Solve the system of equation algebraically
6x-3y=-33
2x+y=-1
To solve the system of equations algebraically, we can use the method of substitution or elimination.
1. Method of Substitution:
We can solve one equation for one variable and substitute it into the other equation.
From the second equation, we can isolate y:
2x + y = -1
y = -2x - 1
Substitute this value of y into the first equation:
6x - 3(-2x - 1) = -33
6x + 6x + 3 = -33
12x + 3 = -33
12x = -36
x = -3
Substitute the value of x into the second equation to find y:
2(-3) + y = -1
-6 + y = -1
y = 5
The solution to the system of equations is x = -3 and y = 5.
2. Method of Elimination:
We can multiply the second equation by 3 and add it to the first equation to eliminate y.
6x - 3y = -33
2x + y = -1
Multiply the second equation by 3:
6x + 3y = -3
Add the multiplied second equation to the first equation:
6x - 3y + 6x + 3y = -33 + (-3)
12x = -36
x = -3
Substitute the value of x into the second equation to find y:
2(-3) + y = -1
-6 + y = -1
y = 5
The solution to the system of equations is x = -3 and y = 5.
Therefore, the system of equations is solved algebraically and the solution is x = -3 and y = 5.