Question

Given a linear system of the form \[\left[\begin{matrix}Ax & + By =C \\ Dx&+Ey=F\end{matrix}\right]\]

Where A,B,C AND E are nonzero real numbers, which of the following is NOT true?

A.)Every sistem of linear equations of this form can be solved using the method of elimination.
B.)The method of elimination can only be used when one of the variables in both of the original equations have coefficients with opposite signs.
C.) It is possible to multiply one equation of a linear system in two variables by a nonzero constant without changing the solution to the system.

Answer 1

B or C ?

Answer 2

@phi @Elsa213 @sweetburger @tkhunny

Answer 3

anyone could help me out i think is C...

Answer 4

or maybe is B..

Answer 5

C is true

if you multiplied every term of the equation of a line by a nonzero number, you would still get the same line when the number is factored out and removed

Answer 6

ok therefore there is only on option B!

Answer 7

before jumping to conclusions, why not a?
i would like to hear your reasoning

Answer 8

I CANT fail so Im going to take the risk

Answer 9

it is not necessary if you think about it, I could multiply a system that has all positive variable by a negative constant in order to eliminate a variable...therefore is not always true

Answer 10

am I right ?

Answer 11

multiply wouldnt make much sense when using the elimination method, if im thinking of the correct method

unless you meant, you can multiply by a negative number to change signs in order to apply the elimination method

but i would agree with B being not true since it states "can only" and using the elimination method is possible with values with the same sign

Answer 12

Is "can be solved" synonymous with "having a solution"?

Answer 13

ok I'm going to answer B

Answer 14

That was what i was thinking for A,

you could solve the system in order to determine it has no solutions, 1 solution, or infinitely many solutions

but if you solve and determine it has no solutions, does that count as solving?

Answer 15

all of you, thanks !

Answer 16

yeah, I did think of that at first but sometimes that is why it is called a tricky question

Answer 17

anyway thanks!