Posted by Donnice on Wednesday, August 10, 2011 at 5:56pm.
Multiply the first equation by 4 and add to the second equation.
If you get 0=0, or 32=32 (i.e. consistent numbers), the system has an infinite set of solution.
Another way to answer the question is to multiply all expressions in the first equation by -32. Then, multiply all expressions in the 2nd equation by 8. You will notice there is no difference in any of the numbers after multiplication, so each equation has the same answer.
In a system of 2 linear equations in two unknowns, each equation represents a straight line.
Three situation can happen:
case 1. When the two lines intersect each other at a point with coordinates x0 and y0.
This means that the equations give a unique solution, x=x0, and y=y0.
This is probably the most common of all cases.
Example:
2x+3y=5
x+4y=5
When solved, x=1 and y=1.
case 2. When the two lines are parallel, but not coincident (i.e. there is a distance between the lines).
This means that there is no solution to the system, since the two parallel lines never meet.
Example:
3x+y=5
3x+y=7
If you subtract one equation from the other, you will get some "silly" results such as 0=2.
Conclusion: no solution.
case 3: when the two lines are parallel and pass through the same points (i.e. lines are coincident), then there is an infinite set of solutions, since any point on the line satisfies both equations.
Example:
2x+3y=6
4x+6y=12
Multiply the first equation by two and subtract the second equation from it, we get 0=0, something that makes sense, but we don't get a unique solution.
So the two lines are coincident and have an infinite set of solutions.
Let
x=t
y=(6-2t)/3
any value of t will result in (x,y) that satisfy both equations.
For the given system, 4 times equation 1 added to equation 2 gives
0=-400, which is not correct, so the answer is case 2: no solution, as Damon has given you before on Wednesday, August 10, 2011 at 5:38pm.