Add the two equations together to get
4x=0, therefore x=0.
Substitute x=0 into the first equation to get y=0 also.
Thus x=0, y=0.
x-2y = 0
If the system has infinitely many solutions, express your answer in the form x=x and y as a function of x
Your answer is
x=? y=?
4x=0, therefore x=0.
Substitute x=0 into the first equation to get y=0 also.
Thus x=0, y=0.
Equation 1: 3x + 2y = 0
Equation 2: x - 2y = 0
We can solve this system of equations by using the method of elimination:
Step 1: Multiply equation 2 by 3.
3(x - 2y) = 3(0)
This simplifies to:
3x - 6y = 0
Step 2: Add equation 1 and the modified equation 2 (from step 1).
3x + 2y + 3x - 6y = 0 + 0
Combining like terms:
6x - 4y = 0
Step 3: Simplify the equation obtained in step 2.
Divide the entire equation by 2:
3x - 2y = 0
Step 4: Now we have two equations: 3x + 2y = 0 and 3x - 2y = 0.
We can see that both equations are exactly the same, just with a different sign for the y-term.
Step 5: Therefore, this system has infinitely many solutions.
We express the solution in the form x = x and y as a function of x:
x = x (Where x can take any real value)
y = -3/2x
So, the answer is x = x and y = -3/2x.
Let's solve the system of equations:
1. Start with the given system:
3x + 2y = 0 ...(Equation 1)
x - 2y = 0 ...(Equation 2)
2. Multiply Equation 2 by 3 to create a cancellation opportunity for the y variable:
3(x - 2y) = 3(0)
3x - 6y = 0 ...(Equation 3)
3. Now, we have two equations with the same value for 3x. Subtract Equation 3 from Equation 1:
(3x + 2y) - (3x - 6y) = 0 - 0
3x + 2y - 3x + 6y = 0
8y = 0
4. Simplify Equation 4:
8y = 0
Divide both sides by 8:
y = 0
5. Substitute the value of y back into Equation 2 to solve for x:
x - 2(0) = 0
x - 0 = 0
x = 0
6. Therefore, the solution to the system of equations is:
x = 0
y = 0
Since we have a unique solution (x=0, y=0), the system does not have infinitely many solutions.