• [ended]Mathematics Tutoring Session #13 (Algebraic Fractions[thurs. 3nd Mar 2016, 3:00 - 4:00pm])

    Mathematics

    Emmanuel Iwara(Tutor)
    03-03-2016 12:03:00 +0000

    Good day all!

    This session has ended for today

    Algebraic Fraction is most foundamental to all who want to know mathematics and often tested by jamb and other exams.

    Under algebraic expressions,fractions have same fundemental approach as done in arithmetical processes.

    An algeraic fraction is the one in which the denominator is an algebraic expression.

    e.g 4/x is an algeraic fraction but x/4 is not an algebraic fraction.

    Simplification of fractions means that the fractions are to be written in their reduced forms, the process involves finding the L.c.m of the denominators,expressing each fraction as a fraction with the common denominator.

    1.Dealing with addition or subtraction of fractions requires a common denominator.

    2. multiply EVERY TERM in the equation by the common denominator. In an equation, (unlike an expression), you may multiply "every term" on both sides of the equal sign by the same value and not change the equation.

    Example 1.Simplify 2x2 - 3ax + 4x2 -6xa -x2

    Solution

    Collection of like terms yield

    2x2 + 4x2-x2 -3ax -6ax

    = 6x2 -x2 -3ax -6ax

    = 5x2 -9ax.

    Example 2. Simplify (2 / x-2) -(3 / x-1)

    Solution

    taking L.C.M = (x -2)(x-1) and simplifying

    Hence (2 / x-2) -( 3/ x-1) = [2(x-1) -3(x-2)] / [(x -2)(x-1)]

    = [2x -2 -3x + 6] / [(x -2)(x-1)]

    = (4 -x) / (x-2)(x-1).

    Example 3.Simplify [3a3 / 3a2 -6ab] + [4b3/ 2b2 - ab]

    Solution

    Factorizing the denominators we have

    [3a3 / 3a(a-2b)] + [4b3/ b(2b-a)], cancelling out common terms

    = (a2) / (a-2b) + (4b2)/(2b-a)

    = (a2)/ (a-2b) -(4b2)/(a-2b)

    Taking L.C.M = (a-2b) and simplifying

    = (a2 -4b2) / (a -2b)

    applying differences of two sqaure in the numerator will yield

    (a+2b)(a-2b) / (a-2b)

    = a + 2b.

    Example 4.Simplify (6-x-x2) / (x2 -4)

    Solution

    (6-x-x2)/(x2 -4)

    = (3 + x)(2-x) / (x+2)(x-2)

    = -(3 + x)(2-x) / (x + 2)(2-x)]

    = -(3 + x) / (x + 2)

    Example 5.Evalute [ab2 -c2]/ 2bc + [a2]/ 2b + c]

    When (a = 2, b = -3 and c = -2)

    Solution

    Substituting the values of a, b,and c into [ab2 -c2]/ 2bc + [a2]/ 2b + c]

    [2(-3)2 -(-2)2] + [22]/ 2(-3) + (-2)]

    = (2 x 9 -4)/ 12 + 4/(-6 -2)

    = (14/12 - 4/8)

    =[(7/6 -1/2)]

    taking L.C.M
    (7-3)/6

    4/6

    2/3

    You may ask your questions if you have any, and i will try the best i can to give solutions to them.

    Now try and see if you can solve the following past jamb and other exams questions.

    1.Simplify 2a2 - a + 6a2-8a2-6a + 5a2 + 10a

    . 2.Simplify (x2 - y2)/(x2 + xy)

    3. Simplify [15 - 2x - x2]/ [x2 -9]

    4.Simplify x/(x + y) + y/(x -y) -(x2)/(x2 -y2)

    5.If a = 2, b = -2 and c = -1/2, evalute (ab2-bc2)(a2c -abc)

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