Elementary Algebra

• Solve one-variable linear equations, including ones with variables on both sides, parentheses, and fractions
• Solve a linear inequality, flipping the sign when multiplying or dividing by a negative, show it on a number line, and translate 'at most' as ≤ and 'at least' as ≥
• Solve absolute value equations by splitting |ax + b| = c into two cases, and recognize when there is no solution
• Use two answer-choice techniques and say when each fits: Pick a Number when the choices contain variables (substitute a simple value, test all four choices, try a second value to break a tie), and Middle Choice First when the choices are numbers in order (test a middle choice, then move up or down)
• Finish on one ACT-hard question and say what made it hard

To solve a linear equation, isolate the variable by performing the same operation on both sides: distribute to clear parentheses, multiply every term by the common denominator to clear fractions, collect the variable terms on one side, then divide. Check by substituting the answer back in. Inequalities are solved the same way with one extra rule: flip the inequality sign when multiplying or dividing by a negative number. On a number line, use an open circle for < or > and a closed circle for ≤ or ≥. For c > 0, absolute value equations |ax + b| = c split into two cases: ax + b = c and ax + b = −c. If c = 0, solve ax + b = 0; if c < 0, there is no real solution because absolute value cannot be negative. Slope and the equation of a line are taught in the Coordinate Geometry lesson, and absolute value inequalities in the Absolute Value Inequalities lesson, so neither is part of today. Linear equations and their applications show up throughout the Math section. Two answer-choice techniques. Pick a number: when the choices contain variables, choose a simple value that fits the question's conditions (avoid 0 and 1, which make many different expressions equal), work out the number the question asks for, then test all four choices; if two choices both match, try a second value. Worked example: which expression is equivalent to 3(x + 2) − x, with choices 2x + 6, 2x + 2, 3x + 6, and 4x + 6? With x = 0 the target is 6 and three choices give 6, which shows why 0 is a poor pick; with x = 2 the target is 3(4) − 2 = 10, and only 2x + 6 gives 10 (the others give 6, 12, and 14). Middle choice first: when the choices are numbers in order, test a middle choice in the question; if it is too big or too small you know which way to move, so at most two tests are needed. Both are slower than algebra when the equation takes one or two clean steps, so save them for messy algebra or abstract setups; for a student aiming at 34 or higher they are a second route and a way to check an abstract answer, not a replacement for algebra. The hardest questions, at the end of the section: 'must be true' and 'could be true' questions about inequalities. An ordinary value (a positive whole number) usually leaves two choices standing, so test a second, unusual value (a negative number, a fraction between 0 and 1, zero, or two equal values) to eliminate the rest. Worked example: if a < b, which must be true: a² < b², 1/a > 1/b, b − a > 0, or a + 1 < b? With a = 1 and b = 2, the first three hold and a + 1 < b fails, so three choices are still standing. With a = −3 and b = 1, a² < b² fails (9 is not less than 1) and 1/a > 1/b fails (−1/3 is not greater than 1), leaving b − a > 0, which is always true because b is larger. Before presenting any must-be-true example of your own, check that exactly one choice survives every value allowed, and never claim a choice fails without showing the value that breaks it.

• If an equation has fractions, multiply every term by the common denominator first. The rest is ordinary algebra.
• For inequality word problems, translate 'at most' as ≤ and 'at least' as ≥.
• If an absolute value equation equals a negative number, there is no solution. The ACT tests this.