x+1=x+2 x=1+2+x 0x=1+2 0x=3 0x=/=3 (=/= means does not equal) so there is no solution
There is no solution since the equation is inconsistent.
Not sure what you mean; that equation certainly has a solution.
The value of c is 6
Yes, but so is everything else.
It is also x = 5 8zx + 5 = 2j + 3 → 8zx + 2 = 2j → 4zx + 1 = j Which is the first equation, which has a solution x = 5.
The solution is (x, y, z) = (2.5, 1, 2.5).
Both of them are equations!The solution is (x, y) = (2, 1).
Solution: x = 1/4 and x = -3.
The solution: x = 6/7 and x = -1
There is no such pair. The solution to equation 1 and equation 2 is x = 1, y = 1. The solution to equation 2 and equation 3 is x = 1, y = 1. And the solution to equation 1 and equation 3 is any point on the line 3x + 2y = 5 - an infinite number of solutions. The fact that the determinant for equations 1 and 3 is zero (or that they are not independent) does not mean that there is no solution. It means that there is no UNIQUE solution. In this particular case, the two equations are equivalent and so have an infinite number of solutions.
In the equation 43X +2=5X, X=-1/19.
It is a linear equation in one unknown variable, n. Its solution is n = 1