Questions & explanations
1. Backsolving: For linear equations with integer answer choices, substitute each option into the equation to see which one works. This is especially useful when the equation is messy or when you want to avoid solving for a variable.
Example: If 3(x-2)+4=2x+5, test options: x=3 gives 3(1)+4=7, right side 2(3)+5=11, no; x=4 gives 3(2)+4=10, right side 13; x=5 gives 3(3)+4=13, right side 15; x=6 gives 3(4)+4=16, right side 17; none? Actually solve: 3x-6+4=2x+5 => 3x-2=2x+5 => x=7. So if 7 not an option, maybe problem different. But backsolving works when answer is among options.
2. When adding or subtracting rational expressions with unlike denominators, find the least common denominator (LCD) by factoring each denominator completely. Multiply each numerator by the missing factors from the LCD, then combine numerators over the LCD. Simplify if possible.
For example, to add \(\frac{1}{x^2 - 1} + \frac{2}{x^2 + 2x + 1}\), factor denominators: \((x-1)(x+1)\) and \((x+1)^2\). LCD = \((x-1)(x+1)^2\). Rewrite: \(\frac{1 \cdot (x+1)}{(x-1)(x+1)^2} + \frac{2 \cdot (x-1)}{(x-1)(x+1)^2} = \frac{x+1 + 2x - 2}{(x-1)(x+1)^2} = \frac{3x - 1}{(x-1)(x+1)^2}\).
3. Strategy: For piecewise functions, evaluate each piece separately. Pay attention to the domain of each piece. When composing with another function, first find the output of the inner function, then determine which piece applies.
For example, if f(x) = { x+1 if x<0; x^2 if x≥0 } and g(x)=2x, then f(g(x)): for x<0, g(x)<0? Actually g(x)=2x, so if x<0 then g(x)<0, so f(g(x))=2x+1; if x≥0 then g(x)≥0, so f(g(x))=(2x)^2=4x^2. So the composition is piecewise.
4. Strategy: For inverse functions, remember that f(f^{-1}(x))=x and f^{-1}(f(x))=x. To find inverse, swap x and y and solve for y. Only one-to-one functions have inverses. For quadratics, restrict domain to make them one-to-one.
For example, f(x)=x^2-4x+3. Complete square: (x-2)^2-1. To have inverse, restrict domain to x≥2 or x≤2. Then inverse: y=(x-2)^2-1 => swap: x=(y-2)^2-1 => (y-2)^2=x+1 => y=2±√(x+1). Choose appropriate sign based on domain.
5. Strategy: For mixture problems, use the 'Alligation' method or set up equations. For example, to mix two solutions of concentrations a% and b% to get c%, the ratio of amounts is (c-b)/(a-c). This is a time-saver. Trap: Ensure you use consistent units (e.g., percent or decimal).
Alligation works for mixtures of two components. It avoids solving two equations. However, be careful with the order: (c-b) corresponds to the amount of the higher concentration if c is between a and b.
6. When multiplying binomials, use the FOIL method (First, Outer, Inner, Last) to ensure all terms are accounted for. For special products, memorize patterns: \((a+b)^2 = a^2 + 2ab + b^2\), \((a-b)^2 = a^2 - 2ab + b^2\), and \((a+b)(a-b) = a^2 - b^2\). These shortcuts save time.
For example, \((2x+3)^2 = (2x)^2 + 2(2x)(3) + 3^2 = 4x^2 + 12x + 9\). For \((x+5)(x-5) = x^2 - 25\). Avoid the common mistake of forgetting the middle term in perfect squares.
7. Strategy: When dealing with function composition, substitute the inner function into the outer function and simplify. For quadratic compositions, complete the square to find vertex or range.
For example, if f(x)=2x+1 and g(x)=x^2-3, then f(g(x))=2(x^2-3)+1=2x^2-5. To find range, complete square if needed. This avoids errors in order of operations.
8. Strategy: To check if a function is even, odd, or neither, substitute -x and simplify. Even if f(-x)=f(x), odd if f(-x)=-f(x). For quadratics, if the function is in vertex form, symmetry axis is x=h; if h=0, it's even.
For example, f(x)=2(x-3)^2+1: f(-x)=2(-x-3)^2+1=2(x+3)^2+1, not equal to f(x) or -f(x), so neither. But if f(x)=2x^2+1, then f(-x)=2x^2+1=f(x), so even.
9. Strategy: For distance problems, use the formula distance = rate × time. When two objects move towards each other, their relative speed is the sum of their speeds. When moving in the same direction, relative speed is the difference. This can simplify calculations. Trap: Remember to account for different start times by adjusting the distance or time.
For example, if one leaves earlier, calculate the head start distance and subtract from total distance before using relative speed.
10. Plug in numbers for absolute value inequalities: To solve |ax+b| < c, rewrite as -c < ax+b < c. For |ax+b| > c, rewrite as ax+b < -c or ax+b > c. Avoid the common mistake of forgetting the 'or' case.
Example: |2x-3| > 5 means 2x-3 < -5 or 2x-3 > 5 => 2x < -2 or 2x > 8 => x < -1 or x > 4. The solution is two separate intervals.
11. Strategy: Use the 'Work' formula for combined rates. When two or more workers work together, their combined rate is the sum of their individual rates. For example, if A can complete a job in 3 hours and B in 6 hours, their combined rate is 1/3 + 1/6 = 1/2 job per hour, so together they take 2 hours. Always express rates as 'jobs per time unit'.
This strategy helps solve work problems quickly by converting to rates. Trap: Do not average the times; always add rates.