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Algebra - Addition and Subtraction of Algebraic Expressions

Grade 7ICSE

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

πŸ”‘Concepts

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Algebraic Terms and Coefficients: An algebraic expression is made of terms. A term is a product of a number (the numerical coefficient) and variables (literal factors). For example, in the term βˆ’5x2y-5x^{2}y, βˆ’5-5 is the coefficient and x2yx^{2}y is the literal part. Visually, you can imagine a term as a single 'package' where the number is the quantity and the variables are the type of item.

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Like and Unlike Terms: Like terms are terms that have the exact same literal factors with the same exponents. For instance, 3ab3ab and βˆ’7ab-7ab are like terms, while 3a3a and 3b3b are unlike. Visually, think of like terms as identical fruits (like apples); you can only combine items if they are the same 'fruit' type.

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Addition of Algebraic Expressions: To add expressions, we only combine the numerical coefficients of the like terms while keeping the literal part unchanged. For example, 2x+3x=5x2x + 3x = 5x. In your mind, visualize grouping all identical variable 'shapes' together on a desk before counting the total number for each shape.

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Subtraction and the Additive Inverse: Subtracting an algebraic expression is the same as adding its additive inverse. To find the additive inverse, change the sign of every term within the expression. Visually, imagine a 'sign-flipper' gate: as terms like (2xβˆ’3y)(2x - 3y) pass through a subtraction gate, they transform into βˆ’2x+3y-2x + 3y.

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The Horizontal Method: This method involves writing all expressions in a single horizontal line. You use brackets to separate expressions and then rearrange the terms so that like terms are adjacent to each other. Visually, this looks like a long string of terms that gradually shrinks as you 'shrink' groups of like terms into single terms.

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The Column Method: In this method, expressions are written in rows, one below the other. You must align the terms such that like terms fall in the same vertical column (e.g., all xx terms in one column, all yy terms in another). Visually, this looks like a standard multi-digit addition or subtraction problem, where each column represents a specific variable 'category'.

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Rules for Brackets (BODMAS): When expressions involve brackets, always simplify the terms inside the innermost brackets first. If a bracket is preceded by a minus sign, every term inside the bracket must have its sign changed when the bracket is removed. For example, 5βˆ’(2x+3)=5βˆ’2xβˆ’35 - (2x + 3) = 5 - 2x - 3.

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Combining Multiple Terms: When an expression has many terms, the final result is reached when no more like terms remain. The result is often arranged in descending powers of the variable (e.g., x2x^{2} then xx then constants) to keep it visually organized and easy to read.

πŸ“Formulae

Addition Rule: ax+bx=(a+b)xax + bx = (a + b)x

Subtraction Rule: axβˆ’bx=(aβˆ’b)xax - bx = (a - b)x

Sign Change (Positive): +(aβˆ’b)=aβˆ’b+(a - b) = a - b

Sign Change (Negative): βˆ’(a+b)=βˆ’aβˆ’b-(a + b) = -a - b

Sign Change (Double Negative): βˆ’(aβˆ’b)=βˆ’a+b-(a - b) = -a + b

Additive Inverse: Expression+(AdditiveΒ Inverse)=0Expression + (\text{Additive Inverse}) = 0

πŸ’‘Examples

Problem 1:

Add the following expressions: 3x+7yβˆ’5z3x + 7y - 5z, 2xβˆ’3y+2z2x - 3y + 2z, and βˆ’x+y+4z-x + y + 4z.

Solution:

Step 1: Write the expressions in a row: (3x+7yβˆ’5z)+(2xβˆ’3y+2z)+(βˆ’x+y+4z)(3x + 7y - 5z) + (2x - 3y + 2z) + (-x + y + 4z) \ Step 2: Group the like terms together: (3x+2xβˆ’x)+(7yβˆ’3y+y)+(βˆ’5z+2z+4z)(3x + 2x - x) + (7y - 3y + y) + (-5z + 2z + 4z) \ Step 3: Add the coefficients of each group: (3+2βˆ’1)x+(7βˆ’3+1)y+(βˆ’5+2+4)z(3 + 2 - 1)x + (7 - 3 + 1)y + (-5 + 2 + 4)z \ Step 4: Simplify: 4x+5y+1z4x + 5y + 1z or 4x+5y+z4x + 5y + z.

Explanation:

We use the Horizontal Method to group terms with the same variables (xx, yy, and zz) and then perform simple arithmetic on their numerical coefficients.

Problem 2:

Subtract 4a2βˆ’5ab+3b24a^{2} - 5ab + 3b^{2} from 7a2βˆ’2abβˆ’4b27a^{2} - 2ab - 4b^{2}.

Solution:

Step 1: Set up the subtraction: (7a2βˆ’2abβˆ’4b2)βˆ’(4a2βˆ’5ab+3b2)(7a^{2} - 2ab - 4b^{2}) - (4a^{2} - 5ab + 3b^{2}) \ Step 2: Remove the brackets and change the signs of the second expression: 7a2βˆ’2abβˆ’4b2βˆ’4a2+5abβˆ’3b27a^{2} - 2ab - 4b^{2} - 4a^{2} + 5ab - 3b^{2} \ Step 3: Group like terms: (7a2βˆ’4a2)+(βˆ’2ab+5ab)+(βˆ’4b2βˆ’3b2)(7a^{2} - 4a^{2}) + (-2ab + 5ab) + (-4b^{2} - 3b^{2}) \ Step 4: Simplify the coefficients: 3a2+3abβˆ’7b23a^{2} + 3ab - 7b^{2}.

Explanation:

Since we are subtracting 'from' the first expression, the second expression is the subtrahend. We carefully flip the signs of 4a24a^{2}, βˆ’5ab-5ab, and 3b23b^{2} before combining them with the first expression.