how many solutions does the following systems of equations have: x + y = -9 and -x - y = 9
5 answers
There is only one solution for this system of equations. The solution is x = -9 and y = 0.
it is infinite
Apologies for the confusion, but the system of equations you provided cannot have an infinite number of solutions.
Given the equations:
1. x + y = -9
2. -x - y = 9
By adding the two equations together, we get 0 = 0, which means that the two equations are actually dependent (essentially the same equation), and the system is inconsistent. This implies that there are no solutions for this system of equations.
Given the equations:
1. x + y = -9
2. -x - y = 9
By adding the two equations together, we get 0 = 0, which means that the two equations are actually dependent (essentially the same equation), and the system is inconsistent. This implies that there are no solutions for this system of equations.
i know it if infinite
I apologize for the confusion earlier. You are correct. The system of equations x + y = -9 and -x - y = 9 does indeed have an infinite number of solutions.
By solving one equation for x, we get x = -9 - y. Substituting this expression for x into the second equation -x - y = 9, we get -(-9 - y) - y = 9, which simplifies to y = -9.
Therefore, x = -9 - y = -9 - (-9) = 0.
This means that the system has infinitely many solutions in the form of (0, -9 + n), where n can be any real number. Thank you for pointing out the error, and I appreciate your understanding.
By solving one equation for x, we get x = -9 - y. Substituting this expression for x into the second equation -x - y = 9, we get -(-9 - y) - y = 9, which simplifies to y = -9.
Therefore, x = -9 - y = -9 - (-9) = 0.
This means that the system has infinitely many solutions in the form of (0, -9 + n), where n can be any real number. Thank you for pointing out the error, and I appreciate your understanding.
1 answer