📝 SAT Math: Heart of Algebra Study Notes
The Heart of Algebra section on the SAT Math test focuses on a core set of algebraic concepts, primarily involving linear equations, inequalities, functions, and systems. Mastering these topics is essential for solving a significant portion of the math problems.
🔢 Solving Linear Equations in One Variable
A linear equation in one variable involves an unknown quantity (often \(x\)) raised to the power of 1. The primary goal is to isolate the variable using inverse operations.
💡 Pro Tip: Inverse Operations
- To undo addition, use subtraction.
- To undo subtraction, use addition.
- To undo multiplication, use division.
- To undo division, use multiplication.
Example: Solve \(5x - 8 = 17\)
\[5x - 8 = 17\] \[5x - 8 + 8 = 17 + 8\] \[5x = 25\] \[\frac{5x}{5} = \frac{25}{5}\] \[x = 5\]
Equations can also involve variables on both sides or require distributing terms.
Example: Solve \(2(x + 3) = 4x - 6\)
\[2x + 6 = 4x - 6\] \[6 + 6 = 4x - 2x\] \[12 = 2x\] \[\frac{12}{2} = x\] \[x = 6\]
⚖️ Solving Linear Inequalities in One Variable
Linear inequalities are solved similarly to linear equations, but with one critical difference: if you multiply or divide both sides by a negative number, you must reverse the inequality sign.
📌 Key Takeaway: Inequality Sign Reversal
If \(a > b\), then \(-a < -b\). If \(ax < b\) and \(a < 0\), then \(x > \frac{b}{a}\).
Example: Solve \(-3x + 5 \le 17\)
\[-3x + 5 \le 17\] \[-3x \le 17 - 5\] \[-3x \le 12\] \[\frac{-3x}{-3} \ge \frac{12}{-3} \quad \text{(Sign reversed!)}\] \[x \ge -4\]
📊 Linear Equations in Two Variables and Functions
These equations represent straight lines when graphed on a coordinate plane. They often appear as functions, where \(y = f(x)\).
Forms of Linear Equations:
- Slope-Intercept Form: \(y = mx + b\)
- \(m\) represents the slope (rate of change).
- \(b\) represents the y-intercept (the point \((0, b)\) where the line crosses the y-axis).
- Standard Form: \(Ax + By = C\)
- \(A\), \(B\), and \(C\) are constants.
- Useful for finding intercepts quickly (set \(x=0\) for y-intercept, set \(y=0\) for x-intercept).
- Point-Slope Form: \(y - y_1 = m(x - x_1)\)
- \(m\) is the slope.
- \((x_1, y_1)\) is a known point on the line.
- Useful for writing an equation when given a point and the slope.
Calculating Slope (\(m\)):
Given two points \((x_1, y_1)\) and \((x_2, y_2)\), the slope is:
\[m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}\]
Interpreting Linear Functions:
In real-world contexts, the slope represents the rate of change (e.g., miles per hour, cost per item). The y-intercept represents the initial value or a fixed cost.
📝 Systems of Linear Equations
A system of linear equations consists of two or more linear equations with the same variables. The solution to a system is the set of values for the variables that satisfy all equations simultaneously. Graphically, this is the point(s) of intersection.
Types of Solutions:
- One Solution: The lines intersect at exactly one point. (Different slopes)
- No Solution: The lines are parallel and never intersect. (Same slope, different y-intercepts)
- Infinitely Many Solutions: The lines are identical (coincident). (Same slope, same y-intercept)
Methods for Solving Systems:
| Method | Description |
|---|---|
| Substitution | Solve one equation for a variable, then substitute that expression into the other equation. |
| Elimination | Multiply equations by constants to make coefficients of one variable opposites, then add the equations. |
| Graphing | Graph both lines; the intersection point is the solution. (Less precise for non-integer solutions) |
Example (Substitution): Solve the system:
\[y = 2x - 3\] \[4x + y = 9\]
Substitute \(2x - 3\) for \(y\) in the second equation:
\[4x + (2x - 3) = 9\] \[6x - 3 = 9\] \[6x = 12\] \[x = 2\]
Substitute \(x = 2\) back into \(y = 2x - 3\):
\[y = 2(2) - 3\] \[y = 4 - 3\] \[y = 1\]
The solution is \((2, 1)\).
📈 Systems of Linear Inequalities
Solving a system of linear inequalities involves finding the region on a graph that satisfies all inequalities. The solution is the overlapping shaded region.
Steps to Solve by Graphing:
- Graph each inequality as if it were an equation (a line).
- Use a solid line for \(\le\) or \(\ge\).
- Use a dashed line for \(<\) or \(>\).
- Choose a test point (e.g., \((0,0)\) if not on the line) for each inequality to determine which side of the line to shade.
- The solution to the system is the region where all the shaded areas overlap.
➕ Absolute Value Equations and Inequalities
Absolute value equations and inequalities often lead to two separate linear equations or inequalities.
Absolute Value Equations:
If \(|ax + b| = c\) where \(c > 0\), then \(ax + b = c\) OR \(ax + b = -c\).
Example: Solve \(|2x - 4| = 6\)
\[2x - 4 = 6 \quad \text{OR} \quad 2x - 4 = -6\] \[2x = 10 \quad \text{OR} \quad 2x = -2\] \[x = 5 \quad \text{OR} \quad x = -1\]
Absolute Value Inequalities:
- If \(|ax + b| < c\) (or \(\le c\)), then \(-c < ax + b < c\) (or \(-c \le ax + b \le c\)). This is an "and" statement.
- If \(|ax + b| > c\) (or \(\ge c\)), then \(ax + b > c\) OR \(ax + b < -c\) (or \(ax + b \ge c\) OR \(ax + b \le -c\)). This is an "or" statement.
Example: Solve \(|x + 3| < 5\)
\[-5 < x + 3 < 5\] \[-5 - 3 < x < 5 - 3\] \[-8 < x < 2\]