Learning objectives
Solve linear equations containing brackets, fractions, decimals or the variable on both sides
Write a linear equation from a context, solve it and interpret the solution
Decide whether an equation has one solution, no solution or infinitely many, and find the constant that produces each case
Rearrange a formula for a chosen variable and evaluate a related expression without solving fully
Key definitions
Linear equation
An equation in which the variable appears only to the first power, such as 3(x − 2) = 5x + 4; it can always be rearranged to the form ax = b.
Identity
An equation that is true for every value of x, such as 2(x + 3) = 2x + 6; it has infinitely many solutions.
Contradiction
An equation with no solution, such as 2x + 5 = 2x + 1: the x-terms cancel and leave a false statement.
Literal equation
A formula containing several letters; solving it for one letter (making it the subject) uses the same inverse operations as a numerical equation.
Explanation
Solving. Undo operations in reverse order, doing the same thing to both sides. Clear fractions first by multiplying every term by the lowest common denominator, clear decimals by multiplying by 10 or 100, and expand brackets carefully — a minus sign in front of a bracket changes the sign of every term inside it. Then collect the variable on the side with the larger coefficient to avoid negatives, and divide. Example: (x + 1)/3 − (x − 2)/4 = 2. Multiply by 12: 4(x + 1) − 3(x − 2) = 24, so 4x + 4 − 3x + 6 = 24 and x = 14. Check by substitution: 15/3 − 12/4 = 5 − 3 = 2. SAT strategy: on multiple-choice items you can back-solve by substituting the answer choices, and on any item you can graph y = left side and y = right side in the built-in calculator and read the x-coordinate of the intersection.
Writing equations from context. Define the variable with its units, translate each phrase into algebra, set the expression equal to the stated total and solve. "A bike-hire shop charges a $12 fee plus $7.50 per hour, and a customer paid $49.50" becomes 12 + 7.5h = 49.5, so 7.5h = 37.5 and h = 5: the bike was hired for 5 hours. Then answer the question actually asked — the SAT often wants a related quantity, such as the cost of a longer hire or the difference between two amounts, rather than the variable itself. Finally, sense-check the result: a negative time, a fractional number of people or a price larger than the whole budget signals a set-up error, usually a sign slip or a rate attached to the wrong quantity.
How many solutions? Simplify both sides fully and write the equation as px + q = rx + s. If p ≠ r, there is exactly one solution. If p = r but q ≠ s, the x-terms cancel and leave a false statement such as 5 = 9, so there is no solution. If p = r and q = s, both sides are identical, the equation reduces to 0 = 0 and there are infinitely many solutions. Constant questions use the same comparison. For 3(2x − k) = 6x + 12 to have infinitely many solutions, 6x − 3k must equal 6x + 12, so −3k = 12 and k = −4; any other value of k gives no solution. For ax + 7 = 5x − 2 to have no solution, a must equal 5 (the constants 7 and −2 already differ). You never need to solve for x in these questions — you only compare coefficients.
Rearranging and related expressions. To make a variable the subject of a formula, treat the other letters as numbers and undo the operations: from P = 2l + 2w, w = (P − 2l)/2. If the target letter appears in two terms, collect those terms on one side and factor it out: from ax + b = cx, x(a − c) = −b, so x = b/(c − a). The SAT also asks for the value of an expression rather than x itself. If 2x − 3 = 7, then 4x − 6 = 2(2x − 3) = 14 immediately, with no need to find x = 5 first. Before solving, always look for a multiple of the given expression. To verify a proposed solution, substitute it into the original equation — checking a rearranged line can hide an earlier slip.
Formulae
ax + b = c ⇒ x = (c − b)/a (a ≠ 0)
px + q = rx + s: p ≠ r → one solution; p = r, q ≠ s → no solution; p = r, q = s → infinitely many
If ax + b = c, then k(ax + b) = kc for any constant k
Worked examples
q: For what value of k does the equation 4(x − 3) + 2x = kx − 12 have infinitely many solutions?
steps: Expand and simplify the left side: 4x − 12 + 2x = 6x − 12.
Compare 6x − 12 with kx − 12: the constants already match (−12 and −12).
For infinitely many solutions the x-coefficients must also match, so k = 6.
Check: with k = 6 both sides are 6x − 12, an identity true for every x. Answer: k = 6.
In context
Pricing plans: a taxi fare of $3.50 plus $2.25 per mile gives 3.5 + 2.25m = total, and solving finds the distance traveled for a known fare.
Converting temperatures: rearranging F = 1.8C + 32 to C = (F − 32)/1.8 is making C the subject of a linear formula.
Common misconceptions
Distributing a negative sign to only the first term inside a bracket: −(x − 5) is −x + 5, not −x − 5.
Multiplying only some terms by the common denominator when clearing fractions — every term on both sides must be multiplied.
Thinking that an equation reducing to 0 = 0 has the solution x = 0; it means every value of x works.
Study tips
Underline what the question asks for (x, 2x, a cost, a number of hours) before solving — many SAT errors come from solving correctly and then answering a different quantity.
For "no solution" or "infinitely many solutions" items, expand both sides fully and compare coefficients instead of trying to solve.
Summary
Clear fractions, expand, collect x on one side, divide — then check by substitution.
Different x-coefficients → one solution; same x-coefficients, different constants → none; identical sides → infinitely many.
Look for a multiple of the given expression before solving for x.
Check your understanding
Question 1: If 5x − 2 = 3x + 8, what is the value of x?
1. 5
2. 3
3. −5
Question 2: How many solutions does 2(x + 4) = 2x + 9 have?
1. None
2. Infinitely many
3. Exactly one