Solve the System of Equations...?

Question

Can anyone help me solve this system of equations and teach me how they solved it.

3(r-3q)= -4(1+p)
2(p-3q+4r)= -9
3(2p+q)+5r= -1

Answer 1

O.K. Here is how you solve it. Please have patience, it will take some time. (I had to deal with serious family news.)

First expand parentheses, re-group and write the eqns with all variables in a chosen standard order, line them up, and number each eqn:

4p - 9q + 3r = - 4 ............(1)
2p - 6q + 8r = - 9 ............(2)
6p +3q + 5r = - 1 ............(3)

Looking at the q and r columns, notice that eqns (1) + (3) - (2) make all terms in q and r vanish. That is great! So now do that same process for the remaining p's and the right hand side (RHS), also. Then:

8p = 4, so p = 1/2.

Now we're getting somewhere! (These equations were unusually tailored to get us quickly to this stage. Notice that the value of p just dropped right into our lap!)

Rewrite eqns (1) - (3), substituting the newly found value of p, and re-naming the equations so produced with a prime ('). (I find that that easily reminds me where the amended eqn came from, should I need to re-check this.) Then:

- 9q +3r = - 6, or (dividing by 3):

- 3q + r = - 2 ........(1') But (3) now ---> (3'), where
+3q +5r = - 4 ........(3')

Notice that one can now add (1') and (3') to get rid of q; then:

6r = - 6, so that r = - 1.

Now that we've got both p and r, substitution back into any equation containing q will determine that. (Your choice.) The result is q = 1/3.

It's always a good idea to check back that each of the original equations is satisfied. I did that; so should you; they are!

So: what worked well with these equations was that it was relatively easy to spot that the coefficients of both q and r would vanish by the same manipulation of the original 3 equations. One is rarely that lucky! (But notice how quick the whole process was, and how few lines of actual math it took if you ignore my detailed verbal descriptions of what I was doing and why.)

If something like that can't be spotted, you just have to grit your teeth and doggedly subtract some multiple of one of the equations from another; then do it again with a different pair, to successively eliminate one of the variables TWICE. At that stage you'll then have two equations in two variables; a similar process eliminates one of those variables; solve for the remaining variable, and then back-substitute all the way back. There's no getting past the fact that it's a very tedious business, in general!

The method used below is an example of the dogged determination, gritted teeth approach. That's because that respondent didn't spot the possibility of using (1) + (3) - (2) to remove both q and r together, in one fell swoop. That can be the secret to success in problems like this. Once you've got the value of one of the variables, you're in the home stretch; you can smell success.

Good luck. Live long and prosper.

Answer 2

3(r - 3q) = - 4(1+p)
2(p - 3q + 4r) = - 9
3(2p + q) + 5r = - 1
First let's get the variables in some kind of order:
3r - 9q = - 4 - 4p

2p - 6q + 8r = - 9
6p + 3q + 5r = -1
4p - 9q + 3r = - 4
there don't seem to be any easy integer substitutions, so
6p - 18q + 24r = - 27
-6p - 3q - 5r = +1
- 21q + 19r = - 26

4p - 12q + 16r = - 18
-4p + 9q - 3r = + 4
- 3q + 13r = - 14

- 21q + 91r = - 98
+ 21q - 19r = + 26
72r = - 72
r = - 1

- 3q - 13 = - 14
3q = 14 - 13
3q = 1
q = 1/3

2p - 6*1/3 - 8 = - 9
2p = 2 - 9 + 8
2p = 1
p = 1/2

6*1/2 + 3*1/3 - 5 =? -1
3 + 1 - 5 = -1

p = 1/2
q = 1/3
r = - 1

Answer 3

The answers are r = -1, q = 1/3, and p = 1/2. Sorry, I tried to explain it but it's really hard to do [the explaining, not the problem] and I think it would just confuse you further. Sorry I couldn't be of more assistance.

Answer 4

5x + 2y = 15 -5x - 2y = -5 _____________ 0 = 10 you remedy utilizing combining words, you upload both equations. 0=10 potential this equipment of equation has no answer in any respect, because the slopes of both lines are a similar, in different words are parallel lines, so there is not any longer element of interception

Answer 5

The first step is to use distributive property:
3r-9q=-4-4p
2p-6q+8r=-9
6p+3q+5r=-1

Add 4p to both sides:
4p-9q+3r=-4
2p-6q+8r=-9
6p+3q+5r=-1
==========

Here's the next step:
4p-9q+3r=-4
-2(2p-6q+8r=-9)
============
4p-9q+3r=-4
-4p+12q-16r=18
============
3q-13r=14

-3(2p-6q+8r=-9)
6p+3q+5r=-1
============
-6p+18q-24r=27
6p+3q+5r=-1
============
21q-19r=26

-7(3q-13r=14)
21q-19r=26
==========
-21q+91r=-98
21q-19r=26
==========
72r=-72
r=-1

21q-19(-1)=26
21q+19=26
21q=7
q=7/21=1/3

4p-9(1/3)+3(-1)=-4
4p-3-3=-4
4p-6=-4
4p=2
p=2/4=1/2

(1/2,1/3,-1)

Check:
3(-1-3(1/3))=-4(1+1/2)
3(-1-1)=-4(3/2)
3(-2)=-4(3/2)
-6=-6

2((1/2-3(1/3)+4(-1))=-9
2(1/2-2/2-8/2)=-9
2(-9/2)=-9
-9=-9

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

Answer 6

re-arrange each equation into standard form:
ap+bq+cr=d
(there will be a different a, b, c, d for each equation)

See if you can add/subtract some pair of equations to eliminate 2 variables.

If not, express p from one equation: p=(d-bq-cr)/a, subtitute into two other equations. you'll get 2 eqautions with 2 varibales (q and r), express one of them from some equation, substitue into remain equation.

You have one equation with 1 variable, solve it. Use your expressions to back out other variables.

Answer 7

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