# 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!ðŸ‘‡ðŸ‘‡ðŸ‘‡ $7 x + 2 y = 12$ { $5 x + 2 y = 8$ That's the question and the answer to that is $2 , - 1$. So palease help! I have one more question like that, too.: $x + y + z = 6$ $x + 2 y + 3 z = 14$ $x + 3 y + 7 z = 28$ 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...)

Sep 6, 2017

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:
$A \overline{x} = \lambda$ then ${A}^{-} 1 \lambda = \overline{x}$ where ${A}^{-} 1$ is the inverse of matrix $A$ and $\overline{x}$ 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:

$7 x + 2 y = 12$ [A]
$5 x + 2 y = 8$ [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}: $\to 2 x + 0 y = 4$

$x = 2$

From [A} with $x = 2$: $y = \frac{1}{2} \left(12 - 7 \cdot 2\right) = - \frac{2}{2} = - 1$

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

Q2:

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

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

[C] - [A]: $\to 2 y + 6 z = 22$ [D]

[C - [B]: $\to y + 4 z = 14$ [E]

$2 \times$[E] - [D]: $\to 2 z = 6 \to z = 3$

From [E] with $z = 3 \to y + 12 = 14 \to y = 2$

From [A] with $y = 2 , z = 3 \to x + 2 + 3 = 6 \to x = 1$

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