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

solve the system of equations Thank you

### 3 Answers by Expert Tutors

Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (7 lesson ratings) (7)
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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
Shelly J. | Excellent Maths Tutoring for academic successExcellent Maths Tutoring for academic su...
5.0 5.0 (248 lesson ratings) (248)
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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
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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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?]