2. $$ \left\{\begin{array}{l}y=4 x+5 \ y-2=x^{2}+3\end{array} ight. $$

Question

2. $$ \left\{\begin{array}{l}y=4 x+5 \ y-2=x^{2}+3\end{array}ight. $$

UpStudy AI · Accepted Answer

We are given the system of equations:
$$
\begin{cases}
y = 4x + 5,\[5mm]
y - 2 = x^2 + 3.
\end{cases}
$$

**Step 1.** Express the second equation in terms of $$ y $$.

Starting with
$$
y - 2 = x^2 + 3,
$$
add $$2$$ to both sides:
$$
y = x^2 + 3 + 2 = x^2 + 5.
$$

**Step 2.** Substitute the expression for $$ y $$ from the first equation.

We have:
$$
4x + 5 = x^2 + 5.
$$

**Step 3.** Solve for $$ x $$.

Subtract $$5$$ from both sides:
$$
4x = x^2,
$$
which can be rewritten as:
$$
x^2 - 4x = 0.
$$

Factor out $$ x $$:
$$
x(x - 4) = 0.
$$

So, the solutions for $$ x $$ are:
$$
x = 0 \quad \text{or} \quad x = 4.
$$

**Step 4.** Substitute the values of $$ x $$ back into the first equation to find $$ y $$.

- For $$ x = 0 $$:
  $$
  y = 4(0) + 5 = 5.
  $$

- For $$ x = 4 $$:
  $$
  y = 4(4) + 5 = 16 + 5 = 21.
  $$

**Final Answer:**

The solutions to the system are:
$$
(x, y) = (0, 5) \quad \text{and} \quad (x, y) = (4, 21).
$$
The solutions are $$ (0, 5) $$ and $$ (4, 21) $$.

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