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
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.
To determine the number of solutions of the system of equations, we can solve it using a suitable method, such as substitution or elimination.
First, let's rewrite the system of equations:
Equation 1: -3x + y = 5
Equation 2: 3x - 3y = -3
To find the number of solutions, we can check the slopes of the two equations. If the slopes are equal and the y-intercepts differ, the system has no solutions. If the slopes are equal and the y-intercepts are the same, the system has infinitely many solutions. If the slopes are different, the system has exactly one solution.
Let's rearrange Equation 1 for y:
1) y = 3x + 5
Now, let's rearrange Equation 2 for y:
2) y = x + 1
Comparing the two equations, we can see that the slopes are different, which means the system has exactly one solution.
To determine the number of solutions for the given system of equations, we can use the method of elimination or substitution.
First, let's rearrange the equations in standard form:
Equation 1: -3x + y = 5
Equation 2: 3x - 3y = -3
Method 1: Elimination
To eliminate one of the variables, we need to multiply one or both of the equations by a constant so that the coefficients of the variable are the same or their negatives are the same.
In this case, we can multiply Equation 1 by 3 and Equation 2 by -1 to make the coefficients of x in both equations equal:
Multiplying Equation 1 by 3:
9x - 3y = 15
Multiplying Equation 2 by -1:
-3x + 3y = 3
Now, if we add the two resulting equations together, the variable y will cancel out:
(9x - 3y) + (-3x + 3y) = 15 + 3
6x = 18
Dividing both sides of the equation by 6, we get the solution for x:
x = 3
Next, substitute the value of x into one of the original equations (let's use Equation 1):
-3(3) + y = 5
-9 + y = 5
y = 14
So, we have found the values of x and y. Since we found unique values for both variables, the system of equations has exactly one solution.
Therefore, the given system of equations has exactly one solution.
Method 2: Substitution
Instead of using the elimination method, we can solve the system of equations by substitution.
From Equation 1: -3x + y = 5, we can isolate y:
y = 3x + 5
Now, substitute this expression for y in Equation 2:
3x - 3(3x + 5) = -3
3x - 9x - 15 = -3
-6x - 15 = -3
-6x = 12
x = -2
Substitute the value of x into Equation 1:
-3(-2) + y = 5
6 + y = 5
y = -1
Again, we found the values of x and y. Since we found unique values for both variables, the system of equations has exactly one solution.
Therefore, the given system of equations has exactly one solution.