Answer:
The solution for given system of equations is: x = 6 and y = 3
Or
(6,3)
Step-by-step explanation:
Given equations are:
[tex]3x+4y=30\ \ \ Eqn\ 1\\3x-y=15\ \ \ \ Eqn\ 2[/tex]
There are three methods to solve simultaneous equations
We will use the elimination method as the coefficients of x in both equations are already same
Subtracting equation 2 from equation 1
[tex]3x+4y - (3x-y) = 30-15\\3x+4y-3x+y = 15\\5y = 15\\\frac{5y}{5} = \frac{15}{5}\\y = 3[/tex]
Putting y = 3 in equation 2
[tex]3x-3 = 15\\3x = 15+3\\3x = 18\\\frac{3x}{3} = \frac{18}{3}\\x = 6[/tex]
Hence,
The solution for given system of equations is: x = 6 and y = 3
Or
(6,3)
