3y=2x-6
y=1-Cx
In the equation above , x and y are variables and c is a constant. If no ordered pair of numbers (x,y) satisfies both of the equations above , what is the value of C ?
This is definitely a classic SAT question…
You might see the solution more easily if we re-write the equations:
2x - 3y = 6
Cx + y = 1
Now, divide the top equation by -3…you will see why in moment:
(-2/3)x +y = -2
cx + y = 1
Can you see the c-value that will cause these two equations to contradict each other?
@pc keller, elaborate more please
@ethiololita6 pckeller is right in that this is a “classic” SAT question, i.e. it or many similar problems tend to appear on the current SAT.
pckeller’s algebra is correct, so you are left with (-2/3)x + y = -2, and cx + y = 1. There is exactly one real value for c which the system has no solution. Which one is it? (This is not a trick question)
@Didar177 Or graphically :
3y = 2x-6
y = 1-cx
y=(2/3)x - 2
y= -cx + 1
If the answer is still not clear, graph both equations on your TI-** for different values of c.
Better yet (thank you to @pckeller for sharing), type both expressions for y one under another in the top left window, “add slider: c”, and play with it to your heart’s content.
Thank u guys very much for help!
Geez,I feel incredibly stupid. Would some one mind telling me the answer?
@ethiololita6 You have two linear equations, and you want to pick c such that there is no solution, i.e. the lines do not intersect. So what can you say about the slopes? About c? Alternatively, reread @pckeller 's post.
We are in no way trying to make you feel dumb or anything, so keep that in mind. But I would rather have you think about it rather than simply telling you the answer.
I do agree with the spirit of what @MITer94 has said. But on the other hand, sometimes you can stare it something and just not see it – and it drives you nuts…so I will in fact give the answer away. Consider that a spoiler warning.
I had left it like this:
(-2/3)x +y = -2
cx + y = 1
Now suppose that c=(-2/3). You would have one equation that said:
-(2/3)x + y = -2 and another that said
-(2/3)x + y = 1. But the left sides of the two equations are identical. And there is no way for the same thing to be simultaneously equal to -2 and 1!
Here are some other systems that have no solution:
3x - 5y = 8
3x - 5y = 2.5
7x + 6y =1000
7x + 6y = 500
Now you try another. I’ll put a tiny twist on it to make sure you really do understand:
Consider the system:
2x - 5y =11
Px + Qy =9
If the system above has no solutions, determine the value of Q/P.
a) -2/5
b) -5/2
c) 7/2
d) -5/7
Is it B? . I didn’t know that two equations need to intersect inorder to have a solution. Are there situations where they could possibly have infinite solutions?
Yes. And yes.
Consider the following system:
3x + 2y = 5
6x + 4y = 10
Every ordered pair that satisfies the first equation also satisfies the second equation – it’s the same line both times!
@ethiololita6 yep. In general, if you have two linear equations in two variables (say x and y):
If they are parallel, then they do not intersect, and hence no solution.
If they are the same line (E.g. 3x+2y = 5, 6x+4y = 10 as in pckeller’s post), then any point on either line is a solution, so infinitely many solutions.
Otherwise, they intersect at exactly one point, so one solution.
Thanks guys! You are the best!!