Comprehensive Guide to Algebraic Simplification, Evaluation, Expansion, Factorisation, and Formula Applications

Algebraic Expression Simplification via Like Terms

An algebraic term consists of a coefficient multiplied by one or more variables raised to specific powers. Like terms are defined as terms that possess the exact same variable parts, including identical exponents, regardless of their numerical coefficients. Terms with different variables or identical variables with different exponents are unlike terms and cannot be combined through basic addition or subtraction.

To simplify an algebraic expression, like terms are grouped together and their numerical coefficients are combined using standard arithmetic operations. Commutativity allows terms to be rearranged along with their preceding operational signs.

For the expression $7x + 4y + 2x - y$, the like terms containing the variable $x$ are $7x$ and $2x$, while the like terms containing the variable $y$ are $4y$ and $-y$.

$7x + 2x = 9x$

$4y - y = 3y$

Combining these results yields the simplified expression:

$9x + 3y$

For the expression $5a^2 + 3a + 2a^2 - 7$, the quadratic terms containing $a^2$ are $5a^2$ and $2a^2$. The linear term $3a$ and the constant term $-7$ have no matching like terms.

$5a^2 + 2a^2 = 7a^2$

Combining all remaining components yields the simplified expression:

$7a^2 + 3a - 7$

For the expression $9ab - 3a + 2ba + 5a$, the commutative property of multiplication establishes that $ba = ab$, which makes $9ab$ and $2ba$ like terms. Additionally, $-3a$ and $5a$ are like terms containing the single variable $a$.

$9ab + 2ba = 11ab$

$-3a + 5a = 2a$

Combining these results yields the simplified expression:

$11ab + 2a$

Algebraic Evaluation and Substitution

Evaluating an algebraic expression involves replacing each variable with its assigned numerical value and carrying out the required arithmetic operations according to the standard order of operations. Special care must be given to negative numbers, especially when squaring negative terms or subtracting negative quantities.

Given the assigned variable values $a = 4$ and $b = -3$:

For the linear expression $3a + 2b$, substitute $4$ for $a$ and $-3$ for $b$:

$3a + 2b = 3 \times 4 + 2 \times (-3)$

$3 \times 4 = 12$

$2 \times (-3) = -6$

$12 + (-6) = 6$

For the expression $a^2 - b$, substitute $4$ for $a$ and $-3$ for $b$:

$a^2 - b = 4^2 - (-3)$

$4^2 = 16$

$-(-3) = 3$

$16 + 3 = 19$

For the expression $2(a - b)$, substitute $4$ for $a$ and $-3$ for $b$ into the grouped terms first:

$2(a - b) = 2(4 - (-3))$

$4 - (-3) = 4 + 3 = 7$

$2 \times 7 = 14$

Expansion and Factorisation of Algebraic Expressions

Expanding an algebraic expression involves removing parentheses by multiplying the term outside the brackets by every term inside the brackets, applying the distributive law $k(x + y) = kx + ky$. Factorisation is the inverse operation, which transforms an expanded polynomial into a product of factors by identifying and extracting the greatest common factor across all terms.

For the expansion of $4(3x - 5)$, distribute the factor $4$ to each term inside the parentheses:

$4 \times 3x = 12x$

$4 \times (-5) = -20$

Combining these products gives the expanded expression:

$12x - 20$

For the factorisation of $12y + 18$, identify the highest common factor of the numerical coefficients $12$ and $18$. The factors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$. The factors of $18$ are $1$, $2$, $3$, $6$, $9$, and $18$. The greatest common factor is $6$.

Dividing each term by $6$ yields:

$\frac{12y}{6} = 2y$

$\frac{18}{6} = 3$

Factoring out $6$ produces the fully factorised expression:

$6(2y + 3)$

Practical Applications of Formulas: Australian Rules Football Scoring

Algebraic formulas allow real-world quantitative systems to be calculated efficiently by defining relationships between different measurable quantities. In Australian Rules Football, scoring is divided into two distinct scoring events: Goals ($G$) and Behinds ($B$).

Each Goal ($G$) is awarded $6\,\text{points}$, and each Behind ($B$) is awarded $1\,\text{point}$. The overall point sum, denoted as $P$, is governed by the linear algebraic formula:

$P = 6G + B$

To determine the total points scored by a team that kicked $11\,\text{Goals}$ and $5\,\text{Behinds}$, set $G = 11$ and $B = 5$:

$P = 6 \times 11 + 5$

$6 \times 11 = 66$

$P = 66 + 5 = 71\,\text{points}$

The team scored a total of $71\,\text{points}$.