Mark M. answered • 07/27/18

Math Tutor--High School/College levels

x-2y+3z = 4

2x-3y+az = 5

3x-4y+5z = b

Multiply the first equation by -2 and add to the second equation. Multiply the first equation by -3 and add to the third equation.

x - 2y + 3z = 4

y + (a-6)z = -3

2y - 4z = (b-12)

Multiply the second equation by -2 and add to the third equation:

x - 2y + 3z = 4

y + (a-6)z = -3

(-2a+8)z = (b-6)

Infinitely many solutions when -2a+8 = 0 and b-6 = 0

So, a = 4 and b = 6

No solution if -2a+8 = 0 and b-6 ≠ 0

So, a = 4 and b ≠ 6

Unique solution if a ≠ 4 and b = any real number

Sandra A.

I really appreciate your support. Thanks a lot.

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07/27/18

Sandra A.

07/27/18