How do you solve the following system?: $5 x + 3 y = 7, 91 x + 56 y = 0$ $5x +3y =7, 91x +56y = 0$

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Answer 1 · 2018-06-21T08:48:55

$x = 56, y = - 91$ $color(purple)(x = 56, y = -91$

$5 x + 3 y = 7, Eqn (1)$ $5x + 3y = 7, " Eqn (1)"$

$280 x + 168 y = 392, Eqn (1) * 56$ $280x + 168y = 392, "Eqn (1) * 56"$

$91 x + 56 y = 0 . Eqn (2)$ $91x + 56y = 0. "Eqn (2)"$

$273 x + 168 y = 0, Eqn (2) * 3$ $273x + 168y = 0, " Eqn (2) * 3"$

$56 \cdot Eqn (1) - 3 \cdot Eqn (2)$ $56 * " Eqn (1)" - 3 * " Eqn (2)"$ ,

$280 x - 273 x = 392$ $280x - 273x = 392$

$x = \frac{392}{7} = 56$ $x = 392/7 = 56$

Substituting value of x in Eqn (1),

$280 + 3 y = 7$ $280 + 3y = 7$

$3 y = - 273 or y = - 91$ $3y = -273 " or " y = -91$

How do you solve the following system?: 5x+3y=7,91x+56y=05x +3y =7, 91x +56y = 0