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Solving Algebraic Equations

  • An algebraic expression with an equal to sign is called an algebraic equation. For example, x + 8 = 3 is an equation in x.
  • The unknown in an equation is shown by a letter. For example, m is the unknown in the
    equation 2m + 3 = 17.
  • If a value is substituted for an unknown, the equation may be true or false. For example, 2y = 6 is true when y = 3, and is false when y = 5.
  • To solve an equation using the balance method, always do the same to both sides of the equation.
  • When solving an equation, you may cut out steps in the method when you can do them mentally.
  • Check if your solution is correct by substituting the value to see if it makes the equation true.
SIMULTANEOUS EQUATIONS
  • In an equation, such as y = 3x – 4, the unknowns x and y are called variables.
  • Simultaneous linear equations can be solved algebraically; either by substitution or by elimination.
  • Simultaneous linear equations can also be solved graphically by drawing the graphs of the two equations on the same axes; their point of intersection gives the solution.
QUADRATIC EQUATIONS
  • A quadratic equation is equation in the form ax² + bx + c = 0, where a, b, and c are constants
    and a ≠ 0.
  • The highest power in a quadratic equation is 2.
  • The values of x that satisfy the quadratic equation are called the roots of the equation.
  • The roots of a quadratic equation can be found by:
    • factorization:
      • the quadratic equation is simply factorized and set equal to zero. For example:
        • Given the quadratic equation ax² + bx + c = 0, if (dx + e) and (fx + g) are the factors,
        • Then (dx + e)(fx + g) = 0.
        • Then either (dx + e) = 0, i.e., x = –e/d
        • Or (fx + g) = 0, i.e., –g/f
Examples
  1. Say whether the following are true or false:
    1. x + 2 = 5, when x = 3.
      • When x = 3; x + 2 = 3 + 2 = 5
      • Thus, x + 2 = 5 is true when x = 3
    2. 3x + 1 = 10 when x = 2
      • When x = 2; 3x + 1 = (3 × 2) +1
      •                                = 6 + 1
      •                                = 7
      • 7 ≠ 10 (‘≠’ means not equal to)
      • Thus, 3x + 1 = 10 is false when x = 2.
  2. Is the equation x/4 = 5 true when x = 24?
    • When x = 24; x/4 = 24/4 = 6
    • 6 ≠ 5
    • Thus, x/4 = 5 is not true when x = 24.
  3. Solve 18 – x = 7.
    • The problem is to find a number which when taken from 18 gives 7.
    • The number is 11.
    • ∴ x = 11 is the solution.
  4. Solve 1/3y = 7.
    • The unknown y is on the Left Hand Side (LHS).
    • It is divided by 3 to give 1/3y on the LHS.
    • Multiply both sides by 3
    • 1/3y × 3 = 7 × 3
    • ∴ y = 21.
  5. Solve 5x – 6 = 29.
    • The unknown x is contained in the LHS.
    • Add 6 to both sides to leave 5x alone on the LHS: 5x – 6 + 6 = 29 + 6
    • ∴ 5x = 35
    • The unknown ‘x’ is on the LHS, and it is multiplied by 5.
    • Divide both sides by 5 to obtain x on the LHS: 5x/5 = 35/5
    • ∴ x = 7
    • To Check: when x = 7; LHS = 5 × 7 – 6
    •                                              = 35 – 6
    •                                              = 29
    • ∴ 5x – 6 = 29 is true when x = 7
  6. Solve 2x + 7 = 12.
    • Subtract 7 from both sides: 2x + 7 – 7 = 12 – 7
    • ∴ 2x = 5
    • Divide both sides by 2: 2x/2 = 5/2
    • ∴ x = 21/2
    • To check: When x = 21/2
    • LHS = 2 × 21/2 +7
    •         = 5 + 7
    •         = 12
    • ∴ 2x + 7 = 12 when x = 5
  7. Solve the equations (Substitution Method)
    3a + b = 10
    2a + 4b = 0

    • 3a + b = 10       – (i)
      2a + 4b = 0       – (ii)
    • from (i), b = 10 – 3a       – (iii)
    • Substitute (10 – 3a) for b in (ii): 2a + 4(10 – 3a) = 0
    • Clear the brackets: 2a + 40 – 12a = 0
    • Collect like terms: 2a – 12a = -40
    •                                    -10a = -40
    • ∴ a = 4
    • Substitue a = 4 in (iii): b = 10 – (3 × 4)
    •                                         b = 10 – 12
    • ∴ b = -2
    • Therefore, a = 4, and b = -2
  8.  Solve the equations (Elimination Method)
    3f = 4 – 4e
    2e = 5f + 15

    • 3f = 4 – 4e        – (i)
      2e = 5f + 15     – (ii)
    • Arrange the equations so that the unknowns are in alphabetical order on the LHS and the numbers are on the RHS:
    • 4e + 3f = 4       – (iii)
      2e – 5f = 15      – (iv)
    • Multiply (iv) by 2: 4e – 10f = 30       – (v)
    • Subtract (v) from (iii) to eliminate terms in e: 13f = -26f
      ∴ f = -2
    • Substitute -2 for f in (ii): 2e = 5(-2) + 15
    •                                                2e = -10 + 15
    •                                                2e = 5
    • ∴ e = 21/2
  9. Solve the equation x² + 6x – 7 = 0
    • First step: Factorize x² + 6x – 7. Find two numbers such that their product is -7 (+1 × -7) and their sum is +6.
      Factors of -7Sum of Factors
      a.-1 and +7+6
      b.+1 and -7-6
    • Of these, only a gives the required result. Therefore x² + 6x – 7 is the same as x² – x + 7x – 7
    • = x(x – 1) + 7(x – 1)
    • = (x – 1) (x + 7)
    • ∴ (x – 1) (x + 7) = 0
    • Second Step: Equate each bracket to zero and solve for x.
    • ⇒ x – 1 = 0
    • ∴ x = 1
    • ⇒ x + 7 = 0
    • ∴ x = -7
    • ∴ x = 1 and – 7
  10. Solve 3y² + 13y + 12 = 0.
    • Factors which when multiplied would give 36 (3 × 12) and when summed up would give 13 are +9 and + 4
    • ∴ 3y² + 13y + 12 is the same as 3y² + 9y + 4y + 12
    • ∴ 3y² + 9y + 4y + 12 = 0
    • ⇒ 3y(y + 3) + 4(y + 3) = 0
    • (3y + 4) (y + 3) = 0
    • ⇒ 3y + 4 = 0
    • ∴ y = –4/3
    • ⇒ y + 3 = 0
    • ∴ y = -3
    • ∴ y = -3 and –4/3
  11. Solve 4 – 3z – z²
    • Factors which when multiplied would give -4 (4 × -1) and when summed would give -3 are -4 and +1.
    • ∴ 4 – 3z – z² is the same as 4 – 4z +z – z²
    • ∴ 4 – 4z +z – z² = 0
    • ⇒ 4(1 – z) + z(1 – z) = 0
    • ∴ (4 + z) (1 – z) = 0
    • ⇒ 4 + z = 0
    • ∴ z = -4
    • ⇒ 1 – z = 0
    • ∴ z = 1
    • ∴ z = 1 and -4
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