View Full Version : Solving using the elimination method
990123e
May 14, 2009, 02:06 PM
I don't understand solving using for the elimination method. Algebra is like chinese to me can some help me... here is an example of a problem...
4x-2y=-2
4x+3y=-12
:mad::mad::mad:
:confused::confused::confused:
Perito
May 14, 2009, 02:11 PM
4x-2y=-2
4x+3y=-12
4x+3y=-12
4x-2y=-2
Subtract the second equation from the first
4x+3y-(4x-2y)=-12-(-2)
5y=-10 Viola! X is eliminated!
y=-\frac {10}{5} = -2
Substitute this back in to either of the equations to solve for x.
If the coefficients of the variable you're going to eliminate aren't the same, multiply the entire equation -- or both equations, with constants so that when added, one of the variables will be "eliminated".
990123e
May 14, 2009, 02:14 PM
4x+3y=-12
4x-2y=-2
Subtract the second equation from the first
4x+3y-(4x-2y)=-12-(-2)
y=-10 Viola! x is eliminated!
Substitute this back in to either of the equations to solve for x.
If the coefficients of the variable you're going to eliminate aren't the same, multiply the entire equation -- or both equations, with constants so that when added, one of the variables will be "eliminated".
so it's that easy because how do i no to subtract sometimes there are questions like
8x-y=20
5x+y=-8
Perito
May 14, 2009, 02:19 PM
8x-y=20
5x+y=-8
multiply 8x-y=20 by 5
multiply 5x+y=-8 by -8
40x -5y = 100
-40x -8y = 64
add them
-13y = 164
y = -164/13 ≈ 12.616
You could also just add them directly since the y-coefficients are the same (in absolute value). This would eliminate y
8x-y=20
5x+y=-8
13x = 12
x = 12/13
You can select constants to multiply each equation with. It'll still be an equation.
990123e
May 14, 2009, 02:21 PM
multiply 8x-y=20 by 5
multiply 5x+y=-8 by -8
40x -5y = 100
-40x -8y = 64
add them
-13y = 164
y = -164/13 ≈ 12.616
You can select constants to multiply each equation with. It'll still be an equation.
how do I still find x and for the first equation I plugged in x and got 8. is that right? (8,-10)
Perito
May 14, 2009, 02:30 PM
I made a sign error in my first post. :mad: :rolleyes: Y is -2 in the first problem. X will turn out to be -3/2
Yes, you just plug that into either equation and solve for x. To check it, insert the values for x and y into the other equation and it has to be an equality. That's how I figured out that I had made a mistake.