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Algebra Skills You Must Master Before Calculus, Algebra Mistakes Killing Your Calculus Score (Must Master)

Calculus concepts are often simple; it's the algebra required to solve the resulting equations that causes students to fail. You must be comfortable solving rational equations, non-linear inequalities, and absolute values.

Core Theorem
To solve rational inequalities like P(x)Q(x)>0\frac{P(x)}{Q(x)} > 0, you cannot simply cross-multiply if the sign of Q(x)Q(x) is unknown. You must find critical points and test intervals. example

Practice Exercises

Example 01Easy
Solve the rational equation: 2xx+1=2x−1x\frac{2x}{x+1} = \frac{2x-1}{x}.
NEED A HINT?
Find a common denominator or cross-multiply since this is an equation (not an inequality), then solve the resulting quadratic equation. example
SHOW DETAILED EXPLANATION

Step 1: Cross Multiply

Multiply both sides by x(x+1)x(x+1) to get: 2x(x)=(2x−1)(x+1)2x(x) = (2x-1)(x+1).

Step 2: Expand and Simplify

2x2=2x2+2x−x−12x^2 = 2x^2 + 2x - x - 1, which simplifies to 2x2=2x2+x−12x^2 = 2x^2 + x - 1.

Step 3: Solve for x

Subtract 2x22x^2 from both sides: 0=x−10 = x - 1, so x=1x = 1. Check if the denominator is zero at x=1x=1 (it is not). example
Example 02Easy
Solve the compound inequality: −2<2x−3x+1≤1-2 < \frac{2x-3}{x+1} \le 1.
NEED A HINT?
Break this into two separate inequalities and solve them independently by moving everything to one side. Do not multiply by (x+1)(x+1) as its sign is unknown. example
SHOW DETAILED EXPLANATION

Step 1: Split the Inequality

Solve −2<2x−3x+1-2 < \frac{2x-3}{x+1} AND 2x−3x+1≤1\frac{2x-3}{x+1} \le 1 separately.

Step 2: Solve Left Part

2x−3x+1+2>0⇒2x−3+2(x+1)x+1>0⇒4x−1x+1>0\frac{2x-3}{x+1} + 2 > 0 \Rightarrow \frac{2x-3+2(x+1)}{x+1} > 0 \Rightarrow \frac{4x-1}{x+1} > 0. Critical points are −1,1/4-1, 1/4.

Step 3: Solve Right Part

2x−3x+1−1≤0⇒2x−3−(x+1)x+1≤0⇒x−4x+1≤0\frac{2x-3}{x+1} - 1 \le 0 \Rightarrow \frac{2x-3-(x+1)}{x+1} \le 0 \Rightarrow \frac{x-4}{x+1} \le 0. Critical points are −1,4-1, 4.

Step 4: Find Intersection

Combine the solution sets from both parts to find the valid range for xx. example
Example 03Hard
Solve the absolute value inequality: ∣x−1∣−∣x−10∣≥5|x-1|-|x-10| \ge 5.
NEED A HINT?
You must split the problem into cases based on the critical points of the absolute values (x=1x=1 and x=10x=10). example
SHOW DETAILED EXPLANATION

Step 1: Identify Intervals

The critical points are 1 and 10. Test three intervals: x<1x < 1, 1≤x<101 \le x < 10, and x≥10x \ge 10.

Step 2: Case 2 Analysis

For 1≤x<101 \le x < 10: ∣x−1∣|x-1| is (x−1)(x-1) and ∣x−10∣|x-10| is −(x−10)-(x-10). The inequality becomes (x−1)−[−(x−10)]≥5(x-1) - [-(x-10)] \ge 5.

Step 3: Solve Linear Inequality

x−1+x−10≥5⇒2x−11≥5⇒2x≥16⇒x≥8x - 1 + x - 10 \ge 5 \Rightarrow 2x - 11 \ge 5 \Rightarrow 2x \ge 16 \Rightarrow x \ge 8. Combined with the interval, the solution here is [8,10)[8, 10).
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