When we are given a system of 3 linear equations, with 3 variables, we proceed as follows:
We consider 2 pairs or equations, for example (1, 2) and (2, 3), and eliminate one of the variables in each pair, creating a system of 2 linear equations with 2 unknowns.
Note that the third equation contains -2y which can be used to eliminate easily -6y in the second equation, and -4y in the fourth.
i) consider equations 1 and 3:
-3x-4y-3z=-7
5x-2y+5z=9
multiply the second equation by -2:
-3x-4y-3z=-7
-10x+4y-10z=-18
adding the 2 equations we have -13x-13z=-25
ii) consider equations 2 and 3. Multiply the third equation by -3:
2x-6y+2z=3
-15x+6y-15z=-27
adding the 2 equations we have -13x-13z=-24
So we got -13x-13z is -25, but also -24. this means the system is inconsistent, so it has no solution.
Answer: the system has no solutions
Answer:
x= -1/4
Step-by-step explanation:
suppose ,
The number is x
now,
twice a number = 2 . x = 2x
The product of twice a number and six = 2x . 6 = 12x
[ "product" means the result of multiplication]
Then,
Eleven times the number = 11 . x = 11x
The difference of eleven times the number and 1/4 = 11x - 1/4
= (44x-1)/4
Hence,
12x = (44x-1)/4
=>48x = 44x-1 [multiplied by 4 on both sides]
=>48x-44x=-1 [subtract 44x on both sides]
=>4x=-1 [divide both sides by 4]
=>x= -1/4
Answer:
∀s ∈ D, C(s) - - - > E(s)
∀s ∈ D, C(s) - - - > ~ E(s)
∃s ∈ D such that M(s) ∧ C(s)
Step-by-step explanation:
D = set of all students
M(s) = s math major
C(s) = s Computer science major
E(s) = s Engineering major
Expressing the following using quantifies variables and predicates :
A.) Every computer science student is an engineering student
∀s ∈ D, C(s) - - - > E(s)
b. No computer science students are engineering students
∀s ∈ D, C(s) - - - > ~ E(s)
c. Some computer science students are also math majors
∃s ∈ D such that M(s) ∧ C(s)
∃s = Existential Domain
∀s = universal
∧ = connective and
~ = not
∈ = belongs to
