To solve the system of equations:
[tex]\[
\begin{cases}
y = 30x + 10 \\
y = 5x^2 - 25
\end{cases}
\][/tex]
we'll find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
1. Step 1: Set the equations equal to each other
Since both expressions are equal to [tex]\(y\)[/tex], we can set them equal to each other:
[tex]\[
30x + 10 = 5x^2 - 25
\][/tex]
2. Step 2: Rearrange the equation
Rearrange the equation to bring all terms to one side, forming a quadratic equation:
[tex]\[
5x^2 - 30x - 35 = 0
\][/tex]
3. Step 3: Simplify the quadratic equation
Divide every term by 5 to simplify the quadratic equation:
[tex]\[
x^2 - 6x - 7 = 0
\][/tex]
4. Step 4: Solve the quadratic equation
To solve the quadratic equation [tex]\(x^2 - 6x - 7 = 0\)[/tex], we can factor it:
[tex]\[
(x - 7)(x + 1) = 0
\][/tex]
This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[
x - 7 = 0 \quad \Rightarrow \quad x = 7
\][/tex]
[tex]\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\][/tex]
5. Step 5: Find the corresponding [tex]\(y\)[/tex] values
Substitute [tex]\(x = 7\)[/tex] and [tex]\(x = -1\)[/tex] back into either original equation to find the corresponding [tex]\(y\)[/tex] values. We'll use the first equation [tex]\(y = 30x + 10\)[/tex]:
- For [tex]\(x = 7\)[/tex]:
[tex]\[
y = 30(7) + 10 = 210 + 10 = 220
\][/tex]
- For [tex]\(x = -1\)[/tex]:
[tex]\[
y = 30(-1) + 10 = -30 + 10 = -20
\][/tex]
6. Step 6: Write the solutions
The solutions to the system of equations are:
[tex]\[
(x, y) = (-1, -20) \quad \text{and} \quad (7, 220)
\][/tex]
Thus, the solutions are:
[tex]\[
(-1, -20)
\][/tex]
[tex]\[
(7, 220)
\][/tex]
