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QUESTION 3.1 Simplify the following: 3.1.1 (2a-3)(2a+3) 3.1.2 (p2-1)-2p(p-1) 3.1.3 4x2y-8xy3/xy

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3.1 Simplify the following:

3.1.1 (2a−3)(2a+3)(2a-3)(2a+3)

  1. Use the difference of squares formula, (A−B)(A+B)=A2−B2(A-B)(A+B) = A^2 - B^2.

    (2a−3)(2a+3)=(2a)2−(3)2(2a-3)(2a+3) = (2a)^2 - (3)^2

  2. Simplify the terms.

    =4a2−9= 4a^2 - 9

    The simplified expression is 4a2−9\boxed{\mathbf{4a^2 - 9}}.

    3.1.2 (p2−1)−2p(p−1)(p^2-1)-2p(p-1)

  3. Distribute the −2p-2p into the second parenthesis.

    (p2−1)−2p(p−1)=p2−1−(2p⋅p−2p⋅1)(p^2-1)-2p(p-1) = p^2-1 - (2p \cdot p - 2p \cdot 1) =p2−1−(2p2−2p)= p^2-1 - (2p^2-2p)

  4. Remove the parenthesis and combine like terms.

    =p2−1−2p2+2p= p^2-1-2p^2+2p =(p2−2p2)+2p−1= (p^2-2p^2) + 2p - 1 =−p2+2p−1= -p^2+2p-1

    The simplified expression is −p2+2p−1\boxed{\mathbf{-p^2+2p-1}}.

    3.1.3 4x2y−8xy3xy\frac{4x^2y-8xy^3}{xy}

  5. Factor out the common term from the numerator.

    The common term is 4xy4xy.

    4x2y−8xy3xy=4xy(x−2y2)xy\frac{4x^2y-8xy^3}{xy} = \frac{4xy(x-2y^2)}{xy}

  6. Cancel out the common term xyxy from the numerator and denominator.

    =4(x−2y2)= 4(x-2y^2) =4x−8y2= 4x-8y^2

    The simplified expression is 4x−8y2\boxed{\mathbf{4x-8y^2}}.

    3.2 Factorise the following:

    3.2.1 20x2−45y220x^2-45y^2

  7. Factor out the greatest common factor, which is 5.

    20x2−45y2=5(4x2−9y2)20x^2-45y^2 = 5(4x^2-9y^2)

  8. Recognize the difference of squares pattern, A2−B2=(A−B)(A+B)A^2-B^2 = (A-B)(A+B), where A=2xA=2x and B=3yB=3y.

    =5((2x)2−(3y)2)= 5((2x)^2-(3y)^2) =5(2x−3y)(2x+3y)= 5(2x-3y)(2x+3y)

    The factorised expression is 5(2x−3y)(2x+3y)\boxed{\mathbf{5(2x-3y)(2x+3y)}}.

    3.2.2 2x2−5x−32x^2-5x-3

  9. Find two numbers that multiply to 2×(−3)=−62 \times (-3) = -6 and add to −5-5.

    These numbers are −6-6 and 1.

  10. Rewrite the middle term using these numbers and factor by grouping.

    2x2−5x−3=2x2−6x+x−32x^2-5x-3 = 2x^2-6x+x-3 =2x(x−3)+1(x−3)= 2x(x-3)+1(x-3)

  11. Factor out the common binomial factor (x−3)(x-3).

    =(x−3)(2x+1)= (x-3)(2x+1)

    The factorised expression is (x−3)(2x+1)\boxed{\mathbf{(x-3)(2x+1)}}.

    3.2.3 5x2−15x+12y−4xy5x^2-15x+12y-4xy

  12. Group the terms and factor out common factors from each group.

    (5x2−15x)+(12y−4xy)(5x^2-15x) + (12y-4xy) =5x(x−3)+4y(3−x)= 5x(x-3) + 4y(3-x)

  13. Notice that (3−x)(3-x) is the negative of (x−3)(x-3).

    Rewrite 4y(3−x)4y(3-x) as −4y(x−3)-4y(x-3).

    =5x(x−3)−4y(x−3)= 5x(x-3) - 4y(x-3)

  14. Factor out the common binomial factor (x−3)(x-3).

    =(x−3)(5x−4y)= (x-3)(5x-4y)

    The factorised expression is (x−3)(5x−4y)\boxed{\mathbf{(x-3)(5x-4y)}}.

    3.3 Solve for xx in each of the following equations:

    3.3.1 3⋅5x−21=543 \cdot 5^x-21=54

  15. Add 21 to both sides of the equation.

    3⋅5x=54+213 \cdot 5^x = 54+21 3⋅5x=753 \cdot 5^x = 75

  16. Divide both sides by 3.

    5x=7535^x = \frac{75}{3} 5x=255^x = 25

  17. Express 25 as a power of 5.

    5x=525^x = 5^2

  18. Equate the exponents.

    x=2x=2

    The solution is x=2\boxed{\mathbf{x=2}}.

    3.3.2 2x2−12x=142x^2-12x=14

  19. Move all terms to one side to form a standard quadratic equation.

    2x2−12x−14=02x^2-12x-14=0

  20. Divide the entire equation by 2 to simplify.

    x2−6x−7=0x^2-6x-7=0

  21. Factor the quadratic equation.

    Find two numbers that multiply to −7-7 and add to −6-6. These numbers are −7-7 and 1.

    (x−7)(x+1)=0(x-7)(x+1)=0

  22. Set each factor equal to zero and solve for xx.

    x−7=0orx+1=0x-7=0 \quad \text{or} \quad x+1=0 x=7orx=−1x=7 \quad \text{or} \quad x=-1

    The solutions are x=7 or x=−1\boxed{\mathbf{x=7 \text{ or } x=-1}}.

    3.4 Solve the following inequality and give your solution in interval notation:
    2x−3≥−102x-3 \ge -10

  23. Add 3 to both sides of the inequality.

    2x≥−10+32x \ge -10+3 2x≥−72x \ge -7

  24. Divide both sides by 2.

    x≥−72x \ge -\frac{7}{2}

  25. Express the solution in interval notation.

    [−72,∞)\left[-\frac{7}{2}, \infty\right)

    The solution in interval notation is [−72,∞)\boxed{\mathbf{\left[-\frac{7}{2}, \infty\right)}}.

    3.5 Solve algebraically for xx and yy in the following equations:

    1. 5x−12=3y5x-12=3y
    2. 4x−5y=174x-5y=17

  26. Rewrite equation (1) in standard form Ax+By=CAx+By=C.

    5x−3y=12(Eq 1’)5x-3y=12 \quad \text{(Eq 1')}

  27. Use the elimination method.

    Multiply Eq 1' by 5 and Eq 2 by 3 to make the coefficients of yy opposites.

    5(5x−3y)=5(12)  ⟹  25x−15y=60(Eq 3)5(5x-3y) = 5(12) \implies 25x-15y=60 \quad \text{(Eq 3)} 3(4x−5y)=3(17)  ⟹  12x−15y=51(Eq 4)3(4x-5y) = 3(17) \implies 12x-15y=51 \quad \text{(Eq 4)}

  28. Subtract Eq 4 from Eq 3 to eliminate yy.

    (25x−15y)−(12x−15y)=60−51(25x-15y) - (12x-15y) = 60-51 25x−12x=925x-12x = 9 13x=913x = 9

  29. Solve for xx.

    x=913x = \frac{9}{13}

  30. Substitute the value of xx into Eq 1' (5x−3y=125x-3y=12) to solve for yy.

    5(913)−3y=125\left(\frac{9}{13}\right)-3y=12 4513−3y=12\frac{45}{13}-3y=12

  31. Subtract 4513\frac{45}{13} from both sides.

    −3y=12−4513-3y = 12 - \frac{45}{13} −3y=12×1313−4513-3y = \frac{12 \times 13}{13} - \frac{45}{13} −3y=156−4513-3y = \frac{156-45}{13} −3y=11113-3y = \frac{111}{13}

  32. Divide by -3 to find yy.

    y=11113×(−3)y = \frac{111}{13 \times (-3)} y=−3713y = -\frac{37}{13}

    The solutions are x=913,y=−3713\boxed{\mathbf{x=\frac{9}{13}, y=-\frac{37}{13}}}.

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