Solve each of the following systems, if possible. Indicate whether the system has a unique solution, infinitely many solutions or no solution.
(a) 3y = x – 10
y = x – 2
(b)x + y = 5
3x + 3y = 15
(a) 3y = x - 10
substitute in the 2nd equation
3(x-2) = x - 10
3x - 6 = x - 10
simplifing...
2x = -4 or x = -2 and y = -4
this is a unique solution
(b) this system has infinitely many solutions and cannot be solved without more information.
3x + 3y = 15
divide equation by 3
(3/3)x + (3/3)y = 15/3 or
x + y = 5
so the equations are essentially identical.
To solve the given systems of equations and determine the type of solution, let's start with system (a):
(a) 3y = x – 10
y = x – 2
To solve this system, we can use the method of substitution. Let's substitute the value of y from the second equation into the first equation:
3(x - 2) = x - 10
Simplifying the equation:
3x - 6 = x - 10
2x = -4
x = -2
Now, substitute the value of x back into one of the equations to find y:
y = -2 - 2
y = -4
Therefore, the solution to system (a) is x = -2 and y = -4.
Moving on to system (b):
(b) x + y = 5
3x + 3y = 15
We can solve this system using the method of elimination. Let's multiply the first equation by 3 to make the coefficients of y in both equations equal:
3(x + y) = 3(5)
3x + 3y = 15
Now we have two equations with identical coefficients for y:
3x + 3y = 15
3x + 3y = 15
As the equations are identical, this means there are infinitely many solutions to this system. The two equations represent the same line in the coordinate plane.
In summary,
(a) System (a) has a unique solution: x = -2 and y = -4.
(b) System (b) has infinitely many solutions.
To solve the given systems of equations, we need to find the values of x and y that satisfy both equations simultaneously. We can do this by using different methods such as substitution, elimination, or graphing. Let's solve each of the systems:
(a) 3y = x - 10 and y = x - 2
We can solve this system using substitution. First, let's solve the second equation for x:
y = x - 2
x = y + 2
Now substitute this value of x into the first equation:
3y = (y + 2) - 10
3y = y - 8
3y - y = -8
2y = -8
y = -4
Now substitute this value of y back into either of the equations to find x:
x = y + 2
x = -4 + 2
x = -2
So, the solution to the system is x = -2, y = -4.
Therefore, the system has a unique solution.
(b) x + y = 5 and 3x + 3y = 15
We can solve this system using the method of elimination. First, let's multiply the first equation by 3 to match the coefficients of y in both equations:
3(x + y) = 3(5)
3x + 3y = 15 (same as the second equation)
The two equations are the same, which means they represent the same line. In other words, they are dependent equations. This implies that there are infinitely many solutions to this system.
Therefore, the system has infinitely many solutions.
In summary:
(a) The system has a unique solution: x = -2, y = -4.
(b) The system has infinitely many solutions.