Expressions and Equations

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These are the solutions to MALTA past questions on Expressions and Equations.
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These are the notable notes regarding factoring

Factoring Formulas

$ \underline{Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (1.)\;\;x^2 - y^2 = (x + y)(x - y) \\[5ex] \underline{Difference\;\;of\;\;Two\;\;Cubes} \\[3ex] (2.)\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[5ex] \underline{Sum\;\;of\;\;Two\;\;Cubes} \\[3ex] (3.)\;\; x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[4ex] $

Formulas Relating to Quadratic Expressions and Equations

$ (1.)\;\; Discriminant = b^2 - 4ac \\[5ex] (2.)\;\; \text{Quadratic Formula}:\;\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[6ex] (3.)\;\; \text{Sum of roots} = -\dfrac{b}{a} \\[5ex] (4.)\;\; \text{Product of roots} = \dfrac{c}{a} $

(1.) (a.) Expand and simplify: $2x(4x - 5y) + 6(x^2 - 2y)$

(b.) Make x the subject of the formula: $q = \sqrt{\dfrac{px + y}{x}}$

$ (a.) \\[3ex] 2x(4x - 5y) + 6(x^2 - 2y) \\[3ex] 8x^2 - 10xy + 6x^2 - 12y \\[3ex] 14x^2 - 10xy - 12y \\[5ex] (b.) \\[3ex] q = \sqrt{\dfrac{px + y}{x}} \\[5ex] \text{Square both sides} \\[3ex] q^2 = \left(\sqrt{\dfrac{px + y}{x}}\right)^2 \\[5ex] q^2 = \dfrac{px + y}{x} \\[5ex] LCD = x \\[3ex] \text{Multiply both sides by the LCD} \\[3ex] x \times q^2 = x \times \dfrac{px + y}{x} \\[5ex] xq^2 = px + y \\[3ex] \text{Subtract } px \text{ from both sides} \\[3ex] xq^2 - px = px + y - px \\[3ex] xq^2 - px = y \\[3ex] \text{Factor the LHS by the GCF}, x \\[3ex] x(q^2 - p) = y \\[3ex] \text{Divide both sides by } q^2 - p \\[3ex] \dfrac{x(q^2 - p)}{q^2 - p} = \dfrac{y}{q^2 - p} \\[5ex] x = \dfrac{y}{q^2 - p} $

(2.) (a.) Find the value of $15a - 3b - c$, when $a = 4$, $b = -6$ and $c = 10$.

(b.) Simplify:
(i.) $-2t \times 3t$
(ii.) $3(x + 2) + 5(1 + x)$

(c.) Solve: $6x - 15 = 4x - 3$

$ (a.) \\[3ex] a = 4 \\[3ex] b = -6 \\[3ex] c = 10 \\[3ex] 15a - 3b - c \\[3ex] = 15(4) - 3(-6) - 10 \\[3ex] = 60 + 18 - 10 \\[3ex] = 68 \\[3ex] (b.)(i.) \\[3ex] -2t \times 3t = -6t^2 \\[3ex] (ii.) \\[3ex] 3(x + 2) + 5(1 + x) \\[3ex] 3x + 6 + 5 + 5x \\[3ex] 8x + 11 \\[3ex] (c.) \\[3ex] 6x - 15 = 4x - 3 \\[3ex] 6x - 4x = -3 + 15 \\[3ex] 2x = 12 \\[3ex] x = \dfrac{12}{2} \\[5ex] x = 6 \\[3ex] $ Check

$x = 6$
LHS	RHS
$ 6x - 15 \\[3ex] 6(6) - 15 \\[3ex] 36 - 15 \\[3ex] 21 $	$ 4x - 3 \\[3ex] 4(6) - 3 \\[3ex] 24 - 3 \\[3ex] 21 $

(3.) (a.) Solve: $\dfrac{(4x - 5)}{(8x - 15)} = \dfrac{2x}{4x - 5}$

(b.) Make x the subject of the formula: $q = \sqrt{\dfrac{p(x + y)}{x}}$

$ (a.) \\[3ex] \dfrac{(4x - 5)}{(8x - 15)} = \dfrac{2x}{4x - 5} \\[5ex] \text{Cross Multiply} \\[3ex] (4x - 5)(4x - 5) = 2x(8x - 15) \\[3ex] 16x^2 - 20x - 20x + 25 = 16x^2 - 30x \\[3ex] 16x^2 - 40x + 25 - 16x^2 + 30x = 0 \\[3ex] -10x = -25 \\[3ex] x = \dfrac{-25}{-10} \\[5ex] x = \dfrac{5}{2} \\[5ex] $ Check

$x = \dfrac{5}{2}$
LHS	RHS
$ \dfrac{(4x - 5)}{(8x - 15)} \\[5ex] (4x - 5) \div (8x - 15) \\[3ex] \left[4\left(\dfrac{5}{2}\right) - 5\right] \div \left[8\left(\dfrac{5}{2}\right) - 15\right] \\[5ex] (10 - 5) \div (20 - 15) \\[3ex] 5 \div 5 \\[3ex] 1 $	$ \dfrac{2x}{4x - 5} \\[5ex] 2x \div (4x - 5) \\[3ex] 2\left(\dfrac{5}{2}\right) \div \left[4\left(\dfrac{5}{2}\right) - 5\right] \\[5ex] 5 \div (10 - 5) \\[3ex] 5 \div 5 \\[3ex] 1 $

$ (b.) \\[3ex] q = \sqrt{\dfrac{p(x + y)}{x}} \\[5ex] q^2 = \dfrac{p(x + y)}{x} \\[5ex] xq^2 = px + py \\[3ex] xq^2 - px = py \\[3ex] x(q^2 - p) = py \\[3ex] x = \dfrac{py}{q^2 - p} $

(4.) (a.) (i.) Factorise completely: $3x^2 - 15xy$
(ii.) Use your answer in (i.) to simplify: $\dfrac{3x^2 - 15xy}{6x}$

(b.) Write as a single fraction: $\dfrac{4x}{5} - \dfrac{x + 1}{4}$

$ (a.) \\[3ex] 3x^2 - 15xy \\[3ex] GCF = 3x \\[3ex] = 3x(x - 5y) \\[3ex] (b.) \\[3ex] \dfrac{3x^2 - 15xy}{6x} \\[5ex] = \dfrac{3x(x - 5y)}{6x} \\[5ex] = \dfrac{x - 5y}{2} \\[5ex] (c.) \\[3ex] \dfrac{4x}{5} - \dfrac{x + 1}{4} \\[5ex] LCD = 20 \\[3ex] = \dfrac{4(4x)}{20} - \dfrac{5(x + 1)}{20} \\[5ex] = \dfrac{16x - 5x - 5}{20} \\[5ex] = \dfrac{11x - 5}{20} $

(5.) (a.) Factorise:
(i.) $x^2 + 6x + 8$
(ii.) $x^2 - 16$

(b.) Simplify as a single fraction: $\dfrac{x^2 + 6x + 8}{x^2 - 16} - \dfrac{1}{x + 5}$

$ (a.) (i.) \\[3ex] x^2 + 6x + 8 \\[3ex] (x + 4)(x + 2)...\text{Factor Quadratic Trinomial} \\[3ex] (ii.) \\[3ex] x^2 - 16 \\[3ex] x^2 - 4^2 \\[3ex] (x + 4)(x - 4) ...\text{Difference of Two Squares} \\[3ex] (b.) \\[3ex] \dfrac{x^2 + 6x + 8}{x^2 - 16} - \dfrac{1}{x + 5} \\[5ex] \dfrac{(x + 4)(x + 2)}{(x + 4)(x - 4)} - \dfrac{1}{x + 5} \\[5ex] \dfrac{x + 2}{x - 4} - \dfrac{1}{x + 5} \\[5ex] \dfrac{(x + 2)(x + 5)}{(x - 4)(x + 5)} - \dfrac{x - 4}{(x - 4)(x + 5)} \\[5ex] \dfrac{(x + 2)(x + 5) - (x - 4)}{(x - 4)(x + 5)} \\[5ex] \dfrac{x^2 + 5x + 2x + 10 - x + 4}{(x - 4)(x + 5)} \\[5ex] \dfrac{x^2 + 6x + 14}{(x - 4)(x + 5)} $

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