Question

Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.
minus, 3, x, plus, y, equals, 5
−3x+y=
5
3, x, minus, 3, y, equals, minus, 3
3x−3y=
−3

Answer 1

To determine the number of solutions in the given system of equations, we can use the method of substitution or elimination.

Let's solve the system using the method of elimination:

First, let's rewrite the given system of equations in a standard form:

1) x + y = 5
2) 3x - 3y = -3

To eliminate the y term, we can multiply equation 1 by 3:

3(x + y) = 3(5)
3x + 3y = 15

Now we have the following two equations:

3x + 3y = 15
3x - 3y = -3

Adding these two equations eliminates the y term:

(3x + 3y) + (3x - 3y) = 15 + (-3)
6x = 12
x = 2

Substituting the value of x back into equation 1 gives:

2 + y = 5
y = 3

So, we have found that x = 2 and y = 3, which satisfies both equations in the given system.

Therefore, the given system of equations has exactly one solution: x = 2 and y = 3.