How Do You Solve Systems of Linear Equations with Parameters?

[songoku]

Homework Statement

1. Find a linear equation in the variables x, y and z that has a general solution:
x = 3 - 4p + q
y = p
z = q
where p and q are arbitrary parameters

2. Express a general solution for the equation in part (1) in two other different ways

3. Write down a linear system of two different non zero linear equations such that the system has the same general solution as in part (1)

Relevant Equations

Not sure

1)
x = 3 - 4p + q
x = 3 - 4y + z
x + 4y - z = 3

2) x + 4y - z = 3
(i) let x = a and y = b, so z = a + 4b - 3
General solution:
x = a
y = b
z = a+ 4b - 3

(ii) let x = r and z = t, so y = (3 - r + t) / 4
General solution:
x = r
y = (3 - r + t) / 4
z = t3) I don't understand this part. Is the answer the same as part (1)? Can I just take random equations like x +2y = 1 and 2y - z = 2? Is this the form asked by the questions?

Thanks

 

[LCKurtz]

Strange question. I don't think your two equations in (3) will do it. You have two planes that aren't parallel and their intersection would be a line, not a plane. He is asking for two linear equations (planes) whose intersection would be a plane. Seems to me they have to be the same plane. Maybe $2x+8y-2z = 6,~5x+20y - 5z = 15$? Seems pretty silly to me.

 

[songoku]

Strange question. I don't think your two equations in (3) will do it. You have two planes that aren't parallel and their intersection would be a line, not a plane. He is asking for two linear equations (planes) whose intersection would be a plane. Seems to me they have to be the same plane. Maybe $2x+8y-2z = 6,~5x+20y - 5z = 15$? Seems pretty silly to me.

Not sure, maybe that is the answer. Thank you very much