SAT · Algebra · TOPIC LESSON
Build an equation, preserve equality, and interpret exactly what its solution says.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
A variable names a quantity; an equation asserts that two expressions have equal values. A solution is a value making that assertion true. In a linear equation, after simplification the unknown appears only to the first power. Multiplying out parentheses can reveal that an apparently complicated equation is linear.
Define the unknown with its unit before writing an equation. A fixed starting charge is added once; a per-item charge multiplies the number of items. The real-world domain may restrict an algebraic answer to a nonnegative whole number.
02 / UNDERSTAND & APPLY
Adding or subtracting the same expression on both sides preserves equality. Multiplying or dividing both sides by a known nonzero number also preserves equality. These operations are reversible, which is why the solution set stays unchanged.
Distribute a negative sign to every term. Clear denominators by multiplying every term on both sides by a common multiple. Do not divide by an expression involving the unknown unless the zero case has been checked; otherwise a valid solution may disappear.
03 / UNDERSTAND & APPLY
Simplify each side, move variable terms to one side, move constants to the other, and divide by the remaining coefficient. Equivalent paths can use different intermediate expressions. Substitute your answer into the original equation, not only the last simplified line.
ORIGINAL WORKED EXAMPLE
Solve 4(2x − 3) − 5 = 3x + 8.
Answer x = 5
04 / UNDERSTAND & APPLY
The numerator is a complete group. In (x + 2)/3, both x and 2 are divided by 3. Multiplying the entire equation by a common denominator prevents fragmented operations.
ORIGINAL WORKED EXAMPLE
Solve (x + 2)/3 − (x − 1)/4 = 5/6.
Answer x = −1
05 / UNDERSTAND & APPLY
Verbal order matters: “seven less than twice n” is 2n − 7. If two plans have equal cost, set their total cost expressions equal, not their rates. Include all fixed charges.
ORIGINAL WORKED EXAMPLE
A tool rental costs $18 plus $7 per hour. The total is $67 before tax. How many hours were billed?
Answer 7 hours
06 / UNDERSTAND & APPLY
Write the simplified equation as ax + b = cx + d. If a ≠ c there is one solution. If a = c but b ≠ d, subtraction gives a false numerical statement, so there is no solution. If a = c and b = d, the equation is an identity and every real value is a solution unless the context restricts the domain.
A variable disappearing does not automatically mean x = 0. Read the remaining statement.
ORIGINAL WORKED EXAMPLE
For which k does 3(x + 2) = kx + 6 have infinitely many real solutions?
Answer k = 3
07 / UNDERSTAND & APPLY
A question may ask for a multiple of the unknown or an expression rather than the unknown itself. Treat a repeated expression as a block when that avoids unnecessary arithmetic. Still distinguish that block from x.
ORIGINAL WORKED EXAMPLE
If 5(2x − 1) + 4 = 39, what is 2x − 1?
Answer 7
08 / UNDERSTAND & APPLY
A calculator can evaluate each original side at a proposed solution. A graph of the two sides can locate an intersection, but a rounded screen coordinate is not an exact proof, and a limited viewing window can hide an intersection. Algebra resolves zero-coefficient parameter cases directly.
Watch for a missing distribution, changing only one side, an incorrect sign when subtracting, and a correct x substituted for a different requested expression.
09 / UNDERSTAND & APPLY
Try these before revealing the reasoning.
ORIGINAL PRACTICE · 1
x = 6
Add 9 to get 6x = 36; divide both sides by 6.
ORIGINAL PRACTICE · 2
x = −9
Expand to 6x + 8 = 5x − 1; subtract 5x and 8.
ORIGINAL PRACTICE · 3
x = 23
Multiply the entire equation by 5, giving x − 3 = 20, then add 3.
ORIGINAL PRACTICE · 4
None
Subtracting 7x leaves 2 = −4, a false statement.
ORIGINAL PRACTICE · 5
Infinitely many
Distribution makes the two sides identical for every real x.
ORIGINAL PRACTICE · 6
9 rides
Let r count rides: 12 + 3r = 39, so 3r = 27 and r = 9.
ORIGINAL PRACTICE · 7
15
The left side is twice the requested expression, so divide the full equality by 2.
ORIGINAL PRACTICE · 8
a = 2
Only equal variable coefficients can cancel x; at a = 2 the false statement 5 = 9 remains.
KEEP THE BIG IDEAS
READ WITH CONTEXT
The teaching notes, passages, worked explanations and practice are original. Official references identify the assessment framework and skill scope; they do not endorse this lesson or license official question reproduction.
Official skill taxonomy reference. All teaching prose, examples, diagrams and practice in this lesson are original; this is not a past-question collection.
Factual skill-scope reference only; no question-bank questions or explanations are reproduced.
SAT analytics distinguish official blueprint weights from independently classified linear-practice question occurrences. These are not historical administered-exam frequencies.