The problem presented involves solving a pair of simultaneous equations:

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

Let's solve these step by step:

Step 1: Express $x$ from the second equation

From the second equation, we already have: $x = y^2 - 3$

Step 2: Substitute $x$ into the first equation

Now, substitute $x = y^2 - 3$ into the first equation $5y + 3x = 2y^2 + 5$: $5y + 3(y^2 - 3) = 2y^2 + 5$ Simplify the equation: $5y + 3y^2 - 9 = 2y^2 + 5$ Now move all terms to one side to set the equation to zero: $3y^2 - 2y^2 + 5y - 9 - 5 = 0$ Simplify further: $y^2 + 5y - 14 = 0$

Step 3: Solve the quadratic equation

We now need to solve $y^2 + 5y - 14 = 0$ using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 5$, and $c = -14$: $y = \frac{-5 \pm \sqrt{5^2 - 4(1)(-14)}}{2(1)}$ $y = \frac{-5 \pm \sqrt{25 + 56}}{2}$ $y = \frac{-5 \pm \sqrt{81}}{2}$ $y = \frac{-5 \pm 9}{2}$ So, we have two solutions for $y$:

1. $y = \frac{-5 + 9}{2} = 2$
2. $y = \frac{-5 - 9}{2} = -7$

Step 4: Find corresponding values of $x$

Now that we have two values of $y$, we substitute them back into $x = y^2 - 3$:

• For $y = 2$: $x = 2^2 - 3 = 4 - 3 = 1$

• For $y = -7$: $x = (-7)^2 - 3 = 49 - 3 = 46$

Final Answer:

The solutions to the system of equations are:

1. $(x, y) = (1, 2)$
2. $(x, y) = (46, -7)$

Would you like more details or have any further questions?

Here are 5 related questions to expand this topic:

1. How can you solve simultaneous equations with one linear and one quadratic equation graphically?
2. What is the general form of a quadratic equation, and how does it apply in real-world situations?
3. What other methods can be used to solve simultaneous equations, aside from substitution?
4. How would you check if a solution is correct for simultaneous equations?
5. How can the discriminant $b^2 - 4ac$ help in understanding the nature of the solutions?

Tip: For quadratic equations, always check the discriminant first to know if you will get real or complex solutions.