Solve the following system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions.
-a+b=-4
2a-3b=-2
Multiply first equation by 2, then add the equations.
-2a+2b=-8
2a-3b=-2
-b = -10
Solve for b then a.
To solve the system using elimination, we want to eliminate one variable by manipulating the equations in a way that allows us to subtract one equation from the other, resulting in a new equation with only one variable. Let's start by multiplying the first equation by 2 to make the coefficients of "a" the same in both equations:
-2(a) + 2(b) = -2(-4)
This simplifies to:
-2a + 2b = 8 (Equation 1')
Now we have two equations with matching coefficients for "a," so we can subtract Equation 2 from Equation 1':
-2a + 2b - (2a - 3b) = 8 - (-2)
Simplifying this equation, we get:
-2a + 2b - 2a + 3b = 8 + 2
Combining like terms:
2b + 3b = 10
Simplifying further:
5b = 10
Dividing both sides by 5:
b = 2
Now that we have the value of "b," we can substitute it back into one of the original equations to solve for "a." Let's use the first equation:
-a + 2 = -4
Simplifying:
-a = -6
Multiplying both sides by -1:
a = 6
Therefore, the solution to the system of equations is the ordered pair (a, b) = (6, 2).