Question

How many solutions are there to the following system of equations?

3x – 9y = 0
–x + 3y = –3

A.
0

B.
1

C.
infinitely many

D.
2

Answer 1

If you have two linear equations such as these, you have one of 3 cases:

1) lines intersect in a point, giving one solution (the coordinates of the point)
2) lines are parallel, giving no solutions
3) lines are coincident, giving infinitely many solutions

Answer 2

@KamiBug Could you help me on this one or is this one not in your place? im not sure how to do this one

Answer 3

one approach is to graph the lines represented by the equations, and see what happens.

Answer 4

Another is to substitute one of the equations into the other. Solve one equation for one variable in terms of the other, then substitute the resulting expression in place of that variable in the other equation.

Answer 5

so use the equation to graph them. and use what you gave to determine which one it is

Answer 6

would the answer be no solution?

Answer 7

If you end up with an expression giving you a value for that variable, you've got a solution.

For example:

\[x+y = 3\]
\[2x-y=0\]
Solve the first one for \(x\) in terms of \(y\) and substitute it in the other:
\[x+y - y = 3 - y\]\[x = 3-y\]\[2x-y=0\]\[2(3-y) - y = 0\]\[6-2y-y=0\]\[6-3y=0\]\[6=3y\]\[y=2\]So this system has 1 solution. We can find the value of \(x\):
\[x+y=3\]\[x+2=3\]\[x=1\]So the solution is the point \((1,2)\) where the two lines intersect.

Answer 8

If instead you end up with something like \[0=1\]you have no solutions (the lines are parallel) and if you end up with something like \[0=0\]you have infinitely many solutions.

Answer 9

I believe the answer is No solutions?

Answer 10

Another way to solve the system here is to notice that if we multiply every term in the second equation by 3, the two equations could add together like this:

\[3x-9y=0\]\[-x+3y=-3\]

\[3x-9y=0\]\[3(-x)+3(3y) = 3(-3)\]----------------\[3x-3x-9y+9y=0-9\]\[0=-9\]therefore no solutions is the correct answer

Answer 11

Thank you

Answer 12

if you put both equations in \[y=mx+b\] form you can do it by inspection. If the values of \(m\) are different, they will intersect somewhere, giving 1 solution. If the values of \(m\) are identical, then if the values of \(b\) are also identical, you have two identical lines and infinitely many solutions. If the values of \(m\) are identical but the values of \(b\) are different, you have two parallel lines and no solution