To find the approximate solution for the system of equations:
- Equation 1: y = x^2 + 3x + 1
- Equation 2: y = x + 1
we need to set the two equations equal to each other since they both equal y.
1. **Set the equations equal**:
x^2 + 3x + 1 = x + 1
2. **Rearranging the equation**:
Subtract x + 1 from both sides to rearrange:
x^2 + 3x + 1 – x – 1 = 0
This simplifies to:
x^2 + 2x = 0
3. **Factoring the equation**:
We can factor out an x:
x(x + 2) = 0
4. **Finding the roots**:
Set each factor to zero:
- x = 0
- x + 2 = 0 → x = -2
5. **Finding the corresponding y-values**:
Now that we have values for x, we can substitute them back into either equation to find y. Using the second equation for simplicity:
For x = 0:
y = 0 + 1 = 1
For x = -2:
y = -2 + 1 = -1
6. **Approximate solutions**:
The approximate solutions for the system of equations are:
- Point 1: (0, 1)
- Point 2: (-2, -1)
In summary, by setting the equations equal to each other, rearranging, factoring, and solving, we found two points that satisfy both equations. These points are the approximate solutions for the given system of equations.