SAT Math algebra: linear equations, inequalities and systems

The five skill areas in SAT Algebra

These are the skill/knowledge testing points the College Board lists for the Algebra domain.

Source: College Board Math overview and the Assessment Framework for the Digital SAT Suite, checked 6 October 2026.

The key ideas

A linear relationship changes by the same amount for every equal step. In y = mx + b, m is the rate of change (the slope) and b is the starting value (the y-intercept, the value of y when x = 0). Almost every context question in this domain is really asking which of those two numbers something is. If a gym charges a joining fee plus a monthly fee, the joining fee is b and the monthly fee is m.

A system of two linear equations is two straight lines. They meet once (one solution), never (parallel lines: same slope, different intercepts, so no solution), or are the same line (infinitely many solutions). To find a constant that gives no solution, make the x and y coefficients proportional while the constants are not. To get infinitely many solutions, make all three proportional.

Inequalities behave like equations with one extra rule: multiplying or dividing both sides by a negative number reverses the inequality sign. In context, check whether the answer must be a whole number and whether 'at most' means you round down.

Many SAT questions ask for an expression, not for x. If 3x + 6 = 21 and the question asks for x + 2, divide the whole equation by 3 to get x + 2 = 7 directly. Spotting this saves time and avoids a final step where errors creep in.

Worked example 1: an equation with a fraction

Question: If (2/3)(x - 6) = x - 10, what is the value of x?

1. Multiply both sides by 3 to clear the fraction: 2(x - 6) = 3(x - 10).
2. Expand: 2x - 12 = 3x - 30.
3. Collect terms: -12 + 30 = 3x - 2x, so x = 18.
4. Check: (2/3)(18 - 6) = (2/3)(12) = 8, and 18 - 10 = 8. Correct.

Worked example 2: a system with no solution

Question: In the system ax + 3y = 7 and 4x + 6y = 9, a is a constant. For what value of a does the system have no solution?

1. No solution means parallel lines: the x and y coefficients are in the same ratio, but the constants are not.
2. The y coefficients are 3 and 6, a ratio of 1 to 2. So a must be half of 4: a = 2.
3. Check the constants: 7 and 9 are not in the ratio 1 to 2 (half of 9 is 4.5), so the lines are distinct and parallel.
4. Answer: a = 2. With a = 2 the first equation is 2x + 3y = 7 and the second, halved, is 2x + 3y = 4.5, which can never both be true.

Worked example 3: a linear model in context

Question: A water tank holds 1,200 litres. When a valve is opened, the volume V, in litres, after t minutes is V = 1,200 - 15t. What does 15 represent, and after how many minutes does the tank hold 450 litres?

1. 15 is the coefficient of t, so it is the rate of change: the tank loses 15 litres every minute. 1,200 is the volume at t = 0.
2. Set V = 450: 450 = 1,200 - 15t.
3. Rearrange: 15t = 1,200 - 450 = 750, so t = 50.
4. Answer: 15 litres drain per minute, and the tank holds 450 litres after 50 minutes.

Worked example 4: an inequality with a budget

Question: A delivery service charges a fixed fee of 12 plus 2.5 for each kilometre. A customer will pay at most 50. What is the greatest whole number of kilometres the delivery can cover?

1. Write the inequality: 12 + 2.5d ≤ 50, where d is the distance in kilometres.
2. Subtract 12: 2.5d ≤ 38. Divide by 2.5: d ≤ 15.2.
3. The question asks for a whole number of kilometres, so round down: 15. Rounding up to 16 would cost 12 + 40 = 52, over budget.
4. Answer: 15 km. Check: 12 + 2.5(15) = 49.5, which is within 50.

Common mistakes that cost marks

• Answering for x when the question asks for 2x, x + y or another expression.
• Mixing up slope and intercept in context: the per-unit amount is the slope, the starting or fixed amount is the intercept.
• Forgetting to reverse the inequality sign when dividing by a negative number.
• Rounding the wrong way in 'at most' or 'at least' problems.
• Calling a system with proportional coefficients 'infinitely many solutions' without checking whether the constants are also proportional.
• Losing a minus sign when moving terms across the equals sign.

Exam technique and how a tutor helps

Algebra questions appear throughout both modules, and because questions in each module run from easier to harder, the early ones are often straightforward linear equations. Do those carefully rather than quickly: errors on easy questions in the first module are costly because that module's performance decides whether the second module is the harder or easier version. Underline what the question asks for before solving.

For context questions, label the variables in words before touching numbers: 'C = total cost, n = number of months'. For systems, decide in a few seconds between elimination by hand and graphing in Desmos. For typed-answer questions, fractions such as 7/2 can be entered directly, so there is no need to convert to a decimal.

In one-to-one lessons a tutor reads the student's working on the shared whiteboard and finds the pattern behind lost marks, which in Algebra is usually reading rather than method. Lessons then pair short sets of official-style questions with a habit to practise, such as writing what each number means in context, until the student does it without prompting.

Self-check: can you do these?

• Solve 4(x - 3) = 2x + 10. (Answer: x = 11)
• For what value of k does the system 2x + ky = 5 and 6x + 9y = 4 have no solution? (Answer: k = 3)
• Find the equation of the line through (1, 4) and (3, 10). (Answer: y = 3x + 1)
• What is the greatest integer x for which 7 - 2x > 1? (Answer: 2)
• A phone plan costs C = 20 + 0.5m for m minutes. Say in words what 0.5 represents.