Um there are an infinite number of equations but some simple ones are: X + 1 = 6 X + 2 = 7 123553X = 617765
It's one of the simplest linear equations in 'x' that can be written. Its solution is [ x = -5 ].
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.
You have two unknown variables, x and y. You therefore need at least two independent equations to find a solution.
-10
Um there are an infinite number of equations but some simple ones are: X + 1 = 6 X + 2 = 7 123553X = 617765
7
Tell whether the ordered pair (5, -5) is a solution of the system
Without any equality signs the given expressions can't be considered as equations.
2x + 3 = 7 and x = 2 3y - 5 = 7 and y = 4
That of course will depend on what system of equations are they which have not been given
It's one of the simplest linear equations in 'x' that can be written. Its solution is [ x = -5 ].
Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.
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.
"The three" is wrong. There is not a single solution to this problem, but infinitely many. Take any three multiples of 5 - multiplying 5 by numbers that don't share common factors. For example, three different prime numbers.
The answer will depend on statement 3 5 - whatever that may be!
(-4,-5)