Mathematics Questions and Answers

2x - 3y = -10; 10x - 6y = -5

2x - 3y = -10 x 2

10x - 6y = -5

4x - 6y = -20 .......(i)

10x - 6y = -5.......(ii)

eqn(ii) - eqn(1)

6x = 15

x = \(\frac{15}{6}\)

= \(\frac{5}{2}\)

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

Sub. for x in equ.(ii) 10(\(\frac{5}{2}\)) - 6y = -5

6y = 25 + 5 → 30

y = \(\frac{30}{6}\)

y = 5

\(\frac{2}{2r - 1}\) - \(\frac{5}{3}\) = \(\frac{1}{r + 2}\)

\(\frac{2}{2r - 1}\) - \(\frac{1}{r + 2}\) = \(\frac{5}{3}\)

\(\frac{2r + 4 - 2r + 1}{2r - 1 (r + 2)}\) = \(\frac{5}{3}\)

\(\frac{5}{(2r + 1)(r + 2)}\) = \(\frac{5}{3}\)

5(2r - 1)(r + 2) = 15

(10r - 5)(r + 2) = 15

10r2 + 20r - 5r - 10 = 15

10r2 + 15r = 25

10r2 + 15r - 25 = 0

2r2 + 3r - 5 = 0

(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)

(r - 1)(2r + 5) = 0

r = 1 or \(\frac{-5}{2}\)

2x + y = 4......(i)

x^2 + xy = -12........(ii)

from eqn (i), y = 4 - 2x

= x2 + x(4 - 2x)

= -12

x2 + 4x - 2x2 = -12

4x - x2 = -12

x2 - 4x - 12 = 0

(x - 6)(x + 2) = 0

sub. for x = 6, in eqn (i) y = -8, 8

=(6,-8); (-2, 8)

3x² - 7x + R. Computing the square, we have
x² - \(\frac{7}{3}\) = -\(\frac{R}{3}\)

(\(\frac{x}{1} - \frac{7}{6}\))² = -\(\frac{R}{3}\) + \(\frac{49}{36}\)

\(\frac{-R}{3}\) + \(\frac{49}{36}\) = 0

R = \(\frac{49}{36}\) x \(\frac{3}{1}\)

= \(\frac{49}{12}\)

y = x³ + x and y = x² + 1
\(\begin{array}{c|c} x & -2 & -1 & 0 & 1 & 2 \\ \hline Y = x^3 + x & -10 & -2 & 0 & 2 & 10 \\ \hline y = x^2 + 1 & 5 & 2 & 1 & 2 & 5\end{array}\)
The curves intersect at x = 1