How do you solve 3x+y =-2 and x-y=6?

Question

With whatever method is easiest...substitution or elimination?

Answer 1

You can also cancel -y and +y just by adding this two equations. so when you add it just left with:

4x=4

x=1 

then plug back 1 to one of the equation.

1-y=6

1-6=y

y=-5

Answer 2

I would solve for x with the first equation, then use that to solve for y in the second, then check by inputting values.

3x+y=-2
x-y=6
x=6+y

then...

3(6+y)+y=-2
18+3y+y=-2
18+4y=-2
4y=-20
y=-5

So..

3x+-5=-2
3x=3
x=1

3(1) + -5 = -2  ... yes! :)

Answer 3

In this scenario, the goal of finding the values of both x and y the fastest is to find which variable will be easiest to solve and determine its values, which in this case must be x because you can use elimination or substitution to derive at x.

 

Elimination Process:

3x+y=2 and x-y=6 (adding them together)

4x=8 --> x=2

2-y=6 --> (through common sense) y=-4 since 2-(-4)=2+4=6

Test the solution witht he other equation:

3(2)+(-4)=6-4=2

 

Substitution Method: (as written above margery b.)

 

The key here Angelica is to use your instinct and determine what variable is best to be eliminated first so that you are left with only one variable to solve.

 

If you need any further help on your math, please check my profile or shoot me another email for tons of mathematical magics.

 

-Andy Lai "If adversary smiles, military people smiles back"

Answer 4

This is a system of linear equations that can be solved by elimination and substitution: First we will use elimination to solve for x 3x + y = -2 x- y= 6 Add the 2 equations and the result is: 3x + y = -2 x- y= 6 4x = 4 Divide both sides by 4 (Inverse operation) 4x/4 = 4/4 x= 1 Substitute 1 for x in the 1st equation 3 (1) + y = -2 3 + y = -2 Subtract 3 from both sides (Inverse operation) 3-3 +y = -2 - 3 y = -2 + -3 y = -5 Substitute the solution for x and y into both equations to verify. 3(1) + -5 = -2 3 + -5 = -2 -2 = -2 1-(-5) = 6 1+5 = 6 6=6

Answer 5

SO the first answer is Substitution and the Second answer offered is using Elimination.  I would say in this case the Elimination is easier, but both are quite simple for this problem.