Medium: Algebra

Medium Algebra is where the SAT stops asking you to solve an equation and starts asking you to build one. The equation still turns out linear, but it is hidden in a paragraph about a rental car or a draining tank, or it is a system whose answer is not a point but a count of solutions, or it is an inequality whose answer is the largest whole number that works. The arithmetic gets no harder than in Foundations; the reading and the setup do.

Three skills carry this level. The first is translation: turning phrases like "3 less than twice a number" and "at most" into symbols without swapping the order. The second is elimination, the method that solves any linear system in two lines once you choose the right multiplier. The third is recognizing, from the coefficients alone, whether two lines cross once, never, or everywhere, which is the single most-tested idea in this domain above the Foundations level.

The Advanced Algebra lesson takes the same three skills and adds unknown constants: systems that must have no solution for some value of k, equations true for every x, and the constraints those conditions impose.

Equations with fractions or decimals are cleared before they are solved. Multiply every term on both sides by the least common denominator, so (x + 1)/2 - (x - 3)/5 = 3 becomes 5(x + 1) - 2(x - 3) = 30, which expands to 5x + 5 - 2x + 6 = 30, so 3x = 19 and x = 19/3. A proportion, one fraction equal to another, is cleared by cross-multiplying, which is the same operation. Decimal coefficients are cleared by multiplying through by a power of ten. The one rule is that every term gets multiplied, including the lone constant on the other side.

Building a linear model from a description is a translation exercise. A fixed amount that happens once is the constant; an amount that repeats per unit of the variable is the coefficient. A rental car that costs 35 dollars plus 20 cents per mile gives C = 35 + 0.20m, and the question then asks either for a value (the cost of 150 miles is 65 dollars), for the variable (how many miles for 55 dollars: 100), or for an interpretation (the 0.20 is the cost of each additional mile, the 35 is the cost before any driving). When two quantities change at constant rates and you are asked when they are equal, set the two expressions equal: a tank with 500 gallons draining at 8 per minute and one with 200 gallons filling at 12 per minute hold the same amount when 500 - 8t = 200 + 12t, so t = 15 minutes.

Elimination solves a system by adding or subtracting the equations so that one variable cancels. For 3x + 2y = 16 and 5x - 2y = 8 the y-coefficients are already opposites, so adding gives 8x = 24 and x = 3, then 9 + 2y = 16 gives y = 7/2. When no pair of coefficients is opposite, multiply one or both equations first: for 2x + 3y = 7 and 5x + 2y = 12, multiply the first by 2 and the second by 3 to get 4x + 6y = 14 and 15x + 6y = 36, and subtract to get 11x = 22, x = 2, y = 1. Questions often ask for a combination like x + y or 2x - y rather than the pair, and sometimes the combination can be found by adding the original equations directly without solving.

The number of solutions of a system is a question about the two lines. Write each in slope-intercept form. Different slopes: the lines cross once and the system has exactly one solution. Same slope, different y-intercepts: parallel lines, no solution. Same slope and same intercept: the same line, infinitely many solutions. In standard form, ax + by = c and dx + ey = f, the slopes are equal exactly when a/d = b/e; the system has no solution when that holds but c/f differs, and infinitely many when all three ratios are equal. So 2x - 3y = 5 and 6x - 9y = 15 are the same line (every coefficient tripled) while 2x - 3y = 5 and 6x - 9y = 12 are parallel. Questions at this level give a numeric system and ask how many solutions it has; the Advanced lesson asks for the value of a constant that produces a given count.

Parallel lines have equal slopes. Perpendicular lines have slopes whose product is -1, so each is the negative reciprocal of the other: a line of slope 2/3 is perpendicular to a line of slope -3/2. To write a line through a given point with a given slope, substitute the point into y = mx + b to find b, or use point-slope form y - y₁ = m(x - x₁). A horizontal line has slope 0 and equation y = k; a vertical line has undefined slope and equation x = k. Converting standard form to slope-intercept form, y = (-a/b)x + c/b, is the fastest way to read a slope from an equation like 3x - 4y = 20.

A linear inequality in one variable is solved like an equation, reversing the sign when multiplying or dividing by a negative, and its solution is a half-line. In context the question usually asks for the greatest or least whole number satisfying a constraint: with a 200 dollar budget, items at 12 dollars each and 20 dollars shipping, 12n + 20 is at most 200 gives n at most 15, so 15 items. Systems of two inequalities in two variables describe a region; a point is a solution if it satisfies both, which is checked by substitution. When asked which point lies in the solution set, test each choice in both inequalities rather than sketching.

Multiplying only the fraction terms by the common denominator and forgetting the constant on the other side.

Writing "3 less than twice a number" as 3 - 2n. Subtraction and division reverse the spoken order; addition and multiplication do not.

Eliminating a variable by adding when the coefficients are equal rather than opposite. Equal coefficients call for subtraction.

Declaring a system to have no solution because the slopes match without checking the intercepts: matching intercepts mean infinitely many, not none.

Reading "exactly two solutions" as a possibility for two lines. It never is.

Treating "at least" as a strict inequality, or "no more than" as greater than.

Reporting the boundary of an inequality as the answer without asking which side, or without rounding to a whole number the right way when the question is about items or people.