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Algebraic Identities and Linear Equations

Algebraic Identities

Algebraic identities are equations that are true for all values of the variables involved. They are fundamental tools in simplifying expressions, solving equations, and proving mathematical statements. Understanding and memorizing these identities can significantly speed up problem-solving in competitive exams.

Basic Identities (Type 1: Square of Binomials)

These identities involve the square of a binomial, which is an expression with two terms.

  1. (a + b)2 = a2 + 2ab + b2

    This identity states that the square of the sum of two terms is equal to the sum of the squares of the two terms plus twice their product.

    Example: If a = 3 and b = 4, then (3 + 4)2 = 72 = 49. Using the identity: 32 + 2(3)(4) + 42 = 9 + 24 + 16 = 49.

  2. (a - b)2 = a2 - 2ab + b2

    This identity states that the square of the difference of two terms is equal to the sum of their squares minus twice their product.

    Example: If a = 5 and b = 2, then (5 - 2)2 = 32 = 9. Using the identity: 52 - 2(5)(2) + 22 = 25 - 20 + 4 = 9.

  3. a2 - b2 = (a + b)(a - b)

    This is the difference of squares identity. It states that the difference of the squares of two terms is equal to the product of their sum and their difference. This is also known as the "difference of two squares" factorization.

    Example: 102 - 62 = 100 - 36 = 64. Using the identity: (10 + 6)(10 - 6) = (16)(4) = 64.

Basic Identities (Type 2: Square of Trinomials)

These identities extend the concept to expressions with three terms.

  1. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

    The square of the sum of three terms is the sum of the squares of each term plus twice the product of each pair of terms.

    Example: Consider (1 + 2 + 3)2 = 62 = 36. Using the identity: 12 + 22 + 32 + 2(1)(2) + 2(2)(3) + 2(3)(1) = 1 + 4 + 9 + 4 + 12 + 6 = 36.

Basic Identities (Type 3: Cubes of Binomials)

These identities involve the cube of a binomial.

  1. (a + b)3 = a3 + b3 + 3ab(a + b)

    This can also be written as: (a + b)3 = a3 + 3a2b + 3ab2 + b3. The cube of the sum of two terms is the sum of their cubes plus three times their product times their sum.

    Example: (2 + 3)3 = 53 = 125. Using the identity: 23 + 33 + 3(2)(3)(2 + 3) = 8 + 27 + 18(5) = 35 + 90 = 125.

  2. (a - b)3 = a3 - b3 - 3ab(a - b)

    This can also be written as: (a - b)3 = a3 - 3a2b + 3ab2 - b3. The cube of the difference of two terms is the difference of their cubes minus three times their product times their difference.

    Example: (4 - 1)3 = 33 = 27. Using the identity: 43 - 13 - 3(4)(1)(4 - 1) = 64 - 1 - 12(3) = 63 - 36 = 27.

Sum and Difference of Cubes

These identities deal with the sum or difference of two cubic terms.

  1. a3 + b3 = (a + b)(a2 - ab + b2)

    The sum of two cubes can be factored into the product of the sum of the terms and a trinomial.

    Example: 23 + 33 = 8 + 27 = 35. Using the identity: (2 + 3)(22 - (2)(3) + 32) = (5)(4 - 6 + 9) = (5)(7) = 35.

  2. a3 - b3 = (a - b)(a2 + ab + b2)

    The difference of two cubes can be factored into the product of the difference of the terms and a trinomial.

    Example: 43 - 23 = 64 - 8 = 56. Using the identity: (4 - 2)(42 + (4)(2) + 22) = (2)(16 + 8 + 4) = (2)(28) = 56.

Identities involving a3 + b3 + c3 - 3abc

This is a very important identity, especially in competitive exams.

  1. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

    This identity is crucial. It connects the sum of cubes with a linear factor (a + b + c) and a quadratic factor.

    A special case of this identity is when a + b + c = 0.

    If a + b + c = 0, then a3 + b3 + c3 = 3abc.

    Example: Let a = 2, b = 3, c = -5. Here, a + b + c = 2 + 3 + (-5) = 0. So, a3 + b3 + c3 should be equal to 3abc. LHS: 23 + 33 + (-5)3 = 8 + 27 - 125 = 35 - 125 = -90. RHS: 3 * (2) * (3) * (-5) = 6 * (-15) = -90. The identity holds true.

    Another form of the identity is: a3 + b3 + c3 - 3abc = 1/2 * (a + b + c) * [(a - b)2 + (b - c)2 + (c - a)2] This form is useful when you are given differences between variables.

Other Useful Identities

  1. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

    This is the square of a trinomial we saw earlier, rearranged to highlight the sum of pairwise products.

  2. (a + b)2 - (a - b)2 = 4ab

    This identity is derived from the first two identities: (a2 + 2ab + b2) - (a2 - 2ab + b2) = a2 + 2ab + b2 - a2 + 2ab - b2 = 4ab.

  3. (a + b)2 + (a - b)2 = 2(a2 + b2)

    This identity is also derived from the first two: (a2 + 2ab + b2) + (a2 - 2ab + b2) = a2 + 2ab + b2 + a2 - 2ab + b2 = 2a2 + 2b2 = 2(a2 + b2).

Memory Trick for Identities:

Think of "Square of Sum/Difference" for (a+b)2 and (a-b)2. Think of "Difference of Squares" for a2 - b2. Think of "Cubes of Sum/Difference" for (a+b)3 and (a-b)3. Think of "Sum/Difference of Cubes" for a3 + b3 and a3 - b3. The most complex one, a3 + b3 + c3 - 3abc, often appears with the condition a + b + c = 0, simplifying it to a3 + b3 + c3 = 3abc.

Application of Identities in Problem Solving

Algebraic identities are used extensively to:

  • Simplify complex algebraic expressions.
  • Factorize polynomials.
  • Solve equations that might not be directly solvable.
  • Evaluate numerical expressions quickly without direct calculation.

Example Problem: Evaluate 1012. We can write 101 as (100 + 1). Using the identity (a + b)2 = a2 + 2ab + b2, where a = 100 and b = 1: 1012 = (100 + 1)2 = 1002 + 2(100)(1) + 12 = 10000 + 200 + 1 = 10201.

Example Problem: If x + 1/x = 5, find x2 + 1/x2. We know that (x + 1/x)2 = x2 + 2(x)(1/x) + (1/x)2 = x2 + 2 + 1/x2. Given x + 1/x = 5, so (x + 1/x)2 = 52 = 25. Therefore, x2 + 2 + 1/x2 = 25. Subtracting 2 from both sides, we get x2 + 1/x2 = 25 - 2 = 23.

Example Problem: Simplify (87 + 13)2 - (87 - 13)2. This matches the form (a + b)2 - (a - b)2 = 4ab, where a = 87 and b = 13. So, the expression simplifies to 4 * 87 * 13. 4 * 87 * 13 = 348 * 13. 348 * 10 = 3480 348 * 3 = 1044 3480 + 1044 = 4524.

Linear Equations

A linear equation is an equation in which the highest power of the variable is 1. It represents a straight line when plotted on a graph. Linear equations can have one, two, or more variables.

Linear Equations in One Variable

These are equations of the form ax + b = 0, where 'a' and 'b' are constants, and 'a' is not equal to 0. The goal is to find the value of the variable 'x' that satisfies the equation.

Steps to Solve:

  1. Isolate the term containing the variable on one side of the equation.
  2. Move all constant terms to the other side.
  3. Solve for the variable by dividing or multiplying as needed.

Example Problem: Solve 3x + 7 = 22. 1. Subtract 7 from both sides: 3x + 7 - 7 = 22 - 7 3x = 15 2. Divide both sides by 3: 3x / 3 = 15 / 3 x = 5 So, the solution is x = 5.

Example Problem: Solve (2y + 5) / 3 = y - 1. 1. Multiply both sides by 3 to eliminate the denominator: 3 * [(2y + 5) / 3] = 3 * (y - 1) 2y + 5 = 3y - 3 2. Move terms with 'y' to one side and constants to the other. Subtract 2y from both sides: 2y + 5 - 2y = 3y - 3 - 2y 5 = y - 3 3. Add 3 to both sides: 5 + 3 = y - 3 + 3 8 = y So, the solution is y = 8.

Linear Equations in Two Variables

These are equations of the form ax + by + c = 0 or ax + by = c, where 'a', 'b', and 'c' are constants, and 'a' and 'b' are not both zero. A single linear equation in two variables has infinitely many solutions, which can be represented as ordered pairs (x, y). To find a unique solution, we need a system of two linear equations with two variables.

System of Linear Equations: A set of two or more linear equations involving the same set of variables. For two variables, a system looks like: Equation 1: a1x + b1y = c1 Equation 2: a2x + b2y = c2

Methods to Solve a System of Two Linear Equations:

  1. Substitution Method:
    1. Solve one of the equations for one variable in terms of the other.
    2. Substitute this expression into the second equation.
    3. Solve the resulting equation for the single variable.
    4. Substitute the value found back into the expression from step 1 to find the other variable.
    Example Problem: Solve: x + y = 5 (Eq 1) 2x - y = 4 (Eq 2) From Eq 1, solve for x: x = 5 - y. Substitute this into Eq 2: 2(5 - y) - y = 4 10 - 2y - y = 4 10 - 3y = 4 10 - 4 = 3y 6 = 3y y = 2. Now substitute y = 2 back into x = 5 - y: x = 5 - 2 x = 3. Solution: (x, y) = (3, 2).
  2. Elimination Method:
    1. Multiply one or both equations by a constant so that the coefficients of one variable are opposites (or the same).
    2. Add (or subtract) the equations to eliminate one variable.
    3. Solve the resulting equation for the remaining variable.
    4. Substitute the value found back into one of the original equations to find the other variable.
    Example Problem: Solve: x + y = 5 (Eq 1) 2x - y = 4 (Eq 2) Notice that the coefficients of 'y' are +1 and -1. If we add the two equations, 'y' will be eliminated: (x + y) + (2x - y) = 5 + 4 x + y + 2x - y = 9 3x = 9 x = 3. Substitute x = 3 into Eq 1: 3 + y = 5 y = 5 - 3 y = 2. Solution: (x, y) = (3, 2).
  3. Graphical Method:
    1. Graph both linear equations on the same coordinate plane.
    2. The point where the two lines intersect is the solution to the system.
    Example: For x + y = 5 and 2x - y = 4: For x + y = 5: If x=0, y=5 (0,5). If y=0, x=5 (5,0). For 2x - y = 4: If x=0, y=-4 (0,-4). If y=0, 2x=4, x=2 (2,0). Plotting these points and drawing the lines, they will intersect at (3, 2).
  4. Cramer's Rule (Using Determinants): For a system: a1x + b1y = c1 a2x + b2y = c2 The determinant of the coefficient matrix is D = | a1 b1 | = a1b2 - a2b1 | a2 b2 | The determinant for x is Dx = | c1 b1 | = c1b2 - c2b1 | c2 b2 | The determinant for y is Dy = | a1 c1 | = a1c2 - a2c1 | a2 c2 | The solution is x = Dx / D and y = Dy / D, provided D ≠ 0. Example Problem: Solve using Cramer's Rule: x + y = 5 2x - y = 4 Here, a1=1, b1=1, c1=5 a2=2, b2=-1, c2=4 D = | 1 1 | = (1)(-1) - (2)(1) = -1 - 2 = -3 | 2 -1 | Dx = | 5 1 | = (5)(-1) - (4)(1) = -5 - 4 = -9 | 4 -1 | Dy = | 1 5 | = (1)(4) - (2)(5) = 4 - 10 = -6 | 2 4 | x = Dx / D = -9 / -3 = 3 y = Dy / D = -6 / -3 = 2 Solution: (x, y) = (3, 2).
Interpreting Solutions of Systems of Linear Equations:

For a system of two linear equations in two variables:

  • Unique Solution: The lines intersect at exactly one point. This happens when the ratio of coefficients of x is not equal to the ratio of coefficients of y (a1/a2 ≠ b1/b2).
  • No Solution: The lines are parallel and never intersect. This happens when the ratio of coefficients of x is equal to the ratio of coefficients of y, but not equal to the ratio of constants (a1/a2 = b1/b2 ≠ c1/c2).
  • Infinitely Many Solutions: The lines are coincident (the same line). This happens when the ratios of all coefficients and constants are equal (a1/a2 = b1/b2 = c1/c2).

Linear Equations in Three or More Variables

These equations involve three or more variables, each raised to the power of 1. For example, ax + by + cz = d. A single such equation also has infinitely many solutions. To find a unique solution, you need a system of equations where the number of independent equations equals the number of variables. Solving systems with three or more variables typically involves extension of the elimination method or matrix methods (like Gaussian elimination), which are beyond the scope of basic competitive exam syllabus but understanding the concept is important.

Example of a system with 3 variables: 2x + y - z = 8 -3x - y + 2z = -11 -2x + y + 2z = -3 Solving this system would require techniques like elimination across three equations or using matrices.

Exam Shortcut for Linear Equations:

When faced with a system of two linear equations, quickly check the ratios of coefficients: If a1/a2 ≠ b1/b2 → Unique Solution If a1/a2 = b1/b2 ≠ c1/c2 → No Solution (Parallel Lines) If a1/a2 = b1/b2 = c1/c2 → Infinite Solutions (Coincident Lines)

For word problems involving two unknown quantities, setting up two linear equations is the standard approach. Identify what the two variables represent and form equations based on the given information.

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