Answer:
y = 4x + 2
y = 2(2x - 1)
Zero solutions.
4x + 2 can never be equal to 4x - 2
y = 3x - 4
y = 2x + 2
One solution
3x - 4 = 2x + 2 has one solution
Explanation:
* Lets explain how to solve the problem
- The system of equation has zero number of solution if the coefficients
of x and y are the same and the numerical terms are different
- The system of equation has infinity many solutions if the
coefficients of x and y are the same and the numerical terms
are the same
- The system of equation has one solution if at least one of the
coefficient of x and y are different
* Lets solve the problem
∵ y = 4x + 2 ⇒ (1)
∵ y = 2(2x - 1) ⇒ (2)
- Lets simplify equation (2) by multiplying the bracket by 2
∴ y = 4x - 2
- The two equations have same coefficient of y and x and different
numerical terms
∴ They have zero equation
y = 4x + 2
y = 2(2x - 1)
Zero solutions.
4x + 2 can never be equal to 4x - 2
∵ y = 3x - 4 ⇒ (1)
∵ y = 2x + 2 ⇒ (2)
- The coefficients of x and y are different, then there is one solution
- Equate equations (1) and (2)
∴ 3x - 4 = 2x + 2
- Subtract 2x from both sides
∴ x - 4 = 2
- Add 4 to both sides
∴ x = 6
- Substitute the value of x in equation (1) or (2) to find y
∴ y = 2(6) + 2
∴ y = 12 + 2 = 14
∴ y = 14
∴ The solution is (6 , 14)
y = 3x - 4
y = 2x + 2
One solution
3x - 4 = 2x + 2 has one solution