Question

How do you solve simultaneous linear equations inequalities? (examples are down in the description). Please help!

Here, this is a problem that I know the answer to, but I seriously have NO idea on how to solve it! Seriously!👇👇👇

`7x+2y=12`
{
`5x+2y=8`

That's the question and the answer to that is `2,-1`. So palease help! I have one more question like that, too.:

`x+y+z=6`

`x+2y+3z=14`

`x+3y+7z=28`

Whats the answer, procedure, steps. I don't get it! Please help (I know that I probably said that a hundred times by now, but back to the point...)

Answer 1

Q1: `x=2, y=-1`
Q2: `x=1,y=2,z=3`

Explanation:

Although you ask for methods of solving simultaneous linear inequalities , both you example are linear equalities. So I assume you meant equalities.

There are 3 basic methods of solving systems of simultaneous linear equations.

(i) Substitution : Isolate one variable in terms of the others and replace in another equation. Continue until all are resolved.

(ii) Linear manipulation : Multiply one equation by a constant (could be 1) and add/subtract another equation so as to remove a variable. Continue until all are resolved.

(iii) Matrix inversion : Express the coefficients of the variables as matrix `A` and the constants as vector `lambda` such that:
`Abarx = lambda` then `A^-1 lambda = barx` where `A^-1` is the inverse of matrix `A` and `barx` is the vector variables.

In broad terms, these are increasing in complexity and their use will usually depend on the number of variables in the system.

Q1:

`7x+2y =12` [A]
`5x+2y =8` [B]

Here we have a simple case with only 2 variables. Both methods (i) and (ii) would be appropriate. However, notice that the coefficient of `y` is 2 in both equations. Hence, method (ii) is perfect.

[A] - [B}: `-> 2x+0y=4`

`x=2`

From [A} with `x=2`: `y=1/2(12-7*2) = -2/2 =-1`

Hence, our result: `x=2, y=-1`

Q2:

`x+y+z=6` [A]
`x+2y+3z=14` [B]
`x+3y+7z=28` [C]

Here we have a 3 variable system. In my opinion, method (ii) is most appropriate here too.

[C] - [A]: `-> 2y+6z =22` [D]

[C - [B]: `-> y +4z =14` [E]

`2xx`[E] - [D]: `-> 2z=6 -> z=3`

From [E] with `z=3 -> y +12 = 14 -> y=2`

From [A] with `y=2, z=3 -> x+2+3 =6 -> x=1`

Hence, our result: `x=1,y=2,z=3`