Question

How many solutions does the following system have?

x+y = 3

2x+ 2y = 5

a.one solution
b.infinitely many solutions
c.two solutions
d.no solutions

Answer 1

it would be b

Answer 2

Can you explain how you got B please

Answer 3

b) is incorrect.

These two equations represent parallel lines. Parallel lines do not intersect, and thus there are no points in common between the two lines. Only a point on both lines is a solution.

Answer 4

If you solve this system of equations by substitution:

\[x +y = 3\]\[x = 3-y\]\[2x+2y=5\]\[2(3-y) + 2y = 5\]\[6-2y + 2y = 5\]\[6 = 5\]A nonsense statement like that means there are no solutions.

If we had instead \[x+y=3\]\[2x+2y=6\]Solving in the same fashion\[2(3-y)+2y=6\]6-2y+2y=6\]\[6=6\]A true statement like that which contains none of the variables in the system of equations means we have infinitely many solutions. The two lines are identical, as you can see if you divide the second equation by 2.

Finally, if we had \[x+y = 3\]\[2x+y = 5\]Solving once again\[2(3-y) + y = 5\]\[6-2y+y = 5\]\[-y = -1\]\[y=1\]\[x = 3-y = 3-1 = 2\]This system has 1 solution, and it is the point where the two lines intersect \((2,1)\).

A guided tour of all of the possible cases at no extra charge :-)

Answer 5

Sorry about the formatting slip: the line just above \(6 = 6\) should be \[6 - 2y + 2y = 6\]

Answer 6

thank you very much

Answer 7

so it has one solution?

Answer 8

@whpalmer4

Answer 9

I explained it all. please read the post again if there's a question. The first set I solved is the one for this problem.