# How is this system is solved. Tried for sometime but could not solve it?

### 2 Answers

- Jeff AaronLv 71 month agoFavourite answer
Multiply the first equation by a + 1:

(a+1)x + (a+1)y + 3(a+1)z = a(a + 1)

Add equations 2 and 3:

(a+1)x + (a+1)y + 9z = 4 + a

Subtract those two equations:

(3a + 3 - 9)z = a(a + 1) - (4 + a)

(3a - 6)z = a^2 + a - 4 - a

(3a - 6)z = a^2 - 4

z = (a^2 - 4) / (3a - 6)

z = ((a + 2)(a - 2)) / (3(a - 2))

z = (a + 2) / 3

Substitute that into the first equation:

x + y + 3((a + 2) / 3) = a

x + y + a + 2 = a

x + y + 2 = 0

y = -x - 2

Substitute y and z into the the third equation:

x + a(-x - 2) + 4((a + 2) / 3) = a

x - ax - 2a + ((4a + 8) / 3) = a

Multiply through by 3:

3x - 3ax - 6a + 4a + 8 = 3a

x(3 - 3a) - 2a + 8 = 3a

x(3 - 3a) = 3a + 2a - 8

x(3 - 3a) = 5a - 8

x = (5a - 8) / (3 - 3a)

Substitute x into y = -x - 2 above:

y = -((5a - 8) / (3 - 3a)) - 2

y = ((5a - 8) / (3a - 3)) - 2( (3a - 3) / (3a - 3))

y = (5a - 8 - 2(3a - 3)) / (3a - 3)

y = (5a - 8 - 6a + 6) / (3a - 3)

y = (-a - 2) / (3a - 3)

y = (a + 2) / (3 - 3a)

Note that if 3 - 3a = 0, i.e. if a = 1, then x and y are undefined, so there are no solutions. That answers question (a).

Also note that I cancelled out a - 2 when solving for z, so what happens if that's zero, i.e. a = 2? That makes z indeterminate, i.e. it can be anything, so that means there are infinity solutions, so that answers question (c).

Any other value of "a" will yield a unique solution, so that's the answer to question (b).

- Anonymous1 month ago
a lot of job to do, good luck