The seventh most common SAT Math question type (6.0% of the bank): one unknown, one equals sign, straight lines only. Covers inverse operations in the right order, clearing fractions by multiplying through, distributing before collecting, gathering variables on one side, the no-solution and infinitely-many variants, and the discipline of answering the expression the stem asked for rather than the value of x — with six worked examples, four traps and three checks.
Subject: SAT Prep · 61 slides · applied lesson
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Title
SAT Math · Type 7 of 19
6.0% of the question bank — 101 of 1675 questions
Objectives
One variable, one equals sign, and no exponents or roots anywhere. This is the most elementary mathematics on the test and it is still one question in seventeen. Almost nobody loses these to the algebra. They lose them by solving for x when the question wanted the value of some expression in x, and by not recognising the two variants where the variable disappears entirely.
The habit that matters most here has nothing to do with algebra: read the last five words of the question before you start, and again before you answer.
Naruhodo Tutoring — SAT Math question bank export (sat-question-index.json) — 101 tagged questions of this type in the site's bank
Section
Section 1
Concept
One variable, one equals sign, and no exponents or roots anywhere. This is the most elementary mathematics on the test and it is still one question in seventeen. Almost nobody loses these to the algebra. They lose them by solving for x when the question wanted the value of some expression in x, and by not recognising the two variants where the variable disappears entirely.
You will see it phrased in these ways:
The second phrasing is the one that costs marks. The equation is easy, the extra step is one substitution, and the value of x is always sitting among the choices.
Picture it
The same three-panel card as the survey deck, so the shorthand carries over: what identifies it, what you write first, and what is built to catch you.
Figure (svg): Linear equations in one variable: the tell, the move, and the trap
Notice that the green panel ends with a reading instruction rather than a mathematical one. That is deliberate, and it is where this type is actually decided.
Prediction
Recognition is quick here, but it decides which toolkit applies.
Predict first
Three of these are linear equations in one variable. Which is not?
Correct: x squared minus 4 equals 12
Why: A squared variable makes it nonlinear, which puts it in type 3 and brings a completely different routine — move everything to one side, factor, and expect two solutions. The other three have x to the first power only, so each has exactly one solution unless the variables cancel. The presence of fractions or brackets does not affect linearity at all; they are only obstacles in the way of isolating x.
Concept
Three question shapes sit under this label, and they want different final actions.
| variant | what it wants | the extra step |
|---|---|---|
| Solve for x | the value of the variable | none — but check that is really what was asked |
| Find the value of an expression | 3x, or 2x plus 1, or x over 4 | one substitution after solving |
| Count the solutions | one, none, or infinitely many | compare the coefficients and the constants |
The middle row is the most common source of lost marks on the whole type, and the fix is entirely a reading habit.
Definition probe
Naming the obstacle chooses the opening move.
Sort into buckets
What stands between you and an isolated x?
Discrimination
Six equations. Count them.
Sort into buckets
How many solutions?
Warm-up
Try it, and read the final phrase carefully.
Discussion prompt
If 3x minus 7 equals 14, what is the value of 6x minus 14?
Hint: Look at the target expression before you solve for x.
Answer:
The answer is 28. The target, 6x minus 14, is exactly twice the given expression 3x minus 7. So doubling both sides of the equation gives it immediately: 2 times 14 is 28.
The long route also works: 3x equals 21, so x equals 7, and 6 times 7 minus 14 is 42 minus 14, which is 28.
The trap is answering 7, the value of x. It is correct as a value and it is the wrong quantity, and it will be among the choices.
Notice the shortcut is worth spotting. Whenever the target is a multiple of the given expression, you never need to find x at all.
Pattern
Three steps, and the first and last are reading rather than algebra.
Read the final phrase and write down what is actually wanted: x, or 2x, or x plus 5.
Why: Doing this first tells you whether you can stop at x, and sometimes reveals a shortcut that avoids finding x entirely.
Clear the obstacles in order: fractions first, then brackets, then gather variables on one side.
Why: Multiplying through by a denominator early removes fractions from every later step. Distributing before collecting prevents sign errors.
Solve, then return to step 1 and produce that quantity.
Why: If the question wanted an expression, one substitution remains. If the variables cancelled, read what is left and answer none or infinitely many.
The bookends are the same instruction twice. That repetition is the whole defence against this type's only real trap.
Section
Section 2
Concept
To isolate a variable, undo what was done to it last, first.
Think of it as unwrapping a parcel: you take off the outermost layer first, and the outermost layer is whatever was done last.
Concept
Multiply both sides by a common denominator, and make sure it reaches every term.
Clearing fractions at the very start means no later step involves them, which is worth more than the elegance of keeping them.
Prediction
Clearing a fraction reaches the whole equation.
Predict first
Multiplying x over 3 plus 5 equals 11 by 3 gives which equation?
Correct: x plus 15 equals 33
Why: Every term is multiplied by 3: the fraction becomes x, the 5 becomes 15, and the 11 becomes 33. Solving gives x equals 18, and checking in the original gives 6 plus 5, which is 11. The second choice multiplies only the fraction and the right-hand side, which is the most common version of this error.
Concept
Expand every bracket first, watching the sign in front of it.
Look for the divide-first shortcut whenever the constant on the other side is a multiple of the coefficient outside the bracket.
Prediction
Compare the coefficients, then the constants.
Predict first
How many solutions does 5x minus 3 equals 5x plus 8 have?
Correct: none
Why: Subtracting 5x from both sides leaves negative 3 equals 8, which is false regardless of x. The coefficients match while the constants differ, so there is no solution. A linear equation can never have two solutions — that would require a curve, which is type 3.
Concept
Move all the variable terms to whichever side keeps the coefficient positive, and the constants to the other.
The choice of side is a small thing that removes a common source of error, so make it deliberately rather than by habit.
Concept
When the stem asks for an expression, solving for x is the second-to-last step.
| the stem asks for | after finding x equals 5 | the answer |
|---|---|---|
| the value of x | stop | 5 |
| the value of 2x | substitute | 10 |
| the value of x plus 3 | substitute | 8 |
| the value of 10 minus x | substitute | 5 — which coincidentally equals x |
Underline the target before starting. It is the single highest-value habit on this type and it costs two seconds.
Prediction
Sometimes you never need x at all.
Predict first
If 2x plus 5 equals 17, what is the value of 4x plus 10?
Correct: 34
Why: The target 4x plus 10 is exactly twice the given expression 2x plus 5, so it is twice 17, which is 34. Solving the long way gives x equals 6, and 24 plus 10 is also 34. The 6 is the value of x, offered because stopping there is the standard error, and 17 is the right-hand side copied across.
Concept
If the variables cancel and leave a false statement, no value of x can work.
Students commonly restart when the variable disappears. Recognising the outcome saves a minute and produces the mark.
Concept
If the variables cancel and leave something always true, every value of x works.
The pair of variants is symmetric: a false leftover means none, a true leftover means all. Nothing in between is possible for a linear equation.
Prediction
The sign in front of the bracket reaches both terms.
Predict first
Expanding 12 minus 3(x minus 2) gives which expression?
Correct: 12 minus 3x plus 6
Why: Negative 3 times x is negative 3x, and negative 3 times negative 2 is positive 6. So the expression is 12 minus 3x plus 6, which collects to 18 minus 3x. The second choice fails to change the sign of the second term, and the fourth incorrectly subtracts 3 from 12 before distributing.
Two truths and a lie
Three statements about linear equations are correct. The one left standing is false.
Eliminate the wrong options
Which statement is FALSE?
Survives elimination: b
Why: A linear equation has exactly one solution, no solutions, or infinitely many — never exactly two. Two solutions requires a squared term, which makes it a quadratic and moves it to type 3. This is a useful sanity check on count questions: if two appears among the choices for a linear equation, it can be eliminated without any work. Note the contrast with statement D, which catches the different confusion between the answer zero and no answer at all.
Check
Read the last five words before you start.
Check your understanding
If 4x minus 9 equals 23, what is the value of 8x minus 18?
Answer: A
Why: The target 8x minus 18 is exactly twice the given expression 4x minus 9, so it equals twice 23, which is 46. The long route agrees: 4x equals 32, so x equals 8, and 64 minus 18 is 46.
Three of the four choices are quantities you genuinely computed on the way to the answer. That is what makes this trap so effective under time pressure.
Section
Section 3
Worked example
Solve 4(x minus 3) plus 7 equals 23.
Figure (svg): A balance showing each step keeping the two sides equal
Distribute the 4: 4x minus 12 plus 7 equals 23.
Why: Brackets come out before like terms can be collected.
Collect the constants on the left: 4x minus 5 equals 23.
Why: Negative 12 plus 7 is negative 5.
Add 5 to both sides, then divide by 4: 4x equals 28, so x equals 7.
Why: Undo the subtraction first, then the multiplication — reverse order.
Substituting back into the ORIGINAL equation, not into a later line, is what catches a distribution error.
Verify: check in the original: 4 times (7 minus 3) plus 7 is 16 plus 7, which is 23.
Why: The original equation is satisfied, so the solution is correct.
Answer: x = 7
Worked example
Solve x over 5 minus 3 equals 4.
Figure (svg): A balance showing the fraction cleared by multiplying every term by five
Multiply every term by 5: x minus 15 equals 20.
Why: Five times x over 5 is x, five times 3 is 15, and five times 4 is 20.
Add 15 to both sides: x equals 35.
Why: Undoing the subtraction isolates x.
Note that only the fraction changing would have given x minus 3 equals 20, and x equals 23.
Why: Naming the wrong answer makes the error visible rather than abstract.
The wrong answer 23 is what you get by multiplying only two of the three terms, and it will be among the choices.
Verify: check in the original: 35 over 5 is 7, and 7 minus 3 is 4.
Why: The original equation holds, confirming 35 rather than 23.
Answer: x = 35
Worked example
Solve 9x minus 4 equals 5x plus 16.
Figure (svg): A balance gathering the variable terms on the left
Subtract 5x from both sides: 4x minus 4 equals 16.
Why: Moving the smaller coefficient keeps the surviving one positive, avoiding a sign error.
Add 4 to both sides: 4x equals 20.
Why: Constants gather on the side opposite the variable.
Divide by 4: x equals 5.
Why: The final inverse operation.
Had you subtracted 9x instead, you would get negative 4x minus 4 equals 16 and then need to divide by a negative — a correct route with one more chance to slip.
Verify: check both sides: 45 minus 4 is 41, and 25 plus 16 is 41.
Why: The two sides agree at x equals 5, so the solution is correct.
Answer: x = 5
Step zero
Before any algebra.
Discussion prompt
A question reads: if 7x minus 2 equals 33, what is the value of 7x plus 5? A student solves for x. Was that necessary, and what should they have noticed first?
Hint: Compare the target with the given expression.
Answer:
Not necessary. The target 7x plus 5 is the given expression 7x minus 2 with 7 added, so the answer is 33 plus 7, which is 40.
What to notice first: whenever the target shares the same variable term as the given expression, the whole equation can be shifted or scaled instead of solved.
The long route works too: 7x equals 35, so x equals 5, and 35 plus 5 is 40. It just takes three times as long and passes through 5, which is a wrong answer sitting in the choices.
The general habit: read the target before you start. It tells you whether to solve at all.
Worked example
If 5x plus 3 equals 28, what is the value of 5x minus 3?
Figure (svg): A number line marking x, five x, and five x minus three
Notice the shortcut: the target differs from the given expression by 6.
Why: 5x minus 3 is (5x plus 3) minus 6, so the answer is 28 minus 6.
So the value is 22, with no need to find x at all.
Why: Subtracting 6 from both sides of the original equation gives exactly the target.
The long route confirms it: 5x equals 25, so x equals 5, and 25 minus 3 is 22.
Why: Both routes agree, which is the point of checking.
Both 5 and 25 are correct values of other quantities, and both will be offered. Only 22 answers the question asked.
Verify: check the original at x equals 5: 25 plus 3 is 28.
Why: The equation holds, so 22 is the value of the requested expression.
Answer: 5x - 3 = 22
Worked example
How many solutions does 3(2x plus 4) equals 6x plus 7 have?
Figure (svg): A table comparing the coefficients and constants of the two sides
Expand the left side: 6x plus 12 equals 6x plus 7.
Why: Three times 2x is 6x and three times 4 is 12.
Subtract 6x from both sides: 12 equals 7.
Why: The variable terms are identical, so they cancel completely.
That statement is false for every x, so there is no solution.
Why: A false leftover means no value of the variable can satisfy the equation.
The variable vanishing is the answer arriving, not a sign that you have gone wrong. Read what is left and decide.
Verify: check by comparing coefficients: both sides have 6x, and the constants 12 and 7 differ.
Why: Same slope with different intercepts is precisely the no-solution condition.
Answer: no solution
Faded example
From memory. The order that keeps the algebra clean.
Fill in the blanks
Clear fractions first by multiplying every term. Then expand any brackets. Then gather the variable terms on one side. And if the variables cancel leaving something false, the equation has no solution.
Why: The order matters because each step makes the next one simpler. Clearing fractions first means no later step involves them; distributing before collecting means no like terms are combined across a bracket that has not been opened.
Worked example
For what value of c does 4(x minus 2) equals 4x plus c have infinitely many solutions?
Figure (svg): Bars contrasting the single value of c giving infinitely many with every other value
Expand the left: 4x minus 8 equals 4x plus c.
Why: Distributing the 4 makes both sides directly comparable.
Infinitely many solutions requires the two sides to be identical, so the constants must match.
Why: Identical sides give 0 equals 0 after cancelling, which is true for every x.
Therefore c equals negative 8.
Why: Matching negative 8 on the left with c on the right.
The mirror question is common too: for what value of c is there NO solution? The answer is every value except negative 8.
Verify: check c equals negative 8: both sides read 4x minus 8.
Why: The equation becomes an identity, so every value of x satisfies it.
Answer: c = negative 8
Fill the middle
Compare the two sides after expanding.
Fill in the blanks
If the coefficients of x differ, there is exactly one solution. If the coefficients match but the constants differ, there is no solution. If both match, there are infinitely many.
Why: This is the same three-case rule as the systems type, which is no accident: an equation with variables on both sides really is two linear expressions being compared, and comparing them is the same question as asking where two lines meet.
Estimation
A rough value catches an arithmetic slip.
Predict first
Roughly what is x if 19x plus 7 equals 200?
Correct: about 10
Why: Round 19 to 20 and 200 stays 200, so x is about 200 over 20, which is 10. The exact answer is 193 over 19, about 10.2. This estimate protects against the two common slips: forgetting to subtract the 7 before dividing, and dividing 200 by 7 instead. Both would produce numbers far from 10.
Check
Clear the fraction first, then the bracket.
Check your understanding
Solve: (x plus 6) over 3 equals x minus 2.
Answer: A
Why: Multiply both sides by 3: x plus 6 equals 3(x minus 2), which expands to 3x minus 6. Subtracting x gives 6 equals 2x minus 6, then adding 6 gives 12 equals 2x, so x equals 6. Checking in the original: (6 plus 6) over 3 is 4, and 6 minus 2 is 4.
Note that choice B is the value the two sides share. Recognising that a number you computed is not the number you were asked for is the whole discipline of this type.
Section
Section 4
Trap
The trap. The question asks for the value of 3x, you solve and find x equals 4, and 4 is among the choices.
You select it. The algebra was flawless; the answer was to a question nobody asked.
The test places the value of x among the choices on essentially every question of this shape, precisely because solving feels like finishing.
The fix. Read the final phrase before you begin and write the target at the top of your working.
Then, before selecting, compare your number against that written target rather than against the choices.
Trap
The trap. To clear the fraction in x over 5 minus 3 equals 4, you multiply the fraction and the right-hand side by 5 but leave the 3 alone, writing x minus 3 equals 20.
That gives x equals 23. The correct equation is x minus 15 equals 20, giving x equals 35.
Nothing later will look wrong, because the equation you are now solving is perfectly consistent — it is simply a different equation.
The fix. Multiplying an equation means multiplying every term on both sides, without exception.
Say the multiplier aloud as you apply it to each term in turn.
Error analysis
A student clearing a fraction. One term was missed.
Annotate
On: \( \frac{x}{4} + 6 = 10 \;\xrightarrow{\times 4}\; x + 6 = 40 \)
Any operation applied to an equation must reach every term. A half-applied operation gives you a clean answer to a different question.
Trap
The trap. You subtract and both x terms disappear, leaving 12 equals 7. You assume you have made a mistake and start the question again.
You have not. A false leftover means the equation has no solution, which is one of the answer choices.
The same applies in reverse: a leftover such as 0 equals 0 means infinitely many solutions.
The fix. Treat a vanished variable as an answer arriving in an unfamiliar form.
Read what is left. False means none; always true means infinitely many.
Elimination
The equation is 6(x plus 2) equals 6x plus 12.
Eliminate the wrong options
How many solutions does it have? Three choices can be ruled out by expanding one side.
Survives elimination: b
Why: Expanding the left gives 6x plus 12, which is exactly the right-hand side. The equation is an identity, true for every value of x, so there are infinitely many solutions. Note that choice D can be eliminated on principle before looking at this equation at all, which is worth remembering as a free elimination on every linear count question.
Trap
The trap. You expand 12 minus 3(x minus 2) as 12 minus 3x minus 6.
The negative 3 must multiply both terms, and negative 3 times negative 2 is positive 6. The correct expansion is 12 minus 3x plus 6.
The two versions differ by 12, and both simplify to clean-looking expressions.
The fix. Write both products explicitly before collecting anything.
Treat the sign in front of the bracket as part of the multiplier, and apply it to every term inside.
Counterexample
A rule stated slightly too strongly.
Discussion prompt
A student says: every linear equation has exactly one solution. Give two counterexamples of different kinds.
Hint: What happens when both sides have the same coefficient of x?
Answer:
Counterexample 1: 2x plus 1 equals 2x plus 5. Subtracting 2x leaves 1 equals 5, which is false. This equation has no solution.
Counterexample 2: 2x plus 1 equals 2x plus 1. Subtracting 2x leaves 0 equals 0, which is always true. This equation has infinitely many solutions.
The mechanism: an equation with the same coefficient of x on both sides is really comparing two parallel lines. If the constants differ the lines never meet; if they agree the lines coincide.
The corrected statement: a linear equation in one variable has exactly one solution provided the coefficients of x on the two sides differ.
This is exactly the same three-case structure as a system of two linear equations, because it is the same geometry seen from a different angle.
Edge cases
Reading the target is the habit. When does it change what you do?
Discussion prompt
When can you answer without solving for x at all, and when must you solve first?
Hint: Compare the target expression with the one in the equation.
Answer:
You can skip solving when the target is a shift or a scaling of the given expression. If 3x minus 7 equals 14 and the target is 6x minus 14, double the equation. If the target is 3x minus 2, add 5 to both sides.
You must solve when the target involves the variable differently — for example if the target is x squared, or x over 2 when the equation involves 3x, and the numbers do not divide cleanly.
A useful test: can you get from the given expression to the target by multiplying by a constant and adding a constant? If yes, do that to the whole equation. If no, solve for x.
The edge case worth knowing: even when the shortcut exists, solving for x is never wrong — only slower. So when unsure, solve. The habit that matters is reading the target, not finding the fastest route to it.
Check
Expand both sides and compare.
Check your understanding
For what value of a does 5(2x plus a) equals 10x minus 15 have infinitely many solutions?
Answer: A
Why: Expanding the left gives 10x plus 5a. For infinitely many solutions the two sides must be identical, so 5a must equal negative 15, giving a equals negative 3. Checking: both sides become 10x minus 15, so every value of x works.
The observation that made this quick came before any algebra: the coefficients of x are both 10 whatever a is, so only the constants could decide the count.
Section
Section 5
Matching
Six equations, six results. Expand where you need to.
Match the pairs
Why: The last row is the one that catches people. An answer of x equals 0 is a perfectly ordinary single solution, and it is completely different from an equation having no solution. The words zero and none are not interchangeable here, and the SAT offers both.
Sorting
Each is what remains after the variables have been dealt with.
Sort into buckets
What does each outcome mean?
Four of these six mean stop and answer immediately. Learning to read them saves the minute students spend re-checking work that had already finished.
Comparison
Fill the blanks from memory.
Comparison matrix
| case | what the two sides look like | the leftover |
|---|---|---|
| One solution | different coefficients of x | a value for x |
| No solution | same coefficient, different constants | a false statement such as 5 equals 8 |
| Infinitely many | the two sides are identical | a true statement such as 0 equals 0 |
Every linear equation lands in exactly one of these three rows. There is no fourth case, which is why exactly two solutions can always be eliminated on sight.
Trade off
Fill in when each route wins.
Comparison matrix
| route | when it is faster | the risk |
|---|---|---|
| Scale the whole equation | the target is a multiple of the given expression | none, if the multiple is exact |
| Shift the whole equation | the target differs by a constant | adding to one side only |
| Solve for x, then substitute | always works | passing through x, which is a wrong answer choice |
| Stop at x | only when x is what was asked | the most common error on this type |
The bottom row is not a method — it is the trap, listed alongside the methods so it is visible as a choice rather than an accident.
Real world
One minute on why the simplest type still matters.
Discussion prompt
Solving a linear equation is the most-used piece of algebra there is. Where does the answer-the-right-quantity problem show up outside a maths test?
Answer:
Anywhere a calculation has intermediate results. Working out a total cost gives you a subtotal, a tax amount and a final figure, and quoting the wrong one is the same error in a different setting.
In unit conversion, where you compute a rate on the way to a quantity and it is easy to report the rate.
In any spreadsheet, where a helper column holds a number you needed but did not want.
The general skill is holding onto the question while doing the work, which is harder than it sounds precisely because the work absorbs your attention.
That is why the fix is written rather than mental: putting the target at the top of your working means you do not have to remember it.
Ranking
Solve each, then order the values of x.
Put in order
Why: Solving each gives: (c) x equals 1, (a) 2x equals 4 so x equals 2, (b) x equals 3, and (d) x equals 10. Ordered smallest to largest that is 1, 2, 3, 10. The exercise is deliberately routine — the point is speed and accuracy on the basic solve, because on the real test this arithmetic has to happen almost without attention so that your attention can go to reading the question.
Concept
This type is 6.0 per cent of the section and the algebra is already within reach. Almost all of the available gain is in reading and in the two count variants.
| session | what you do | why |
|---|---|---|
| 1 | Twenty solves with fractions and brackets, timed, aiming for under thirty seconds each. | The algebra must become automatic so attention is free for the question. |
| 2 | Fifteen questions asking for an expression, writing the target at the top before starting. | This is where the marks actually leak, and the fix is a written habit. |
| 3 | Fifteen count questions, including finding a constant for none or infinitely many. | The variant students least expect, and it is entirely rule-based. |
| 4 | Mixed set under time, substituting every answer back into the original equation. | Builds the check that catches partial multiplication and sign errors. |
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Explain it to yourself
Close the deck.
Discussion prompt
Without looking, state the three possible outcomes for a linear equation in one variable, the condition that produces each, and what you see when you solve.
Hint: Compare the coefficients first, then the constants.
Answer:
One solution: the coefficients of x on the two sides differ. Solving leaves a value for x.
No solution: the coefficients match but the constants differ. The variables cancel and leave a false statement.
Infinitely many: the two sides are identical once expanded. The variables cancel and leave something always true.
There is no fourth case, and in particular a linear equation can never have exactly two solutions.
If you also said that an answer of x equals 0 belongs in the first case and not the second, you have the distinction the test most often exploits.
Explain it
Two minutes, out loud.
Discussion prompt
A friend solves every equation correctly and still drops marks on this type. What is happening, and what single change do you give them?
Answer:
Diagnose it: they are answering the value of x when the question asked for an expression. More algebra practice will not help, because the algebra is already right.
Explain why it happens: solving genuinely finishes the mathematical work, so stopping there feels like completing the question. The test relies on that feeling.
Give the change as a physical action: write the target quantity at the top of the working, before the first line of algebra.
Then the closing check: compare your answer to what you wrote, not to the answer choices — because the value of x will be among the choices.
Add the bonus: reading the target first sometimes shows the target is a multiple or a shift of the given expression, in which case you never need to find x at all.
Commit first
Commit before you check.
Predict first
How many solutions does 4(x minus 1) equals 4x minus 4 have?
Correct: infinitely many
Why: Expanding the left gives 4x minus 4, which is identical to the right-hand side. Subtracting gives 0 equals 0, true for every x, so every value is a solution. The near-miss to watch for is 4(x minus 1) equals 4x minus 5, where the constants differ and there would be no solution instead. One digit separates the two answers.
Connect it up
Blank paper.
Draw it
Draw a flow for solving a linear equation. Start with a box labelled READ THE TARGET and write the three things it might be: x, an expression in x, or a solution count. Then the solving chain: clear fractions by multiplying every term, expand brackets watching the sign, gather variables on one side, divide. At the end, draw the chain splitting three ways: a value for x, a false leftover meaning no solution, and a true leftover meaning infinitely many. Close the loop by drawing an arrow from the end back to the READ THE TARGET box, labelled answer that quantity.
Exit ticket
One question before you close the deck.
Predict first
You solve an equation, find x equals 6, and see 6 among the answer choices. What do you do?
Correct: Re-read the stem to check it asked for x
Why: The presence of your number among the choices is not evidence that it is the answer — the value of x is deliberately included on questions that ask for an expression. Substituting back checks your algebra, which was probably fine, and does nothing about answering the wrong quantity. Re-reading the stem is the only action that addresses the actual risk, and it takes about three seconds.
Recap
One type, one discipline: the algebra is easy, so spend the saved attention on the question.
| never do this | do this instead |
|---|---|
| Select x because it appears among the choices | Re-read the stem and compare against your written target |
| Multiply only the fraction when clearing it | Multiply every term on both sides |
| Restart when the variable disappears | Read the leftover: false means none, true means infinitely many |
| Expand a leading minus onto the first term only | Change the sign of every term in the bracket |
| Confuse x equals 0 with no solution | Zero is an answer; none means no answer exists |
| Solve for x when the target is twice the given expression | Double the whole equation instead |
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