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...)

1 Answer
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:
#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#