Simultaneous Equations with 3 Unknows

meano

New member
Joined
Oct 27, 2014
Messages
37
Right, I have hopefully mastered simultaneous equations with 2 unknowns and I am now looking at equations with 3 unknowns. I have worked through several examples and have being able to work them out with minimal referencing of coursework.

I have an equation which I am struggling with slightly.

(EQ1) 2a - 5b + c = 1

(EQ2) a + c = 2

(EQ3) b - 3c = -3

Where should I start? I tried adding EQ2 to EQ3 to make EQ4. After multiplying EQ1 x 1(a) and EQ4 x 2(a) it didn't work out. Help would be appreciated.

Thanks again.
 
Ok, I understand the substitution method but still don't know where to start.

None of the examples I have worked through (with solutions and explanations) show an equation like I have got.
 
Well

a+c=2
c=2-a

We put it in equation 3.

b-3c= -3
b-3(2-a)=-3
b-6+3a=-3
b+3a=3
b=3-3a

We now use that knowledge and put everything in equation 1

2a-5b+c=1
2a-5(3-3a)+(2-a)=1
2a-(15-15a)+2-a=1
a-15+15a+2=1
16a-13=1
16a=14
a=14/16
a=7/8

Now we know that a is 7/8 try to solve b and c like i did but remember to put in 7/8 instead of a.
 
That's all I needed thanks very much. I substituted in EQ 3 but didn't think to do it in EQ2 like you did. I'll give it a go now without reading the rest of your post and cheating with the answers!
 
Right, I have hopefully mastered simultaneous equations with 2 unknowns and I am now looking at equations with 3 unknowns. I have worked through several examples and have being able to work them out with minimal referencing of coursework.

I have an equation which I am struggling with slightly.

(EQ1) 2a - 5b + c = 1

(EQ2) a + c = 2

(EQ3) b - 3c = -3

Where should I start? I tried adding EQ2 to EQ3 to make EQ4.
What reason did you have for doing that? Adding EQ2 to EQ3 give a+ b- 3c= -1 which doesn't really help! Never do something without having a reason! And in solving systems of equations your objective should be to eliminate unknowns until you get to "one equation in one unknown". Seeing that EQ2 does not involve "b", I might decide to eliminate b from the other two equations- add 5 times EQ3 to EQ1: (2a- 5b+ c)+ 5(b- 3c)= 2a- 5b+ c+ 5b- 15c= 2a- 14c= -15+ 1 or 2a- 14c= -14 and we have eliminated "b". We can divide through by 2: a- 7c= -7.

So now we have
(EQ2) a+ c= 2

(EQ4)a- 7c= -7

Do you see what to do now?

After multiplying EQ1 x 1(a) and EQ4 x 2(a) it didn't work out. Help would be appreciated.

Thanks again.
 
What reason did you have for doing that? Adding EQ2 to EQ3 give a+ b- 3c= -1 which doesn't really help! Never do something without having a reason! And in solving systems of equations your objective should be to eliminate unknowns until you get to "one equation in one unknown". Seeing that EQ2 does not involve "b", I might decide to eliminate b from the other two equations- add 5 times EQ3 to EQ1: (2a- 5b+ c)+ 5(b- 3c)= 2a- 5b+ c+ 5b- 15c= 2a- 14c= -15+ 1 or 2a- 14c= -14 and we have eliminated "b". We can divide through by 2: a- 7c= -7.

So now we have
(EQ2) a+ c= 2

(EQ4)a- 7c= -7

Do you see what to do now?

Hi, yes that is what I was trying to do, eliminate one of the unknowns. I couldn't see it for looking!!

I could see b wasn't in EQ2 but couldn't see how to eliminate it. Now you have pointed out multiplying x 5 I can see it. b x 5 = 5b then adding to -5b in EQ1 eliminates b.

Thanks for that I appreciate it
 
Top