to be less than 12

Linear inequalities don't give the exact value of a variable but show the range in which the variable might exist.

Examples of Linear Inequalities

Example 1:

2x - 3 < 6x + 9 < 18 - 3x

  1. 2x - 3 < 6x + 9

    -12 < 4x —> x > -3

  2. 6x + 9 < 18 - 3x

    9x < 9 —> x < 1

Solution: -3 < x < 1

Example 2:

3x + 11 < 7x + 3 < 9 - x

  1. 3x + 11 < 7x + 3

    8 < 4x —> x > 2

  2. 7x + 3 < 9 - x

    8x < 6 —> x < 3/4

Solution: 2 < x < 3/4

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