y=2x-1
2x-y-1=0
a)this sytem has excatly onw solution
b)this system has indinetly many solutions
c)this system has no solution
d)(1,1) and (0,0)
mmmmhhh?
Isn't the second equation the same as the first?
And how many points would satisfy the first?
So what do you think?
ohhhh so just one solution right?
If you mean infinite not indinetly than the answer is C.
2x-y-1=0 subtract 2x from each side
-y-1 = -2x + 0 add 1 to each side
-y = -2x + 1 multiply everything by -1
y = 2x-1
The equations are equal, so the system has infinite solutions.
oh ok thanks alot
To determine the number of solutions for the given system of equations, we can solve the system and analyze the result.
First, let's solve the system of equations:
Equation 1: y = 2x - 1
Equation 2: 2x - y - 1 = 0
To find the solutions, we can solve one equation for one variable and substitute it into the other equation. Let's use Equation 1 to solve for y:
Substituting y = 2x - 1 into Equation 2:
2x - (2x - 1) - 1 = 0
2x - 2x + 1 - 1 = 0
0 = 0
Notice that we end up with 0 = 0, which means the equation is always true regardless of the values of x and y. This indicates that the two equations represent the same line and are coincident.
Now, let's analyze the number of solutions based on the coincident lines:
- If the two lines representing the equations overlap completely (i.e., they coincide), then the system has infinitely many solutions. Every point on one line will also satisfy the other line.
- If the two lines representing the equations are distinct and parallel, then the system has no solution. The lines will never intersect.
Since the given system represents coincident lines, it means that option b) "this system has infinitely many solutions" is the correct answer. Any point on the line y = 2x - 1 is a solution to the system.
The point (1,1) and (0,0) satisfy the given system of equations y = 2x - 1 and 2x - y - 1 = 0. So, option d) "(1,1) and (0,0)" is also correct.
Therefore, the correct answer is option b) "this system has infinitely many solutions" and option d) "(1,1) and (0,0)".