Mistake Master
Solving linear equations
A linear equation puts the variable to the first power only: no squares, no roots, no variables multiplied together. Solving one means getting that variable alone while keeping both sides equal. The method is short and it is the same every time. Almost every point lost here is a slip inside that method, or answering one half-step before the question does.
§1
What this topic is about
▸
Two moves carry the whole topic: keep the equation balanced, and isolate the variable by undoing what was done to it, one operation at a time. Everything from Topic 1.2 on leans on doing these cleanly under time.
§2
Balance: the one rule
▸
An equation is a claim that two sides are equal. You can add, subtract, multiply, or divide, but whatever you do to one side you do to the other, or the claim stops being true.
Take $2x+5=11$. The goal is to peel the $5$ and the $2$ off the $x$. Subtract $5$ from both sides, then divide both sides by $2$. Do the same thing to both sides at every step and the equation stays true the whole way down.
- Undo addition with subtraction, and multiplication with division.
- Work from the outside in: clear what is added or subtracted first, then deal with what multiplies the variable.
- One operation per step keeps the signs visible.
§3
Distribute and combine
▸
Before you can isolate $x$, the side it sits on often needs cleaning up: a factor multiplied across a parenthesis, and like terms gathered.
- A factor outside parentheses multiplies every term inside, so $3(x+2)$ is $3x+6$, not $3x+2$.
- A negative in front flips the sign of every term it reaches, so $-(x-4)$ is $-x+4$, not $-x-4$.
- Only terms with the same variable part combine. $3x$ and $4$ stay apart, so $3x+4+2x$ is $5x+4$, not $9x$.
Worked example. Solve $3(x+2)=2x+11$.
The $3$ multiplies both terms inside: $$3(x+2)=3x+6,$$ so the equation reads $3x+6=2x+11$.
Subtract $2x$ from both sides: $$x+6=11.$$
Subtract $6$ from both sides: $$x=5.$$ Multiplying only the first term, or keeping the sign on the $-4$ in a problem like $-(x-4)$, are the two ways this cleanup step goes wrong.
§4
Isolate the variable
▸
Once the sides are clean, isolating $x$ is a sequence of inverse operations applied to both sides. The frequent slip is doing the operation to one side only, or to one term instead of the whole side.
Worked example. Solve $x/3 + 2 = 7$.
Take $2$ off both sides: $$x/3 = 5.$$
Multiply both sides by $3$: $$x = 15.$$ Stopping at $x/3=5$, or multiplying by $3$ before subtracting the $2$, are the two ways this lands wrong.
-
Do every operation to the whole side.Dividing $2x+5$ by $2$ gives $x + 5/2$, not $x+5$. The division reaches all of it, both terms.
-
Undo in the right order.Clear what is added or subtracted first, then divide by the coefficient. Reversing that order forces fractions and invites slips.
§5
Answer the question that was asked
▸
This is the pattern that catches strong solvers. You do the algebra perfectly, land on $x$, and report it, when the item asked for something built from $x$: $2x$, $x+3$, or the value of an expression.
The SAT writes a wrong answer choice equal to $x$ on purpose, so the correct-but-early stop lands on a real option. The fix is mechanical: before you bubble or type, reread the last line of the question and check that your number answers that.
Worked example. If $5x-3=2x+9$, what is the value of $2x$?
$5x-3=2x+9$, so $3x=12$, so $x=4$.
The question wants $2x$, not $x$. Double it: $2x=8$. Why $8$ and not $4$? Because $4$ answers a question one step earlier than the one printed.
Do the same thing to both sides, distribute to every term, combine only like terms, and report the exact thing asked.
§6
Five patterns that cost real points
▸
Five patterns recur on this topic. They are the same ones the diagnostic routes on.
The factor reaches only the first term.
$3(x+2)$ becomes $3x+2$. The number out front multiplies every term inside the parentheses, not just the one it sits next to.
Fix. Multiply the outside factor onto each term in turn: $3(x+2)=3x+6$. Say "and" between the terms so the second one never gets skipped.
A leading minus flips one term, not all of them.
$-(x-4)$ becomes $-x-4$. A negative in front changes the sign of every term it reaches, so the $-4$ turns into $+4$ and the line is $-x+4$.
Fix. Read $-(x-4)$ as $-1$ times each term: $-x+4$. The same care applies when a subtraction sits in front of parentheses, as in $5-2(x-3)$.
Unlike terms get merged.
$3x+4$ collapses to $7x$, or the $2$ in $2x$ gets treated as a loose constant. A number attached to the variable and a number on its own are different kinds of thing.
Fix. Combine only terms that share the exact variable part. $3x+4+2x=5x+4$: the $x$-terms join, the $4$ rides along untouched.
One side moves, or the order slips.
An operation lands on one side only, or on a single term instead of the whole side, and the equation quietly stops being true. A close cousin: dividing out the coefficient before clearing the added number, which forces fractions you did not need.
Fix. Apply each step to both whole sides. Clear what is added or subtracted first, then divide by the coefficient.
You solve for $x$ and report it, but the item asked for more.
The algebra is right, you land on $x$, and you commit it, when the question wanted $2x$, or $x+3$, or some expression built from $x$. The test writes a wrong choice equal to $x$ so the early stop lands on a real option.
Fix. Before you commit, reread the final line of the question and confirm your number answers that one, not the step before it.
Ten quick checks across the five patterns: distributing, signs, combining like terms, isolating, and answering the exact question. Pick or type your answer, then check. Progress is saved.