Elimination (Algebra) Method Help?

Systems of equations can be added by their coefficients.

4. x-y + x + y = 16

2x = 16

x = 8

y = 12 - 8 = 4

Sometimes you need to multiply both sides of an equation to cancel terms.

5. -3x +6y = 36 (Multiplied the first equation by 3)

3x -5y = 10

-3x + 6y +3x - 5y = 22

y = 22

x = 2y - 12 = 32

4)

x-y=4

x+y=12

First you have to eliminate one of the unknowns(either x or y whichever is easier to eliminate) by either adding or subtracting the two equations.

If the unknown is negative in one equation and positive in another equation, subtract. If the unknown is positive or negative in both equations, then add them together.

You should add or subtract element by element that is x from x, y from y and the number from number.

(x-y=4) +(x+y=12)

That would be [(x+x) +( -y+y) = (4+12)]

This will give you 2x=16 (y has been eliminated)

then x=16/2

hence x=8

When x=8, substitute this one of the equations

x+y=12

8+y=12

y=12-8

y=4

So, answers are x=8 and y=4.

I know it is a bit confusing but once you know how, you will find it to be very easy.

5) is a bit more complicated as you have to multiply one the equation;s.

-x+2y=12

3x-5y=10

(-x+2y=12) *3

it will become -3x+6y=36

then take it from there and do elimintion as shown above

Why refer to 4) and 5) rather than 1) or 2) ?

1)

2x = 16

x = 8

y = 4

2)

-3x + 6y = 36

3x - 5y = 10______ADD

y = 46

92 - 12 = x

x = 80

x = 80 , y = 46

(4) Since x = 4+y (from the first equation),

ELIMINATE x from the second equation by writing

(4+y) + y = 12

So 2y = 8, y = 4, and x = 8.

(5) Since x = 2y - 12 (from the first equation),

ELIMINATE x from the second equation by writing

3(2y - 12) - 5y = 10

6y - 36 - 5y = 10

y = 46

x = 80

4. Add both eqs. You get,

x - y + x + y = 4 + 12,

ie, 2x = 16, therefore x = 8

Use that in 2nd eq. you get,

8 + y = 12, therefore y = 12 - 8 = 4.

5. Multiply 1st eq. by 3 and add to 2nd eq. You get,

-3x+6y+3x-5y=36+10

ie, y = 46

Use that in 1st eq. you get,

-x + 92 = 12

ie x = 80