# How to solve this algebra II problem (solving systems of linear equations in 3 variables)?

Which values should be given to a, b, and c so that the linear system shown has (-1, 2, -3) as its only solution?

x + 2y - 3z = a

- x - y + z = b

2x + 3y - 2z = c

thanks in advance! ♥

### 3 Answers

- GMTLv 67 years agoFavorite Answer
Given that

(-1, 2, -3) = ((x, y, z)

x + 2y - 3z = a

-x - y + z = b

2x + 3y - 2z = c

x + 2y - 3z = a =-1 + 4 + 9 = 12 = a

-x - y + z = b = 1 - 2 - 3 = -4 = b

2x + 3y - 2z = c = -2 + 6 + 6 = 10 = c

a = 12

b = -4

c = 10

I hope this helps!

- David NuttallLv 67 years ago
Sub in the given values into the system of equations and then determine a, b and c.

-1 + 2(2) - 3(-3) = -1 + 4 + 9 = 12 = a

etc.

- currenLv 43 years ago
First, use the Distributive Property on both equations: .5x + .5y + 1 = 0 .5x - .5y = 2 Now, take both equation--i'm going to use the bottom one--to remedy for a variable (i will decide upon x). .5x - .5y = 2 .5x = 2 + .5y Divide all sides via half. X = 1 + y Now substitute x into the primary equation: .5x + .5y + 1 = 0 .5(1 + y) + .5y + 1 = 0 Use the Distributive Property. .5 + .5y + .5y + 1 = zero Add like terms. 1.5 + y = zero y = -1.5 Now alternative y into the primary equation: .5x + .5y + 1 = zero .5x + .5(-1.5) + 1 = 0 .5x - .75 + 1 = zero Add like phrases. .5x + .25 = 0 .5x = -.25 Divide either side through half of. X = -.5 ok! So now you've gotten y = -1.5 and x = -.5!