Question

Substitute this simultaneous linear equation
$3x {}^{ - 1} - {4y}^{ - 1} = 5 \\ {2x}^{ - 1} + {3y}^{ - 1} = 7$
​

Answer 1

Answer:

${3x}^{ - 1} - {4y}^{ - 1} = 5 \: ....(1) \\ \\ {2x}^{ - 1} + {3y}^{ - 1} = 7 \: ....(2)$

__o__o__

$\frac{1}{3x} - \frac{1}{4y} = 5 \: ....(1) \\ \\ \\ \frac{1}{2x} + \frac{1}{3y} = 7 \: ....(2)$

From equation (1) ;

$\frac{1}{3x} - \frac{1}{4y} = 5 \\ \\ \frac{1}{4y} = \frac{1}{3x} - 5 \\ \\ \frac{1}{4y} = \frac{1 - 15x}{3x} \\ \\ \\ 4y(1 - 15x) = 3x \\ \\ 4y = \frac{3x}{1 - 15x} \\ \\ y = \frac{3x}{4(1 - 15x)} \\ \\ \\ y = \frac{3x}{4 - 60x}$

$\frac{1}{2x} + \frac{1}{3y} = 7 \\ \\ \\ \frac{1}{2x} + ( \frac{1}{3( \frac{3x}{4 - 60x} )} ) = 7 \\ \\ \\ \frac{1}{2x} + ( \frac{1}{ \frac{9x}{4 - 60x} } ) = 7 \\ \\ \\ \frac{1}{2x} + \frac{4 - 60x}{9x} = 7 \\ \\ \\ \frac{9}{18x} + \frac{2(4 - 60x)}{18x} = 7 \\ \\ \\ \frac{9}{18x} + \frac{8 - 120x}{18x} = 7 \\ \\ \\ \frac{9 + 8 - 120x}{18x} = 7 \\ \\ \frac{17 - 120x}{18x} = 7 \\ \\ 18x \times 7 = 17 - 120x \\ \\ 126x = 17 - 120x \\ \\ 126x + 120x = 17 \\ \\ 246x = 17 \\ \\ x = \frac{17}{246}$

$y = \frac{3x}{4 - 60x} \\ \\ y = \frac{3( \frac{17}{246}) }{4 - 60( \frac{17}{246} )} \\ \\ y = - \frac{17}{12}$