How do solve the following linear system?: $9 x + 1 = - 4 y, 7 x - 2 y - 13 = 0$ $9x + 1 = -4y, 7x - 2y - 13 = 0$ ?

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Answer 1 · 2018-03-05T16:26:43

$x = - \frac{62}{5}, y = \frac{27}{5}$ $color(blue)(x = -62/5), color(green)(y = 27/5)$

$9 x + 1 = - 4 y$ $9x + 1 = -4y$ Eqn (1)

$y = \frac{9 x + 1}{- 4}$ $y = (9x + 1) / -4$

$7 x - 2 y = 13$ $7x - 2y = 13$ Eqn (2)

Substituting value of yin Eqn (2),

$7 x - (2 \cdot \frac{9 x + 1}{- 4}) = 13$ $7x - (2 * (9x + 1)/-4) = 13$

$- 28 x + 18 x + 2 = - 52$ $-28x + 18x + 2 = -52$

$- 10 x = - 54$ $-10x = - 54$ or $x = \frac{27}{5}$ $x = 27/5$

Substituting value of x in Eqn (1),

$y = \frac{9 \cdot (\frac{27}{5}) + 1}{- 4} = \frac{243 + 5}{- 20} = - \frac{62}{5}$ $y = (9 *(27/5) + 1) / -4 = (243 + 5) / -20 = -62/5$

How do solve the following linear system?: 9x + 1 = -4y, 7x - 2y - 13 = 0 9x+1=−4y,7x−2y−13=09x + 1 = -4y, 7x - 2y - 13 = 0 ?