How do I solve a system of equations with two unknowns? (such as the one below)?

Question

-3 = x + 2y
-6 = x + 5y

I thank you both!

Answer 1

You substitute one into the other. Turn (-3=x+2y) into ( x = -2y-3) and now that you know what x is equal to you can plug it into the other equation. Now you have the equation (-6= (-2y-3) + 5y). You simplify that equation to get the value of y. When you have that you plug it into one of the original equations and solve for x.

Answer 2

Two Methods: Pick one - Don't mix them up

Method 1
Pick a variable to eliminate and get a new equation and solve for the other varible and plug answer back in.

ex: I choose to elim x

-1(-3 = x + 2y)
-6 = x + 5y
-------------------
-3 = + 3y

-1 = y <-- ans to plug back in any equation and get x

-3 = x + 2(-1)

-3 = x - 2

-1 = x and -1 = y


Method 2 : Solve one equation for y and plug that equation into the other. I using to 1st equation

-3 - x = 2y

-3 - x
------ = y <- plug into eq two
2

-6 = x + 5( (-3 -x)/2)
-6 = x + (-15 - 5x)/2
2(-6 = x + (-15 - 5x)/2) <-- elim fractions

-12 = 2x -15 - 5x
3 = -3x
-1 = x

plug back in to get y
(-3 -(-1) ) / 2 = y

-2 / 2 = y
-1 = y

There you go, that simple. remember pick one method. you might ask which one is easiest, well which ever you find easy to understand.

Answer 3

- {x+2y = -3 (Saymontlously)
{ x+5y = -6
----------------
-x+x-2y+5y=-3
0+3y = -3
3y=-3
y=-3/3
y=-1

and x+2y =-3
x+2(-1)=-3
x-2 =-3
x=-3+2
x =-1
check x+2y =-3
-1+2(-1)=-3
-1-2 =-3
-3=-3

Answer 4

when u solve the above eqn. by simultaneous eqn. rules ..that is
subtract the two eqn. then u get -3y=3, which gives y=-1
then substitute y=-1 in any one of the above eqns.(im substituting in 1st eqn...then we get
-3=x+2(-1)
-3=x-2
x=-1
therefore x=-1,y=-1.......ok did u understand