To solve the system of equations using substitution, we start with the equations:
- Equation 1: y = x² – 3x + 7
- Equation 2: 3x + y = 2
Since Equation 1 is already solved for y, we can substitute the expression for y in Equation 2.
Substituting y from Equation 1 into Equation 2 gives us:
3x + (x² – 3x + 7) = 2
Now, simplify the equation:
- Combine like terms: x² – 3x + 7 + 3x = 2
- This reduces to: x² + 7 = 2
Next, we need to rearrange this equation:
x² + 7 – 2 = 0
x² + 5 = 0
This means:
- x² = -5
Since x² cannot equal a negative number in the set of real numbers, we conclude that there are no real solutions for x. However, if we consider imaginary numbers, we get:
- x = ±√(-5) = ±i√5
Now, we can find the corresponding values for y by substituting these x values back into Equation 1:
y = (±i√5)² – 3(±i√5) + 7
y = -5 – 3(±i√5) + 7
This simplifies to:
y = 2 – 3(±i√5)
Therefore, the solutions for the system of equations are:
- x = i√5, y = 2 – 3i√5
- x = -i√5, y = 2 + 3i√5
In conclusion, the system does not have real solutions, but it does have complex solutions.