Question

Resolva os sistemas propostos.

A)
2x + 3y = -7
3x + 4y = -13

B)
5x + 5y = -10
2x + 3y = -13

C)
-14x + 2y = -26
3x + y = -3

D)
4x + 3y = -7
3x + 5y = -25

Answer 1

Explicação passo-a-passo:

a)

$2x + 3y = -7$

$(2). x +3y+7=0$

$x= \frac{(-3y+7)}{2}$

$x = - \frac{3}{2} y - \frac{7}{2}$

$3x + 4y = -13$

$3x+4y+13=0$

$(3) . x + (4y+13)=0$

$x=\frac{-4y+13}{3}$

$x=-\frac{4}{3} y-\frac{13}{3}$

b)

$5x + 5y = -10$

$5.x+(5y+10)=0\\\\x= \frac{(-5y+10)}{5} \\x= \frac{5.(y+2)}{5} \\x= -y -2$

$2x + 3y = -13\\2x+3y+13=0\\2.x+(3y+13)=0\\x= \frac{-(3y+13)}{2} \\x=-\frac{3}{2} y - \frac{13}{2}$

c)

$-14x + 2y = -26\\-14x+(2y+26)=0\\(-14).x+(2y+26)=0\\x= \frac{-(2y+26)}{-14} \\x= \frac{2.(y+13)}{2.(-7)} \\x= \frac{1}{7}y+\frac{13}{7}$

$3x + y = -3\\3x+y+3=0\\(3).x+(y+3)=0\\x= \frac{-(y+3)}{3} \\x= -\frac{1}{3}y -1$

d)

$4x + 3y = -7$

$4x+3y+7=0\\(4).x+(3y+7)=0\\x= \frac{-(3y+7)}{4} \\x= - \frac{3}{4} -\frac{7}{4}$

$3x + 5y = -25\\3x+5y+25=0\\(3).x+(5y+25)=0\\x= \frac{-(5y+25)}{3} \\x=-\frac{5}{3}y - \frac{25}{3}$