Mathematics Questions and Answers

\(\frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2}\) = \(\frac{(2a)^2 - (7b)^2}{(a + b)(2a - 7b)}\)

= \(\frac{(2a + 7b)(2a - 7b)}{(a + b)(2a - 7b)}\)

= \(\frac{2a + 7b}{a + b}\)

\(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) = \(\frac{1}{(x + 1)(x + 2)}\) + \(\frac{1}{(x + 1)(x + 1)}\)

\(\frac{(x + 1)+ (x + 3)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{x + 1 + x + 3}{(x + 1)(x + 2)(x + 3)}\)

\(\frac{2x + 4}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2(x + 2)}{(x + 1)(x + 2)(x + 3)}\)

= \(\frac{2}{(x + 1)(x + 3)}\)

px² + qx + b = 0

Using almighty formula

\(\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}\).........(i)

Where a = p, b = q and c = b

substitute for this value in equation (i)

= \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\)

3x - 5y = 3, 2y - 6x = -5

-5y + 3x = 3........{i} x 2

2y - 6x = -5.........{ii} x 5

Substituting for x in equation (i)

-5y + 3(\(\frac{19}{24}\)) = 3

-5y + 3 x \(\frac{19}{24}\) = 3

-5y = \(\frac{3 - 19}{8}\)

-5 = \(\frac{24 - 19}{8}\)

= \(\frac{5}{8}\)

y = \(\frac{5}{8 \times 5}\)

y = \(\frac{-1}{8}\)

(x, y) = (\(\frac{19}{24}, \frac{-1}{8}\)

-1 < 5 - 2x \(\geq\) 7 = -1 < 5 -2x and 5 - 2x \(\leq\) 7

= 2x < 5 + 1 and 5 - 7 \(\leq\) 2x = x < 3 and -1 \(\leq\) x

Integral value of x are -1, 0, 1, 2