solve the system of equations Thank you

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

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# 3 Answers

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?]

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?]