Simultaneous Equations:
(3x + 5y = 7)×(-3)
(4x + 3y = 2)×(5)
-9x - 15y = -21
20x+15y =10 (-15y and and+15y cancels out)
11x = -11
\(\frac{11x}{11}\) = \(\frac{-11}{11}\) (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
\(\frac{5y}{5}\)=\(\frac{10}{5}\) (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
