krit.club logo

Algebra - Introduction to Algebra

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

๐Ÿ”‘Concepts

โ€ข

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like xx or yy), and operators (like ++, โˆ’-, ร—\times, and รท\div).

โ€ข

Terms are parts of an expression separated by ++ or โˆ’- signs. For example, in 3x+53x + 5, 3x3x and 55 are terms.

โ€ข

A coefficient is the numerical factor of a term containing a variable. In the term 7x27x^2, 77 is the coefficient.

โ€ข

Like terms are terms that have the same variables raised to the same powers. Only like terms can be added or subtracted to simplify an expression (e.g., 2x+4x=6x2x + 4x = 6x).

โ€ข

Expansion (or multiplying out) is the process of removing brackets. The term outside the bracket multiplies every term inside the bracket: a(b+c)=ab+aca(b + c) = ab + ac.

โ€ข

Factorization is the inverse of expansion. It involves finding the highest common factor (HCF) of all terms and placing it outside a bracket.

โ€ข

Substitution is the process of replacing variables with given numerical values to find the value of an expression.

๐Ÿ“Formulae

a(b+c)=ab+aca(b + c) = ab + ac

(a+b)(c+d)=ac+ad+bc+bd(a + b)(c + d) = ac + ad + bc + bd

xaร—xb=xa+bx^a \times x^b = x^{a+b}

xaxb=xaโˆ’b\frac{x^a}{x^b} = x^{a-b}

(xa)b=xab(x^a)^b = x^{ab}

๐Ÿ’กExamples

Problem 1:

Simplify the expression: 4x+7yโˆ’2x+3y+54x + 7y - 2x + 3y + 5

Solution:

2x+10y+52x + 10y + 5

Explanation:

Group the like terms together: Group 1 (xx terms): 4xโˆ’2x=2x4x - 2x = 2x Group 2 (yy terms): 7y+3y=10y7y + 3y = 10y Group 3 (constants): 55 Combine them to get 2x+10y+52x + 10y + 5.

Problem 2:

Expand and simplify: 3(2xโˆ’4)+2(x+5)3(2x - 4) + 2(x + 5)

Solution:

8xโˆ’28x - 2

Explanation:

First, expand both brackets: 3ร—2x=6x3 \times 2x = 6x 3ร—โˆ’4=โˆ’123 \times -4 = -12 2ร—x=2x2 \times x = 2x 2ร—5=102 \times 5 = 10 This gives: 6xโˆ’12+2x+106x - 12 + 2x + 10. Now, collect like terms: 6x+2x=8x6x + 2x = 8x โˆ’12+10=โˆ’2-12 + 10 = -2 Result: 8xโˆ’28x - 2.

Problem 3:

Factorize completely: 12x2yโˆ’18xy212x^2y - 18xy^2

Solution:

6xy(2xโˆ’3y)6xy(2x - 3y)

Explanation:

Identify the Highest Common Factor (HCF) for the coefficients and the variables: HCF of 1212 and 1818 is 66. HCF of x2x^2 and xx is xx. HCF of yy and y2y^2 is yy. The total HCF is 6xy6xy. Divide each term by 6xy6xy to find the terms inside the bracket: 12x2y6xy=2x\frac{12x^2y}{6xy} = 2x 18xy26xy=3y\frac{18xy^2}{6xy} = 3y Result: 6xy(2xโˆ’3y)6xy(2x - 3y).

Problem 4:

Evaluate the expression 2a2โˆ’3b2a^2 - 3b when a=3a = 3 and b=โˆ’4b = -4.

Solution:

3030

Explanation:

Substitute the values into the expression: 2(3)2โˆ’3(โˆ’4)2(3)^2 - 3(-4) Calculate the square first: 32=93^2 = 9 2(9)โˆ’3(โˆ’4)2(9) - 3(-4) Multiply: 18โˆ’(โˆ’12)18 - (-12) Subtracting a negative is the same as adding: 18+12=3018 + 12 = 30.

Problem 5:

Solve for xx: 5xโˆ’7=135x - 7 = 13

Solution:

x=4x = 4

Explanation:

To isolate xx, first add 77 to both sides: 5x=13+75x = 13 + 7 5x=205x = 20 Next, divide both sides by 55: x=205x = \frac{20}{5} x=4x = 4.