Question

x+6y=2 (1)
x=4-6y (2)
someone help please

Answer 1

no solutions

Answer 2

how do you determine that

Answer 3

solve as a system, right?

Answer 4

sub for x by putting (2) into (1)
(4-6y)+6y=2
4=2
which, because that is impossible, means the system has no solution at all (the lines of their graphs don't intersect). So the answer is "NO SOLUTION"

Answer 5

each equation represents a line
the lines are parallel, dont intersect
--> no solution (i.e. no pair of x,y values ) that satisfies both simultanously

http://www.wolframalpha.com/input/?i=solve%28x%2B6y%3D2+%3B+x%3D4-6y%29

Answer 6

(4-6y)+6y=2....
4=2... so no solution..

Answer 7

rewrite the equations in the form of ax+by+c=0

x+6y-2=0 (1)
x+6y-4=0 (2)

if
\[\frac{a1}{a2}=\frac{b1}{b2}=\frac{c1}{c2}(infinte solutions)\]
if
\[\frac{a1}{a2}\neq \frac{b1}{b2}(unique solutions)\]
if
\[\frac{a1}{a2}=\frac{b1}{b2}\neq\frac{c1}{c2}(no solutions)\]
in this case
\[\frac{a1}{a2}=\frac{1}{1},\frac{b1}{b2}=\frac{\cancel6}{\cancel6}=\frac{1}{1},\frac{c1}{c2}=\frac{-2}{-4}=\frac{1}{2}\]
\[\frac{1}{1}=\frac{1}{1}\neq \frac{1}{2}\]
\[\therefore,no solutions\]

Answer 8

Now that I didn't know :)