Let’s solve the given simultaneous equations step-by-step.

The equations are:

1. $y = 3x + 4$
2. $y = x^2 - 7x + 5$

Step 1: Substitute $y = 3x + 4$ into $y = x^2 - 7x + 5$

Since both expressions are equal to $y$, we equate them: $3x + 4 = x^2 - 7x + 5$

Rearranging all terms to one side: $0 = x^2 - 7x - 3x + 5 - 4$ $x^2 - 10x + 1 = 0$

Step 2: Solve the quadratic equation

The equation to solve is: $x^2 - 10x + 1 = 0$

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -10$, and $c = 1$. Substituting these values: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(1)}}{2(1)}$ $x = \frac{10 \pm \sqrt{100 - 4}}{2}$ $x = \frac{10 \pm \sqrt{96}}{2}$ $x = \frac{10 \pm 4\sqrt{6}}{2}$ $x = 5 \pm 2\sqrt{6}$

So, the two solutions for $x$ are: $x_1 = 5 + 2\sqrt{6}, \quad x_2 = 5 - 2\sqrt{6}$

Approximating these values to two decimal places: $x_1 \approx 9.90, \quad x_2 \approx 0.10$

Step 3: Find corresponding $y$-values

Substitute each $x$ value into $y = 3x + 4$:

1. For $x_1 = 9.90$: $y_1 = 3(9.90) + 4 = 29.70 + 4 = 33.70$

2. For $x_2 = 0.10$: $y_2 = 3(0.10) + 4 = 0.30 + 4 = 4.30$

Final Answer:

$(x_1, y_1) \approx (9.90, 33.70)$ $(x_2, y_2) \approx (0.10, 4.30)$

Would you like me to verify or explain any part in more detail?

Here are 5 related questions you might consider:

1. What happens if the equations are non-linear and cannot be solved using substitution?
2. How does the discriminant ($b^2 - 4ac$) of a quadratic equation determine the nature of solutions?
3. Can simultaneous equations with one quadratic and one linear equation always have two solutions?
4. What would be the steps if both equations were quadratic?
5. How do you verify the solutions graphically?

Tip: Always check your solutions by substituting them back into both original equations to confirm accuracy.