2G

[naved_s9994]

Ex 2G Q8 in MM Essential CAS 3/4
Cant get it at all?

Techniques, stratergies to solve? Thanks!

[olly_s15]

whats the question?

[naved_s9994]

Cant use latex..Sorry

Find the value of p for which the system of equations

3x + 2y -z =1
x + y + z = 2
px + 2y -z = 1

has more than one solution.

Thats exactly how its worded in the text...

[dcc]

Looking at the equations, is there any way that you can obviously reduce this to a situation where you KNOW there will be an infinite number of solutions? (Remember, two "independent" equations in three variables will have an infinite number of solutions)

[Studyinghard]

The top equation and bottom equation are the same hence p = 3.

[stonecold]

do you have to know how to find the general solution by hand?

[naved_s9994]

The top equation and bottom equation are the same hence p = 3.

Lol, yea I got that bit..  (Easy)

But the answer in the worked sols is

z= 3 + lamda
    ________
        3

[naved_s9994]

do you have to know how to find the general solution by hand?

Yea...   

[stonecold]

yeah, and then you just let lambda =0,1,2,3...... to get as many solutions as you want.

that is the general solution, like in trig....

[Studyinghard]

man my mum did all this stuff when i told her about this question and she got an answer like with normal substitutions etc.

[stonecold]

^  your mum knows maths!  that is awesome.  i wish someone in my house knew maths...

[Studyinghard]

Ill try to do a short version of her working out.
(ceebs latex)

Make 3x + 2y -z = 1 and make x the subject so you get

x = ( 1-2y+z ) / 3

now sub x into the second equation and you get.

[ (1-2y + z) /3 ] + y + z = 2

get rid of the 3 from the denominator and you get..

1-2y+z+3y+3z = 6
y = 5 - 4z


now sub y into first equation where you made x the subject.

In this one :

x = (1 -2y + z) / 3

sub in y .

x = (1- 10 + 8z +z) /3

= 3z -3

now sub both x and y into the second equation which x + y + z = 2

3z - 3 + 5 - 4z + z = 2
-z = -2
z = 2

So z = 2.
sub that into x = 3z -3 and x = 3

sub that into y = 5 - 4z = -3

NOW you sub that into px + 2y - z = 1

3p +2(-3) + 2 = 2
3p -6 - 2 = 1
3p = 9
p = 3


anyone get that ?

[stonecold]

yeah it made perfect sense, thanks!

but do you have to be able to find the general solution, like the one the cas calculator gives, and then you sub in integers to get different solution sets, or not?

[the.watchman]

yeah it made perfect sense, thanks!

but do you have to be able to find the general solution, like the one the cas calculator gives, and then you sub in integers to get different solution sets, or not?

I don't think you need to show different solution sets, just the general solution in terms of, say,

[stonecold]

^okay, thanks!