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

2x-y=3

x+2y=-6

multiply the second equation by -2

-2x-4y=12

add the first equation and the new second equation

2x-y=3

-2x-4y=12

-5y=15

y=-3

substitute into the first equation

2x-(-3)=3

2x+3=3

2x=0

x=0

solution:(0,-3)

 

using substitution...

2x-y=3

x+2y=-6

x=-2y-6

substitute into the first equation

2(-2y-6)-y=3

-4y-12-y=3

-5y-12=3

-5y=15

y=-3 again,

substitute into the first equation

2x-(-3)=3

2x+3=3

2x=0

x=0 again

these are just two ways to solve the problem; you can solve it in other ways:you can multiply the first equation by 2 and take it from there or solve for y in the first equation and then substitute into the second equation

Hi Roseland,

 

2x-y=3    equation 1

 

x+2y=-6  equation 2

 

solving equation 1, we get

2x-3=y   

 

substitute the value 0f y=2x-3 in equation 2

 

x+2y=-6

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

 

x+4x-6=-6

 

5x=-6+6

 

5x=0

x=0

 

y=2x-3=2(0)-3=0-3=-3

 

 

The solution to the given system of equations is x=0 and y=-3

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

R1 2, -1, 3
R2 1, 2, -6

R1 > R1 - 2*R2

R1 0, -5, 15
R2 1, 2, -6

R1 > R1 / (-5)

R1 0, 1, -3
R2 1, 2, -6

R2 > R2 - 2*R1

R1 0, 1, -3
R2 1, 0, 0

0x + y = -3
x + 0y = 0

y = -3
x = 0

check:

2(0) - (-3) = 3 √
(0) + 2(-3) = -6 √

[Why keep writing the variables when all the information is in the coefficients and constants?]