Question

1278 Найдите решение систем уравнений
{ 4х+9у-9=0
5х-6у-17=0

Answer 1

Ответ:

$(3;-\frac{1}{3} )$

Объяснение:

$\left \{ {{4x + 9y-9=0} \atop {5x - 6y -17=0}} \right.$ $\left \{ {{4x + 9y-9=0|*5} \atop {5x - 6y -17=0|*4}} \right.$ $\left \{ {{20x + 45y-45=0} \atop {20x - 24y -68=0}} \right.$ $\left \{ {{20x=45 - 45y} \atop {20x=68 + 24y}} \right.$ ⇒ 45 - 45y = 68 + 24y

45 - 45y = 68 + 24y

-23 = 69y|:(69)

$y = -\frac{23}{69} = -\frac{23}{23 * 3} = -\frac{1}{3}$

20x = 45 - 45y ⇒ $x = \frac{45 - 45y}{20} = \frac{5(9 - 9y)}{5 * 4} = \frac{9 - 9y}{4} = \frac{9 + \frac{9}{3} }{4} = \frac{12}{4} = 3$

$(3;-\frac{1}{3} )$