(vi) 2x − 3y = 14; 5x + 2y = 16

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Solution

$2 x - 3 y = 14 \Rightarrow x = \frac{14 + 3 y}{2} . . . (i) 5 x + 2 y = 16 . . . . (ii) Substituting eq (i) in eq (ii), we get : 5 \times (\frac{14 + 3 y}{2}) + 2 y = 16 \Rightarrow 70 + 15 y + 4 y = 32 \Rightarrow 19 y = - 38 \Rightarrow y = - 2 Substituting y = - 2 in eq (i), we get : x = \frac{14 + 3 \times (- 2)}{2} ∴ x = 4 Hence, x = 4, y = - 2 is the solution of the given equations .$

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2 x - 3 y = 14; 5 x + 2 y = 16

I. 2x + 3y = 14

II. 4x + 2y = 16

\frac{1}{2 (2 x + 3 y)} + \frac{12}{7 (3 x - 2 y)} = \frac{1}{2}

\frac{7}{2 x + 3 y} + \frac{4}{3 x - 2 y} = 2

2 x + 3 y \neq 0

3 x - 2 y \neq 0

Expand each of the following, using suitable identities:

(i) (ii)

(iii) (iv)

(v) (vi)

12 (2 x - 3 y)^{2} - 16 (3 y - 2 x)

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