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February 1, 2015

February 1, 2015

Posted by **Tiffany** on Thursday, February 10, 2011 at 1:09am.

The two numbers chosen for my solution will be 3 and 6.

Solving the system of equations by the Addition/Subtraction method

Step 1 – Write both equations in the form of ax+ by= c

Equation 1: 4x-y = 3

Equation 2: x+y = 6

Step 2 –Multiply one or both equations by numbers so that the absolute values of the either of the coefficients of the x terms or the y terms are alike, I will multiply my equation by -4.

x + y = 6

-4(x + y) = -4(6)

-4x -4y = -24

Step 3 – Eliminate one of the variables by adding the equations if the signs of the coefficients of the variable are different. Subtract the equations if the signs of the coefficients of the variable are the same.

(4x – y = 3) + (-4x -4y = -24)

-3y = -12 --> y = 3

Step 4 – Plug back into equation 1 to solve for x:

4x – y = 3

4x – (3) = 6

2x = 12 --> x = 6

Step 5

CHECK using equation 2:

x+y = 6+ 3 = 9

So the solution is (6

- Math -
**Reiny**, Thursday, February 10, 2011 at 9:04amThis looks like a lot of

"Much Ado About Nothing"

If your equations are

4x-y=3 and

x+y = 6 , why not just add them as they are?

5x=9

x=9/5 or 1.8

sub into x+y=6

you can get

y = 4.2

your solution of x=6, y=3 work in your second equation, but not in the first.

So your solution is incorrect.

The error was when you added your equations

4x-y=3

-4x-4y=-24

------------

-5y = -21

y = -21/-5 = 4.2

etc.

- Math -
**renos**, Monday, November 21, 2011 at 3:21pmx=9/5

y=21/5

solve one equation for x and plug the answer to the other equation and solve it

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