(Algebra) Solving Linear Equations-Elimination?

Question

Find the common solution for each system of equations using elimination and/or substitution.

1) x + y = 20 x - y = 10
2) x + y = 8 x - y = -2
3) x + y = 5 x - y = -3
4) x - y = 4 3x + 2y = 7
5) 5x + 5y = -5 3x - y = -7
6) 4x - y = 1 2x + y = 17
7) 2x - y = 4 2x + 4y = 4
8) -x + 3y = 14 x + 22 = 5y
9) 8x - 6y = 2 2x + 3y = 2
10) 2x - 5y = 10 3x - 2y = -7

Answer 1

Okay, I'll do the first one for you and then hopefully that will give you a hint to how the others are done.

x+y = 20
x-y = 10

Add the equations together. You get
2x = 30

So x = 15. Now plug 15 back in for x in the equations. You'll see that y = 5.

The others are done in a similar way.

Answer 2

You add the equations together to eliminate one of the variables, then solve for the remaining variable. In order to eliminate on the variables, their coefficients have to be opposites. For example: 2x -3y -x
-2x 3y x

Sometimes you have to multiply one or both of the equations in order to get one of the variables to be eliminated when you add them.
5)
5x + 5y= -5
3x - y = -7

You can eliminate y from the equations if you multiply the second one by 5.

(3x - y = -7)(5)----------------->15x -5y= -35

Now add the first equation and your new second equation, and solve for x.
5x + 5y = -5
15x - 5y = -35
-------------------
20x = -40
x=-2
Now you have x, so put that into one of your original equations and solve for y.

3x - y = -7
3(-2)-y = -7
-6 - y = -7
-y = -1
y = 1

So your solution is: (-2, 1)

You can eliminate either x or y, whichever is more convienient.

Answer 3

Multiply #a million via 2 >>4x - 6y = 12 Subtract #2 from this >> x = 4 replace this into #a million >> 8 - 3y = 6 upload 3y to the two sides & deduct 8 from the two sides >> 8 - 6 = 3y i.e.3y = 2 Divide the two sides via 3 >> y = 2/3