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x + y = 2 --- eqn I 3x - 2y = 1 --- eqn II Multiply eqn I by 2 2x + 2y = 4 --- eqn III add eqn II to eqn III 5x = 5 => x = 1 Substitute 1 for x in eqn I 1 + y = 2 => y = 1

x + y = 2........... (i) 3x - 2y = 1......(ii) From (i): x = 2 - y ..........(iii) subt. (2 - y) for x in (ii) 3(2 - y) 2y = 1 6 - 3y - 2y = 1 -5y = 1 - 6 = -5 y = \(\frac{-5}{-5}\) = 1 subst. (i) for y in (iii) x = 2 - 1 = 1 x = 1, y = 1

(a) \(\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\) = \(\frac{(x - 4)(x - 4)}{(x - 4)(x - 3)}\) = \(\frac{(x - 4)}{(x - 3)}\) (b) Arithmetic Progression: \(\frac{1}{2}, \frac{1}{x}, \frac{1}{3}\) \(\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}\) \(\frac{2 - x}{2x} = \frac{x - 3}{3x}\) \(\frac{2 - x}{x - 3} = \frac{2x}{3x}\) \(\implies \frac{2 - x}{x - 3} = \frac{2}{3}\)

(a) \(T_{n} = a + (n - 1)d\) (terms of an A.P) 3rd term = -9 ; 8th term = 26 \(\implies T_{3} = a + 2d = -9 ... (1)\) \(\implies T_{8} = a + 7d = 26 ..... (2)\) (i) Solving (2) - (1), we have \(7d - 2d = 26 - (-9)\) \(5d = 35 \implies d = 7\) (ii) Putting d = 7 in (1) above, we have \(a + 2(7) = -9\) \(a + 14 = -9\) \(a = - 9 - 14 = -23\) (b) \(\frac{XP}{XY} = \frac{2}{5}\) (similar triangles) \(\frac{24}{\text{Area of \Delta XYZ} = \frac{2^{2}}{5^{2}}\) \(\text{Area of } \Delta XYZ = \frac{24 \times 25}{4} = 150 cm^{2}\) \(\therefore \text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\)

\(\begin{pmatrix}3 & -2 \\ -7 & 5 \end{pmatrix} + 2 \begin{pmatrix}-2 & 4 \\ 3 & -1 \end{pmatrix}\) \(2 \begin{pmatrix}-2 & 4 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix}2 \cdot -2 & 2 \cdot 4 \\ 2 \cdot 3 & 2 \cdot -1\end{pmatrix} = \begin{pmatrix}-4 & 8 \\ 6 & -2\end{pmatrix}\) \(\begin{pmatrix}3 & -2 \\ -7 & 5 \end{pmatrix} + \begin{pmatrix}-4 & 8 \\ 6 & -2\end{pmatrix} = \begin{pmatrix}3 + (-4) & -2 + 8 \\ -7 + 6 & 5 + (-2)\end{pmatrix}\) \(= \begin{pmatrix}-1 & 6 \\ -1 & 3\end{pmatrix}\) Thus, the final result is: \(\begin{pmatrix}-1 & 6 \\ -1 & 3\end{pmatrix}\)

I = \(\begin{pmatrix}1 & 0 \\0 & 1 \end{pmatrix}\) Then, 3x - 2I becomes 3p - 2I (since we are finding f(p)) 3\(\begin{pmatrix}2 & 1 \\-1 & 0 \end{pmatrix}\) - 2\(\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}\) = \(\begin{pmatrix}6 & 3 \\-3 & 0 \end{pmatrix}\) - \(\begin{pmatrix}2 & 0 \\ 0 & 2 \end{pmatrix}\) = \(\begin{pmatrix}4 & 3 \\-3 & -2 \end{pmatrix}\)

dy/dx = 2x + 7 y = ∫2x + 7 y = x2 + 7x + C at (2,0) 0 = 22 + 7(2) + C 0 = 4 + 14 + C 0 = 18 + C C = -18 ∴ The equation is y = x2 + 7x - 18

The set of points equidistant from two parallel lines can be determined by the perpendicular distance from a point to each line. Given the lines are of the form ax + by + k\(_1\) = 0 and ax + by + k\(_2\) = 0, the locus of points equidistant from these lines is another line parallel to them. To find this line, which represents the equation of the locus, we need to find the average k\(_1\) and k\(_2\), which we shall be denoted by q This line will have the same coefficients for x and y as the original lines. Thus, the equation of the locus is: ax + by + q = 0

\(\frac{x+4}{3}-\frac{x-3}{2}\) < 4 \(\frac{(2(x+4))-(3(x-3))}{6}\) < 4 2(x+4) – 3(x-3) < 4 x 6 2x + 8 – 3x+9 < 24 -x + 17 < 24 -x < 24 – 17 -x < 7 x > -7

y = x\(^2\) - x - 12 = (x - 4)(x + 3) ∴ x = 4 or x = -3 Checking the cases for y \( \geq \) 0 We check values on the range x - 4 \(\geq\) 0; x + 3 \(\leq\) 0; x - 4 \(\leq\) 0 and x + 3 \(\geq\) 0 for the range which satisfies the inequality x\(^2\) - x - 12 \(\geq\) 0. We find that the inequality is satisfied on the range x \(\leq\) -3 and x \(\geq\) 4.

3, - 6, +12, -24 a = 3, r = -2 \(Sn = \frac{a(1-r^n)}{1-r}\\ ∴1-2^8 = \frac{3(1-(-2^{n}))}{1-(-2)}\\ 1-2^8 = \frac{3(1-(-2^{n}))}{3}\) 1 - 2\(^8\) = 1 - (-2)\(^n\) Since 1 - 2\(^8\) = 1 - (-2)\(^n\)

2p x 1 + 8 x 2 = 24 \(\to\) 4p = 24 - 8 = 16, p = 4 3 x 1 + -5q x 2 = -17 \(\to\) -10q = -17 - 3 -10q = -20 q = 2

2(1) - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4 2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1 2(1) - 2 = 0 2(0) - (-1) = 1 2(1) - (1) = 1 \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0. First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1. Then, a\(_0\) which is the coefficient void of x = 6. The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6. The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\). = \(\pm (1, 2, 3, 6)\). Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0. Therefore x = 1 is a root of the equation. Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6. x\(^2\) - x - 6 = (x - 3)(x + 2). x = -2, 3. \(\therefore\) The roots of the equation = 1, -2 and 3.

The value for which the equation is impossible 2x = 10 (if y = 0) x = 5