Subjects algebra

Ig1 Cycle Test

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Ig1 Cycle Test


1. Factorise fully. (i) Factorise $27y^2 - 3$ - Take out common factor 3: $27y^2 - 3 = 3(9y^2 - 1)$ - Recognize difference of squares: $9y^2 - 1 = (3y - 1)(3y + 1)$ - Final factorisation: $3(3y - 1)(3y + 1)$ (ii) Factorise $2m - pk + 2k - pm$ - Group terms: $(2m - pm) + (2k - pk)$ - Factor common terms in each group: $m(2-p) + k(2-p)$ - Factor out common $(2-p)$: $(2-p)(m+k)$ 2. Solve the equation $\frac{x - 1}{x + 1} - \frac{6}{x - 1} = 1$ - Multiply both sides by $(x+1)(x-1)$ to clear denominators: $(x-1)(x-1) - 6(x+1) = (x+1)(x-1)$ - Expand: $(x-1)^2 - 6(x+1) = x^2 - 1$ - Expand and simplify: $x^2 - 2x + 1 - 6x - 6 = x^2 - 1$ $x^2 - 8x - 5 = x^2 - 1$ - Subtract $x^2$ from both sides: $-8x - 5 = -1$ - Add 5 to both sides: $-8x = 4$ - Divide by $-8$: $x = -\frac{1}{2}$ 3. (a) Solve $3m + 12 \leq 8m - 5$ - Subtract $3m$ from both sides: $12 \leq 5m - 5$ - Add 5: $17 \leq 5m$ - Divide by 5: $\frac{17}{5} \leq m$ or $m \geq 3.4$ (b) Solve $\frac{2x + 5}{3 - x} = \frac{14}{15}$ - Cross multiply: $15(2x + 5) = 14(3 - x)$ - Expand: $30x + 75 = 42 - 14x$ - Add $14x$: $44x + 75 = 42$ - Subtract 75: $44x = -33$ - Divide by 44: $x = -\frac{33}{44} = -\frac{3}{4} = -0.75$ 4. (c) Solve simultaneous equations: $y = 4 - x$ $x^2 + 2y^2 = 67$ - Substitute for y: $x^2 + 2(4 - x)^2 = 67$ - Expand: $x^2 + 2(16 - 8x + x^2) = 67$ $x^2 + 32 -16x + 2x^2 = 67$ $3x^2 -16x + 32 = 67$ - Simplify: $3x^2 - 16x - 35 = 0$ - Use quadratic formula $x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot (-35)}}{2 \cdot 3}$ $= \frac{16 \pm \sqrt{256 + 420}}{6} = \frac{16 \pm \sqrt{676}}{6} = \frac{16 \pm 26}{6}$ - Solutions for x: $x = \frac{16 + 26}{6} = 7$, $x = \frac{16 - 26}{6} = -\frac{5}{3}$ - Find corresponding y: For $x=7$, $y=4-7=-3$ For $x=-\frac{5}{3}$, $y=4 - \left(-\frac{5}{3}\right) = 4 + \frac{5}{3} = \frac{17}{3}$ 5. y is inversely proportional to $(x + 1)^2$ with $y=50$ at $x=0.2$. (a) Write $y$ in terms of $x$. - Because $y \propto \frac{1}{(x+1)^2}$, write $y = \frac{k}{(x+1)^2}$ - Use $x=0.2$, $y=50$ to find $k$: $50 = \frac{k}{(1.2)^2} = \frac{k}{1.44}$ $k = 50 \times 1.44 = 72$ - Final formula: $y = \frac{72}{(x+1)^2}$ (b) Find $y$ when $x=0.5$: $y = \frac{72}{(1.5)^2} = \frac{72}{2.25} = 32$ 6. (a) Factorise $x^2 - 3x - 10$ - Find factors of $-10$ that sum to $-3$: $-5$ and $2$ - Factorise: $(x - 5)(x + 2)$ (b)(i) Show that $\frac{x+2}{x+1} + \frac{3}{x} = 3$ simplifies to $2x^2 - 2x - 3 = 0$. - Multiply both sides by $x(x+1)$ to clear denominators: $x(x+1) \left( \frac{x+2}{x+1} + \frac{3}{x} \right) = 3x(x+1)$ - Distribute: $x(x+2) + 3(x+1) = 3x(x+1)$ - Expand: $x^2 + 2x + 3x + 3 = 3x^2 + 3x$ - Simplify left side: $x^2 + 5x + 3 = 3x^2 + 3x$ - Move all terms to one side: $0 = 3x^2 + 3x - x^2 - 5x - 3 = 2x^2 - 2x - 3$ - Hence, the equation reduces to: $2x^2 - 2x - 3 = 0$