Question bank — Linear equations in one variable — solving, transposition method

False. "Linear" means the highest power of the variable is $1$; here $x$ is squared, so it is a quadratic. One variable, yes — but not linear.

No. A linear equation needs $a\ne 0$; with $a=0$ the variable vanishes and there is no $x$ to solve for — it is just the true statement $4=4$.

True. The "flip the sign" shortcut is exactly the same-thing-to-both-sides rule in disguise; the balance is never broken.

True. Multiplying both sides of the first by $4$ gives $x=8$; multiplying is a reversible (invertible) operation, so no solutions are gained or lost.

False in spirit. It gives the true-but-useless $0=0$, destroying the equation. Only reversible operations (multiply/divide by a non-zero number, add/subtract) preserve the solution.

False. Transposing gives $-3=2$, a contradiction, so there is no solution. Equal coefficients but unequal constants means the two "lines" are parallel — see 3.2.04-Graphical-solution-of-linear-equations.

False. Both sides are identical for every $x$, so there are infinitely many solutions — it is an identity, not an equation to pin one value down.

False. Dividing by the variable is forbidden: if $x=0$ you'd divide by zero, and even otherwise it turns an easy problem into a messier one. Divide by the known coefficient $6$ instead.

The $+3$ was moved but its sign not flipped. Crossing the equals turns $+3$ into $-3$: $5x=13-3=10$, so $x=2$.

Only the left denominator was removed; the right side must be multiplied by $2$ too. Correct: $2x+4=10$, giving $x=3$.

The coefficient is $-2$, not $2$. Dividing by $-2$: $x=\frac{-8}{-2}=4$. Negative divided by negative is positive.

The constant $-3$ was moved but not flipped. Crossing turns $-3$ into $+3$: $3x=9+3=12$, so $x=4$.

Multiplying by $3$ must hit every term, including the $+2$: $x+6=15$. Better: transpose the $+2$ first, get $\frac{x}{3}=3$, then $x=9$.

The $3$ was distributed over $x$ only, not over $-2$. Distributing fully: $3x-6=9$, so $x=5$.

$-x$ means $-1\cdot x$; multiplying (or dividing) both sides by $-1$ flips the sign: $x=-5$.

Because solving needs dividing by $a$ ($x=\frac{c-b}{a}$), and division by zero is undefined. Also, with $a=0$ the term $ax$ disappears and there is no variable left to solve for.

The equation was built like PEMDAS — multiply first, then add. To peel it apart we reverse that order: strip the added constant first, then the multiplying coefficient, just like removing socks before shoes... the last thing on comes off first.

Because it is literally applying the inverse operation to both sides at once. Adding the opposite quantity to each side keeps the balance-scale level, and the term simply reappears on the other side with its sign reversed.

Its graph $y=ax+b$ is a straight line; a line meets a horizontal target height at most once (unless it lies on top of it, giving infinitely many). So the count is always zero, one, or infinite — never two.

If the variable could be $0$, that value is silently discarded (you can't divide by $0$), so a genuine solution may vanish. Transposing instead keeps every possibility on the table.

Once a real-world sentence is turned into $ax+b=cx+d$, the algebra is blind to its origin — combine, separate, transform, obtain. The story only decides what the numbers are.

Every value of $x$ works — the equation is an identity (infinitely many solutions). Graphically the two lines are the same line.

No value of $x$ works — the equation is a contradiction (no solution). The two lines are parallel and never meet.

The denominator $a-c$ becomes $0$, and dividing by zero is undefined — exactly the borderline where the equation switches to either an identity ($b=d$) or a contradiction ($b\ne d$).

Every real number is a solution, because $0=0$ holds no matter what $x$ is — infinitely many.

None. The left side is always $0$, which can never equal $5$, so no $x$ satisfies it — a contradiction.

With two unknowns and one equation there are infinitely many pairs $(x,y)$ that fit; you've stepped into 2.1.08-Linear-equations-in-two-variables, a different game needing a second equation.

When you balance a chemical equation you adjust coefficients so atoms match on both sides — the same "keep both sides equal" principle, see 2.5.02-Balancing-chemical-equations. Replacing $=$ with $<$ or $>$ gives 2.3.01-Linear-inequalities, where one extra sign-flip rule appears.