Respuesta :

Simultaneous Equations:
(3x + 5y = 7)×(-3)
(4x + 3y = 2)×(5)

-9x - 15y = -21
20x+15y =10        (-15y and and+15y cancels out)

11x = -11
[tex] \frac{11x}{11} [/tex] = [tex] \frac{-11}{11} [/tex] (11 and 11 cancels out)
x= -1
substitute the x we found into either original expressions

3(-1) + 5y = 7
-3 + 5y = 7
+3 -3+ 5y = 7+ 3
5y = 10
[tex] \frac{5y}{5} [/tex]=[tex] \frac{10}{5} [/tex] (5 and 5 cancels out)
y=2 
Check it!
4(-1)+3(2) = 2
-4 + 6 = 2
2 = 2

3(-1) + 5(2) = 7
-3 + 10 = 7
7 = 7

x= -1
y= 2

Abrax
i think there was a something wrong in you question because the answer does not be the same as you want