28. Solve the equations. $$ \begin{array}{r}x+3 y=13 \ x^{2}+3 y^{2}=43\end{array} $$

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UpStudy AI · Accepted Answer

We start with the system
$$
\begin{aligned}
x + 3y &= 13, \
x^2 + 3y^2 &= 43.
\end{aligned}
$$

**Step 1.** Express $$ x $$ in terms of $$ y $$ from the first equation:
$$
x = 13 - 3y.
$$

**Step 2.** Substitute $$ x = 13 - 3y $$ into the second equation:
$$
(13-3y)^2 + 3y^2 = 43.
$$

**Step 3.** Expand the square and simplify:
$$
(169 - 78y + 9y^2) + 3y^2 = 43.
$$
$$
169 - 78y + 12y^2 = 43.
$$

Subtract $$ 43 $$ from both sides:
$$
12y^2 - 78y + 126 = 0.
$$

Divide the entire equation by $$ 6 $$ to simplify:
$$
2y^2 - 13y + 21 = 0.
$$

**Step 4.** Solve the quadratic equation $$ 2y^2 - 13y + 21 = 0 $$ using the quadratic formula:
$$
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$
where $$ a=2 $$, $$ b=-13 $$, and $$ c=21 $$. First, compute the discriminant:
$$
b^2 - 4ac = (-13)^2 - 4(2)(21) = 169 - 168 = 1.
$$

Now, substitute into the formula:
$$
y = \frac{13 \pm \sqrt{1}}{4}.
$$

Thus, the two solutions for $$ y $$ are:
$$
y = \frac{13 + 1}{4} = \frac{14}{4} = \frac{7}{2} \quad \text{and} \quad y = \frac{13 - 1}{4} = \frac{12}{4} = 3.
$$

**Step 5.** Find $$ x $$ using $$ x = 13 - 3y $$.

For $$ y = \frac{7}{2} $$:
$$
x = 13 - 3\left(\frac{7}{2}\right) = 13 - \frac{21}{2} = \frac{26}{2} - \frac{21}{2} = \frac{5}{2}.
$$

For $$ y = 3 $$:
$$
x = 13 - 3(3) = 13 - 9 = 4.
$$

**Final Answer:**
$$
\boxed{\left(\frac{5}{2}, \frac{7}{2}\right) \quad \text{and} \quad (4,\,3)}
$$
The solutions are $$ \left(\frac{5}{2}, \frac{7}{2}\right) $$ and $$ (4, 3) $$.

Solve the equations.

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