Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.
-x+5y=
−x+5y=
\,\,-7
−7
-3x+14y=
−3x+14y=
\,\,-18
−18
To determine the number of solutions for the given system of equations, we need to solve the equations and analyze the results.
Equation 1: -x + 5y = -7
Equation 2: -3x + 14y = -18
To solve this system, we can use the method of substitution or elimination. Let's use the method of elimination to find the value of x and y.
Multiply Equation 1 by 3 to make the coefficients of x equal:
Equation 1: -3x + 15y = -21 (multiplied by 3)
Equation 2: -3x + 14y = -18
Now, subtract Equation 2 from Equation 1:
(-3x + 15y) - (-3x + 14y) = -21 - (-18)
-3x + 15y + 3x - 14y = -21 + 18
y = -3
Now, substitute the value of y = -3 into one of the original equations. Let's substitute it into Equation 1:
-x + 5(-3) = -7
-x - 15 = -7
-x = -7 + 15
-x = 8
x = -8
Therefore, the solution to the system of equations is x = -8 and y = -3.
Since we have found a unique solution (one set of values for x and y), the system of equations has exactly one solution.
To determine if the system of equations has no solutions, infinitely many solutions, or exactly one solution, we need to solve the system of equations and check the consistency of the solution.
Let's write the system of equations in matrix form as:
⎡-1 5 ⎤ ⎡-7⎤
⎢ ⎥ ⎢ ⎥
⎢-1 5 ⎥ ⋅ ⎢-7⎥ = ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣-3 14⎦ ⎣-18⎦
To solve the system, we can use an augmented matrix and perform row reduction operations:
⎡-1 5 -7 ⎤
⎢ ⎥
⎢-1 5 -7 ⎥
⎢ ⎥
⎣-3 14 -18⎦
Performing row reduction operations:
R2 = R2 - R1
R3 = R3 - 3R1
⎡-1 5 -7 ⎤
⎢ ⎥
⎢0 0 0 ⎥
⎢ ⎥
⎣0 -1 3 ⎦
Now we can analyze the row echelon form:
The last row represents the equation 0y = 3, which is not possible. This means there is no solution to the system of equations. Therefore, the system has no solutions.
5 y = x - 7
if x = 0
y = -7/5
if x = 1
y = -6/5
if x = anything
y = (anything -7) /5
put in an infinite number of anythings