Just looking over my Linear Algebra textbook so I know what to expect next semester and I have a quick question. The problem is:
x + 4y - z = 12
3x + 8y - 2z = 4
I haven't seen arrays or anything yet, so the way I do it I get:
z = 4y - 32
y = (z+32) / 4 = (1/4)x+8
x = -20
In the answers, x and y are the same as mine but it says z = Any real number
Can somebody explain to me why.
x + 4y - z = 12
3x + 8y - 2z = 4
I haven't seen arrays or anything yet, so the way I do it I get:
z = 4y - 32
y = (z+32) / 4 = (1/4)x+8
x = -20
In the answers, x and y are the same as mine but it says z = Any real number
Can somebody explain to me why.
-
well, x=-20
and y=(z/4)+8
Here the value of y depends on the value of z. i.e for every distinct value of z you will get a distinct y.
So, z can take any real value and y will change accordingly.
I hope that answers your problem.
.....And a Happy New Year!
and y=(z/4)+8
Here the value of y depends on the value of z. i.e for every distinct value of z you will get a distinct y.
So, z can take any real value and y will change accordingly.
I hope that answers your problem.
.....And a Happy New Year!
-
z = 4y - 32 is the same equation as y = (z+32) / 4.
Stating it twice is not a solution. A solution should not be circular; try to define the variables only in terms of numbers or, if necessary, things you have previously defined.
To solve three variables, you need three different equations. You only have two, so it's impossible to solve it completely. What's you'll learn in Linear Algebra is that the number of variables minus the number of equations is the number of degrees of freedom of the solution. This is how many of the variables can be set to any real number; the other variables can be written in terms of the "free" ones.
Another way to think of this is graphically. The solution set to one linear equation in three variables is a plane. Two planes intersect in a line. Your answer should be a description of that line.
This problem has 3 variables and 2 linear equations, so the solution has 1 degree of freedom. It happens that you can solve for x; that's unusual, usually any of the variables can be the free one. Ignoring x for a moment, the solution set is ALL values of y and z such that y = (z+32) / 4. Either y or z can be equal to any real number. Once you choose a value for z, y is fixed (or once you choose a y, z is fixed).
Stating it twice is not a solution. A solution should not be circular; try to define the variables only in terms of numbers or, if necessary, things you have previously defined.
To solve three variables, you need three different equations. You only have two, so it's impossible to solve it completely. What's you'll learn in Linear Algebra is that the number of variables minus the number of equations is the number of degrees of freedom of the solution. This is how many of the variables can be set to any real number; the other variables can be written in terms of the "free" ones.
Another way to think of this is graphically. The solution set to one linear equation in three variables is a plane. Two planes intersect in a line. Your answer should be a description of that line.
This problem has 3 variables and 2 linear equations, so the solution has 1 degree of freedom. It happens that you can solve for x; that's unusual, usually any of the variables can be the free one. Ignoring x for a moment, the solution set is ALL values of y and z such that y = (z+32) / 4. Either y or z can be equal to any real number. Once you choose a value for z, y is fixed (or once you choose a y, z is fixed).