https://r2.erweima.ai/mathbot/image/17264844650417cfvaazw.jpg

The system of equations shown in the image is:

\[
\begin{aligned}
1) &\ 3x^2 + 7y^2 = 187 \\
2) &\ 3x^2 - 7y = 47
\end{aligned}
\]

Let’s solve this step by step.

### Step 1: Subtract the two equations to eliminate \(x^2\)
If we subtract equation (2) from equation (1), we get:

\[
(3x^2 + 7y^2) - (3x^2 - 7y) = 187 - 47
\]

Simplifying:

\[
3x^2 - 3x^2 + 7y^2 + 7y = 140
\]

\[
7y^2 + 7y = 140
\]

### Step 2: Solve for \(y\)
Factor out the 7 from the left side:

\[
7(y^2 + y) = 140
\]

Divide both sides by 7:

\[
y^2 + y = 20
\]

This is a quadratic equation. To solve for \(y\), we can use the quadratic formula. The general form of a quadratic equation is:

\[
ax^2 + bx + c = 0
\]

For \(y^2 + y - 20 = 0\), we have \(a = 1\), \(b = 1\), and \(c = -20\). The quadratic formula is given by:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values of \(a\), \(b\), and \(c\):

\[
y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-20)}}{2(1)}
\]

\[
y = \frac{-1 \pm \sqrt{1 + 80}}{2}
\]

\[
y = \frac{-1 \pm \sqrt{81}}{2}
\]

\[
y = \frac{-1 \pm 9}{2}
\]

This gives two possible solutions:

\[
y = \frac{-1 + 9}{2} = \frac{8}{2} = 4
\]
or
\[
y = \frac{-1 - 9}{2} = \frac{-10}{2} = -5
\]

So, \(y = 4\) or \(y = -5\).

### Step 3: Substitute \(y\) back to find \(x^2\)
We can now substitute each value of \(y\) into one of the original equations to find \(x^2\). Let’s use equation (2):

\[
3x^2 - 7y = 47
\]

#### Case 1: \(y = 4\)

Substitute \(y = 4\) into the equation:

\[
3x^2 - 7(4) = 47
\]

\[
3x^2 - 28 = 47
\]

Add 28 to both sides:

\[
3x^2 = 75
\]

Divide by 3:

\[
x^2 = 25
\]

Take the square root of both sides:

\[
x = 5 \quad \text{or} \quad x = -5
\]

#### Case 2: \(y = -5\)

Substitute \(y = -5\) into the equation:

\[
3x^2 - 7(-5) = 47
\]

\[
3x^2 + 35 = 47
\]

Subtract 35 from both sides:

\[
3x^2 = 12
\]

Divide by 3:

\[
x^2 = 4
\]

Take the square root of both sides:

\[
x = 2 \quad \text{or} \quad x = -2
\]

### Final Solutions:
The possible pairs \((x, y)\) are:

1. \(x = 5\), \(y = 4\)
2. \(x = -5\), \(y = 4\)
3. \(x = 2\), \(y = -5\)
4. \(x = -2\), \(y = -5\)

Would you like more details on any part of this solution?

### Related Questions:
1. How can we verify the solutions by substituting into both original equations?
2. What is the geometric interpretation of these equations?
3. How does changing the constants in these equations affect the solution?
4. Can these equations be solved using matrix methods?
5. What happens if the system is not solvable or has no real solutions?

### Tip:
Always check your solutions by substituting them back into the original equations to ensure accuracy.

Solve the system of equations: 3x^2 + 7y^2 = 187 and 3x^2 - 7y = 47.

Explore how to solve the quadratic system of equations 3x^2 + 7y^2 = 187 and 3x^2 - 7y = 47 with detailed steps using algebraic methods and the quadratic formula. This tutorial is perfect for high school students learning about quadratic equations and systems of equations.

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