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Algebra - Algorithms

Grade 10IB

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

🔑Concepts

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Algebraic Algorithms involve systematic step-by-step procedures to solve equations, simplify expressions, or rearrange formulae.

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The Distributive Law: The algorithm for expanding brackets where a term outside is multiplied by every term inside, expressed as a(b+c)=ab+aca(b + c) = ab + ac.

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Solving Linear Equations: The algorithm uses inverse operations (addition/subtraction, then multiplication/division) to isolate the variable xx on one side of the equation.

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Factorization Algorithm: The process of finding the Highest Common Factor (HCF) or identifying patterns like the difference of two squares to turn an expression into a product of factors.

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Changing the Subject: A procedural approach to rearrange a formula to solve for a different variable by applying the same operation to both sides of the equals sign.

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Solving Systems of Equations: Algorithms such as the Substitution Method or the Elimination Method are used to find the intersection of two linear paths.

📐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

a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)

x2+(a+b)x+ab=(x+a)(x+b)x^2 + (a+b)x + ab = (x + a)(x + b)

💡Examples

Problem 1:

Solve the linear equation for xx: 4(x−3)=2x+104(x - 3) = 2x + 10

Solution:

4x−12=2x+104x - 12 = 2x + 10 4x−2x=10+124x - 2x = 10 + 12 2x=222x = 22 x=11x = 11

Explanation:

Apply the expansion algorithm to the left side: 4×x4 \times x and 4×−34 \times -3. Then, use the balancing algorithm to group like terms by subtracting 2x2x and adding 1212 to both sides. Finally, divide by the coefficient of xx.

Problem 2:

Factorize the quadratic expression: x2−7x+12x^2 - 7x + 12

Solution:

(x−3)(x−4)(x - 3)(x - 4) patterns.

Explanation:

Use the 'Sum and Product' algorithm. We look for two numbers that multiply to +12+12 (the constant) and add to −7-7 (the coefficient of xx). Those numbers are −3-3 and −4-4. Thus, the factors are (x−3)(x - 3) and (x−4)(x - 4).

Problem 3:

Solve the simultaneous equations using the Elimination Algorithm: x+y=15x + y = 15 x−y=3x - y = 3

Solution:

x+y=15+(x−y=3)2x=18\begin{array}{r} x + y = 15 \\ + (x - y = 3) \\ \hline 2x = 18 \end{array} x=9x = 9 9+y=15⇒y=69 + y = 15 \Rightarrow y = 6

Explanation:

The algorithm involves adding the two equations to eliminate the variable yy. This results in a single-variable equation for xx. Once xx is found, substitute it back into one of the original equations to solve for yy.

Problem 4:

Make hh the subject of the formula: V=13πr2hV = \frac{1}{3} π r^2 h

Solution:

3V=πr2h3V = π r^2 h h=3Vπr2h = \frac{3V}{π r^2}

Explanation:

To isolate hh, apply the inverse operation algorithm. First, multiply both sides by 33 to remove the fraction. Then, divide both sides by the product πr2π r^2 to leave hh alone.