Six equations are listed below.
A. 2x - 2y = 10 B. -2x - 6y = -8 C. 2x + 6y = 8
D. x + 3y = 4 E. x - y = 5 F. -x + y = 4
Using only these equations, write the pairs of equations that form systems with:
a) no solution
A. 2x - 2y = 10
E. x - y = 5
b) one solution
B. -2x - 6y = -8
F. -x + y = 4
c) infinitely many solutions
C. 2x + 6y = 8
D. x + 3y = 4
Did I do this correctly?
2007-11-21 06:44:53 · 2 answers · asked by Madera 2 in Science & Mathematics ➔ Mathematics
Nope, you have two that have infinitely many solutions... notice how if you multiply the bottom equation by 2, you get two equivalent equations, with no contradictions.
To have *no* solution, you should have two equations that are similar but come out with different answers.
NO SOLUTIONS:
E) x - y = 5
F) -x + y = 4
Why? Because you can multiply the bottom equation by -1 and get:
x - y = -4
No way to have the same thing be both 5 and -4.
INFINITELY MANY SOLUTIONS:
B) -2x - 6y = -8
C) 2x + 6y = 8
Why? Multiply either equation by -1 and you get two equations that are exactly the same. So any the variables could be most anything.
ONE SOLUTION:
A) 2x - 2y = 10
D) x + 3y = 4
Why? It's what is left. :-)
2007-11-21 06:54:40 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
No women, go to seek tutoring. You are way off.
2007-11-21 06:59:55 · answer #2 · answered by jonathin l 2 · 0⤊ 2⤋