Taking Algebra 1 Early (before 9th grade) - ppt download - Free Printable

Problem Analysis:

The image presents systems of equations for two different classes: Math 8 and Algebra 1. The goal is to solve these systems step by step and explain the differences in complexity between the two classes.

#### Key Observations:
1. Math 8 deals with linear systems of equations.
2. Algebra 1 introduces quadratic equations and systems with unequal coefficients, increasing the complexity.

Let us solve each example systematically.

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Math 8 Examples

#### Example 1:
Solve the system:
\[
\begin{cases}
y = \frac{1}{3}x - 6 \\
y = 4x + 2
\end{cases}
\]

Step 1: Equate the two expressions for \( y \).
Since both equations are solved for \( y \), we can set them equal to each other:
\[
\frac{1}{3}x - 6 = 4x + 2
\]

Step 2: Solve for \( x \).
First, eliminate the fraction by multiplying through by 3:
\[
x - 18 = 12x + 6
\]

Rearrange terms to isolate \( x \):
\[
x - 12x = 6 + 18
\]
\[
-11x = 24
\]
\[
x = -\frac{24}{11}
\]

Step 3: Substitute \( x = -\frac{24}{11} \) into one of the original equations to find \( y \).
Using \( y = 4x + 2 \):
\[
y = 4\left(-\frac{24}{11}\right) + 2
\]
\[
y = -\frac{96}{11} + \frac{22}{11}
\]
\[
y = -\frac{74}{11}
\]

Solution for Example 1 (Math 8):
\[
\boxed{\left( -\frac{24}{11}, -\frac{74}{11} \right)}
\]

---

#### Example 2:
Solve the system:
\[
\begin{cases}
3x - 6y = -9 \\
3x + 2y = 4
\end{cases}
\]

Step 1: Eliminate one variable using the elimination method.
Subtract the first equation from the second equation to eliminate \( x \):
\[
(3x + 2y) - (3x - 6y) = 4 - (-9)
\]
\[
3x + 2y - 3x + 6y = 4 + 9
\]
\[
8y = 13
\]
\[
y = \frac{13}{8}
\]

Step 2: Substitute \( y = \frac{13}{8} \) into one of the original equations to find \( x \).
Using \( 3x - 6y = -9 \):
\[
3x - 6\left(\frac{13}{8}\right) = -9
\]
\[
3x - \frac{78}{8} = -9
\]
\[
3x - \frac{39}{4} = -9
\]
Multiply through by 4 to clear the fraction:
\[
12x - 39 = -36
\]
\[
12x = 3
\]
\[
x = \frac{1}{4}
\]

Solution for Example 2 (Math 8):
\[
\boxed{\left( \frac{1}{4}, \frac{13}{8} \right)}
\]

---

Algebra 1 Examples

#### Example 1:
Solve the system:
\[
\begin{cases}
y = 2x^2 + 3x - 6 \\
y = 4x + 2
\end{cases}
\]

Step 1: Equate the two expressions for \( y \).
Set the quadratic equation equal to the linear equation:
\[
2x^2 + 3x - 6 = 4x + 2
\]

Step 2: Rearrange into standard quadratic form.
Move all terms to one side:
\[
2x^2 + 3x - 6 - 4x - 2 = 0
\]
\[
2x^2 - x - 8 = 0
\]

Step 3: Solve the quadratic equation using the quadratic formula.
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = -1 \), and \( c = -8 \). Substitute these values:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-8)}}{2(2)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 64}}{4}
\]
\[
x = \frac{1 \pm \sqrt{65}}{4}
\]

So, the solutions for \( x \) are:
\[
x = \frac{1 + \sqrt{65}}{4} \quad \text{and} \quad x = \frac{1 - \sqrt{65}}{4}
\]

Step 4: Substitute the \( x \)-values back into one of the original equations to find \( y \).
Using \( y = 4x + 2 \):

For \( x = \frac{1 + \sqrt{65}}{4} \):
\[
y = 4\left(\frac{1 + \sqrt{65}}{4}\right) + 2
\]
\[
y = 1 + \sqrt{65} + 2
\]
\[
y = 3 + \sqrt{65}
\]

For \( x = \frac{1 - \sqrt{65}}{4} \):
\[
y = 4\left(\frac{1 - \sqrt{65}}{4}\right) + 2
\]
\[
y = 1 - \sqrt{65} + 2
\]
\[
y = 3 - \sqrt{65}
\]

Solution for Example 1 (Algebra 1):
\[
\boxed{\left( \frac{1 + \sqrt{65}}{4}, 3 + \sqrt{65} \right) \text{ and } \left( \frac{1 - \sqrt{65}}{4}, 3 - \sqrt{65} \right)}
\]

---

#### Example 2:
Solve the system:
\[
\begin{cases}
3x - 6y = -9 \\
6x + 2y = 4
\end{cases}
\]

Step 1: Simplify the equations if possible.
Divide the first equation by 3:
\[
x - 2y = -3
\]
The second equation remains:
\[
6x + 2y = 4
\]

Step 2: Use the elimination method.
Add the two equations to eliminate \( y \):
\[
(x - 2y) + (6x + 2y) = -3 + 4
\]
\[
x + 6x = 1
\]
\[
7x = 1
\]
\[
x = \frac{1}{7}
\]

Step 3: Substitute \( x = \frac{1}{7} \) into one of the original equations to find \( y \).
Using \( x - 2y = -3 \):
\[
\frac{1}{7} - 2y = -3
\]
Multiply through by 7 to clear the fraction:
\[
1 - 14y = -21
\]
\[
-14y = -22
\]
\[
y = \frac{11}{7}
\]

Solution for Example 2 (Algebra 1):
\[
\boxed{\left( \frac{1}{7}, \frac{11}{7} \right)}
\]

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Final Answers:

- Math 8, Example 1: \(\boxed{\left( -\frac{24}{11}, -\frac{74}{11} \right)}\)
- Math 8, Example 2: \(\boxed{\left( \frac{1}{4}, \frac{13}{8} \right)}\)
- Algebra 1, Example 1: \(\boxed{\left( \frac{1 + \sqrt{65}}{4}, 3 + \sqrt{65} \right) \text{ and } \left( \frac{1 - \sqrt{65}}{4}, 3 - \sqrt{65} \right)}\)
- Algebra 1, Example 2: \(\boxed{\left( \frac{1}{7}, \frac{11}{7} \right)}\)