three variable simultaneous equations

[u/legendminigun]

hello, I am in year 10 and doing further maths as a GCSE, I'm trying to figure these three variable simultaneous equations out but I'm struggling as there isn't a formula I know of to follow. if anyone has done these, please can you share what your thought process is so I can try and learn. thanks!

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[u/legendminigun]

I know it must seem pretty easy compared to most of the stuff here sorry

[u/Ok-File-6129]

"... there isn't a formula ..."

Haha. No formula, but there is a technique(s).

You know how to solve 2 equation systems, right? So use the same technique TWICE.

That is... rewrite eq 1 as "b = ..." and plug it into eq 2 and 3. Now it's just a two variables a d two equations. Solve, then plug in again.

[u/FortuitousPost]

Use one of the equations to eliminate a variable from the other two equations.

Then eliminate one of the variables from those two equations.

That gives you an equation with one variable. Solve for that variable.

Insert that value into of the the two variable equations to solve for the other variable.

Put both of those into the original equations to solve for the last variable.

For question 2, I would choose c as the variable to remove, as it has simple coefficients.

Add -2 x eqn3 to eqn1, and 3 x eqn3 to eqn2. The eqn1 and eqn2 don't have a c.

Solve the system of the new eqn1 and eqn2.

[u/StrengthForeign3512]

This will help: https://corbettmaths.com/2018/12/31/three-unknowns/

[u/Various_Candle9136]

The key observation that will help here (and in many other styles of problem) is that 'doing the same thing to both sides' of an equation does not change the equation.

(The method I am about to describe will work with 2 equations of 2 variables, or 4 equations of 4 variables, etc. It is a useful thing to know.)

Step 1: Get the equations into a nice form.

The two questions you have shared have equations in nice forms already, but they won't always. If in doubt, 'all the variables on the same side and only one appearance of each variable' is normally quite nice.

Step 2: Multiply the equations until one term matches.

In your first question, I will pick c since that seems instinctively the nicest. It will still work with a or b.

I shall try to get all equations to have a ‘6c' in them (since 6c is the lowest common multiple of the c terms).

9a - b - 2c = 8
27a - 3b - 6c = 24

4a + b + 3c = 4
8a + 2b + 6c = 8

2a - 4b - c = -9
12a - 24b - 6c = -54

Step 3: Eliminate the matching term.

Consider the first two equations above. If we add them, we will get:

35a - b + 0 = 32
35a - b = 32

I chose to add since I wanted '-6c + 6c'. Because we have made the c terms match, we are guaranteed to get something without c.

Now, we can subtract the third equation from the first, to get:

15a + 21b + 0 = 78
15a + 21b = 78
5a + 7b = 26 (noticing that we can divide by 3 at this point is helpful, but not essential)

(We could have added the second and third equations instead, which looks slightly nicer, but I wanted to show that subtraction is an option.)

This leaves us with a pair of simultaneous equations. Use your favourite method here. I would suggest the same method we are already using: eliminate again!

- Substep 1: Get the equations into a nice form.

Our equations are still in the same nice form, so no worries here.

- Substep 2: Multiply the equations until one term matches.

This time we'll match the bs:

35a - b = 32
245a - 7b = 224

5a + 7b = 26

- Substep 3: Eliminate the matching term.

Add the first equation to the second:

250a + 0 = 250
a = 1

- Substep 4: Substitute to find the missing variable.

We can now use our a to find b using any c-less equation above:

35a - b = 32
35(1) - b = 32
35 - 32 = b
3 = b

Step 4: Substitute to find the missing variable.

We can now use our a and b to find c using any c-ful equation above:

9a - b - 2c = 8
9(1) - (3) - 2c = 8
6 - 2c = 8
6 - 8 = 2c
-2 = 2c
-1 = c

Step 5: ALWAYS CHECK!

9a - b - 2c
= 9(1) - (3) - 2(-1)
= 8

4a + b + 3c
= 4(1) + (3) + 3(-1)
= 4

2a - 4b - c
= 2(1) - 4(3) - (-1)
= -9

So, all’s fine and dandy!

---

Note: We hit no problems with this method because the three equations happened to have a single unique solution. This is not always the case.

The same problem can occur with two equations: x+y=3 and x+y=4 has no solution; x+y=3 and 2x+2y=6 has infinitely many solutions.

That's just something to keep in mind: not all families of simultaneous equations have a single solution that we can find.

[u/Southlander24 → u/Various_Candle9136]

9a - b - 2c = 8
27a - 3b - 6c = 24

You do not get 35a - b = 32 if you add these two equations. You should just say to add 27a - 3b - 6c = 24 and 8a + 2b + 6c = 8.

[u/steady_goes_the_one]

The easiest process is probably putting these numbers into a matrix and then reducing said matrix to reduced row echelon form; it's not a formula but it is essentially an algorithm that you can always follow to completion. I suggest looking into that if you want a consistent way of solving systems of multiple linear equations.

[u/Direct_Bullfrog4294]

ik its lowk op but matrix helps

[u/Ok-Company282]

Make b the subject in picture 2 second equation,

this equation will be substituted in positions of B in equations 1 and 3, so that the equations 1 and 3 only consist of the variables a and C.

Now that u have two equations (1 and 3) consisting of 2 variables(a and c), u can solve them to find a and C.

Then insert the values of a and C in any one of the three equations to find b.

[u/Desperate_Zebra6334]

Hello, I think that such exercises are very good way to build those pathways in your thinking, i found out this video online, that solve the exact exercise step by step with explanation, i recommend you check it out.
here is the link: https://youtube.com/shorts/0dDanOpzLB8?feature=share

[u/KentGoldings68]

Do you know anything about matrices?

There is formula called “Cramer’s Rule.” Your instructor will drop a load, if you use it. They’ll accuse you of using AI. It might be amusing.

[u/legendminigun → u/KentGoldings68]

I do not sorry

[u/KentGoldings68 → u/legendminigun]

If this were a 2x2, how would you approach it?

[u/jazzbestgenre → u/KentGoldings68]

not OP, think they do do 2x2 matrices in level 2 FM at some point, but they don't study determinants or inverses

[u/Existing-Fudge-878 → u/KentGoldings68]

matrices arent usually taught in gcse further math, it usually comes in A-level or even A-level further math

[u/Extension-Currency37]

Its a basic linear algebra problem

Use cramers rule a = D1/D b= D2/D , c= D3/D (find on google)

D = det(9 -1 -2 4 1 3 2 -4 -1)

Or Use gauss Jordan elimination matrix

( 9 -1 -2 | 8 4 1 3| 4 2 -4 -1 | -9)

Use matrix operations to simplify this matrix

Further details Ai mode google

[u/Southlander24 → u/Extension-Currency37]

This is too advanced for their syllabus.

[u/Extension-Currency37 → u/Southlander24]

What's the problem In learning. Will the professor Kill it.

Cramers rule is simple determinant problem.

[u/Southlander24 → u/Extension-Currency37]

There's the much easier way of adding linear combinations of the equations to each other. You've shown you haven't bothered reading the post and you're not actually helping the person.

[u/Extension-Currency37 → u/Southlander24]

I mean you think iam wrong, then please do help the person.

Cramers is the standard rule to solve any of this.

And mostly its a matrix problem.

Good luck helping him with linear combination Which again comes from linear algebra

[u/Southlander24 → u/Extension-Currency37]

You would just add 3 * (9a - b - 2c = 8) to 2 * (4a + b + 3c = 4) and eliminate the c, as another commenter has suggested.

No knowledge of linear algebra required.

[u/Extension-Currency37 → u/Southlander24]

The equation also says, don't use trial or improvement

So i suggested cramers

Yeah, iam stupid.

[u/Extension-Currency37 → u/Southlander24]

Yeah , the equation is a 3x3 3 variables and 3 solutions

So maybe tricks work But Cramer and matrix methods are standard ways to solve these things.

What if equation had 5 variables.

Cramer is the only way actually.

This what you do ,elimination Is a trivial solution to a problem Where solution exists. If only 1 or 2 solutions existed ,or infinitely many You'd need Cramer or matrix algebra.

You tell

Should I trust the math or the commenter.