Algebra. Systems of linear equations trouble.?

Question

A system such as -

3/x + 4/y = 5/2
5/x - 3/y = 7/4 can be solved by eliminations. One way to do this is to let p=1/x and q=1/y. Substitute, solve for p and q, and then find x and y. (hint: 3/x = 3 * 1/x = 3p) Use this method to solve the each system.

What the Hell are they talking about? Q and P? Can someone work out this problem the way these directions are explaining it so I have an idea of what the book is asking me to do and so can go on and do the rest of the problems. Thanks.

Answer 1

The idea here is that working with equations like this, where the denominator in a fraction is a variable, can easily get confusing. So to help you keep track of what's going on, your text is proposing a "change of variable", a very common mathematical tool, to help turn this into a type of problem you should have already done before. Let's take the system you gave as an example:

If we let p = 1/x and q = 1/y as the text suggest, we can rewrite your system of equations as
3p + 4q = 5/2
5p - 3q = 7/4
Now, even though our variables are p and q, instead of x and y, like we are used to, it is essentailly the same as a "normal" system of equations problem, since variables are only symbols that represent unkowns, and it doesn't really matter what we call them.

In case you're have trouble with systems of equations in general, it works out as follows.

Let's multiply the top equation by 3 and the bottom equation by 4. This will make the coefficients on q be 12 and -12 repectively, so that we can eliminate the varialbe q by adding the equations together like so:

9p + 12q = 15/2
20p - 12q = 7

Adding these equations yields

29p + 0 = 29/2

or

29p = 29/2

and dividing both sides by 29 yields

p = 1/2

now we have solved for p, which we can use to solve for q, as such

3( 1/2 ) + 4q = 5/2

3/2 + 4q = 5/2

4q = 1

q = 1/4

And since we know that p = 1/x and q = 1/y, it follows that x = 2 and y = 4.

Any easy way to check your answers, is to use them in the origianl equations, and see if they work.

3/2 + 4/4 = 3/2 + 1 = 3/2 + 2/2 = 5/2
and
5/2 - 3/4 = 10/4 - 3/4 = 7/4

so you're good to go.

Answer 2

What they are doing is telling you to let p = 1/x and q = 1/y to make the system easier to solve.

So if p = 1/x then 3/x + 4/y = 5/2 becomes 3p +4/y = 5/2, then if q = 1/y then it becomes 3p + 4q = 5/2. Do that with the other equation, solve for p and q, then you can figure out what x and y are.

Answer 3

"they" introduce a new variable p that is defined as p = 1/x
and q = 1/y

next "they" replace the 3/x with 3p
that is allowed because 1/x = p thus 3/x = 3p

so what you get then is

3p + 4q = 5/2
5p - 3q = 7/4

then it is the task of the reader ( that is you ) top solve p and q.

once you have p = some number
then you also know the x ,
since 1/x = p = some number
thus x = 1/p = 1/ (some number)

likewise the y

that is what "they" tried to explain.

Answer 4

Whenever math talks about 1/x or 1/ any number, they are talking about the inverse. The inverse is not exactly the opposite it is kinda the mirror image though.
by letting p = 1/x and letting q=1/y you are able to get the variables out of the bottom of the eqation, you are taking the inverse. You know that the inverse of 1/x is x/1, right?
so by doing this step (which is called U substitution) You end up with p/3 + q/4=2/5 and p/5 - q/3 = 4/7 and the variables get on the top which is easier to simplify.

Answer 5

substitute p=1/x and q = 1/y into each equation:
3p +4q =5/2
5p -3q =7/4
solve first equation for p
p = (5/2 -4q)/3 substitute this value of p into second eq.
5(5/2 -4q)/3 -3q =7/4
5(5/2 -4q) -9q =21/4
25/2 -20q -9q =21/4
-29q = -50/4 +21/4 =-29/4
therefore, q =1
substitute q = 1 into second equation:
5p -3(1) =7/4
5p =19/4
p = 19/20
since we used p = 1/x then 19/20 = 1/x and x = 20/19
q= 1/y then 1 = 1/y and y =1

Answer 6

they want you to use simpler variables, then go back and plug in your answer at the end
so you could rewrite the problem as
3p+4q=5/2
5p-3q=7/4

then solve using elimination
p= 1/2 and q=0.25
since p=1/x, then x=2
since q=1/y, then y=4

Answer 7

once you row shrink your augmented matrix you'd manage to jot down a corresponding set of equations. note the most suitable variables (pivots) and the others that are the loose variables. Now clean up for the most suitable variables in words of the loose variables.. The loose variables will manage to be assigned an arbitrary cost (t) which determines the cost of the most suitable variables. there'll be infinitely many ideas yet you are able to write a familiar answer in words of t.