Literal Equations and Formulas: Practice

Reduce the equation first.

$kx - 2x = k - 3 \;\Rightarrow\; (k - 2)x = k - 3$

No solution needs the coefficient zero and the constant nonzero. At $k = 2$ the coefficient is $0$ and the constant is $k - 3 = -1 \neq 0$, giving $0 \cdot x = -1$.

Reduce the equation.

$kx - 2x = 3k - 6 \;\Rightarrow\; (k - 2)x = 3(k - 2)$

Both the coefficient $k - 2$ and the constant $3(k - 2)$ vanish at $k = 2$, giving $0 \cdot x = 0$: every real number.

The reduced form is $(k - 2)x = 3(k - 2)$. No solution would need the coefficient zero and the constant nonzero.

$k - 2 = 0 \;\Rightarrow\; 3(k - 2) = 0 \text{ too}$

The moment the coefficient vanishes the constant vanishes with it, so "no solution" never happens.

The coefficient $k - 2$ is identical in both, so it cannot be the difference.

$\text{(i) constant} = -1 \neq 0, \qquad \text{(ii) constant} = 0$

At $k = 2$ the first becomes $0 \cdot x = -1$ (no solution) and the second becomes $0 \cdot x = 0$ (every real number). The right-hand constant decides.

The isolating step divides by $a - c$, so that difference is what must be nonzero.

$a - c \neq 0 \;\Rightarrow\; a \neq c$

For example $a = 0$, $c = 3$ is fine ($a - c = -3$), while $a = 5$, $c = 5$ fails even though neither is zero.

Isolate $\frac{1}{v}$, combine over a common denominator, then take reciprocals.

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{fu} \;\Rightarrow\; v = \frac{fu}{u - f}$

The divisor is $u - f$, so it must be nonzero.

$u - f \neq 0 \;\Rightarrow\; u \neq f$

Physically, an object at the focal length ($u = f$) forms no finite image.

Clear the denominator, gather the $R_2$ terms, factor, and divide.

$R\,R_1 = R_2(R_1 - R) \;\Rightarrow\; R_2 = \frac{R R_1}{R_1 - R}$

Expand and reduce.

$kx - k = 4x - k \;\Rightarrow\; (k - 4)x = 0$

For $k \neq 4$ the only solution is $x = 0$. At $k = 4$ the equation is $0 \cdot x = 0$, solved by every real number, so uniqueness fails there.

The equation reduces to $(k - 2)x = 3(k - 2)$. For $k \neq 2$ cancel the common factor.

$x = \frac{3(k - 2)}{k - 2} = 3$

The solution is $x = 3$ for every $k \neq 2$.

The equation reduces to $(k - 2)x = k - 3$. For $k \neq 2$ divide by the coefficient.

$x = \frac{k - 3}{k - 2}$

Unlike the companion equation, this value genuinely depends on $k$.

Factor the target $P$ out of both terms, then divide by the whole factor.

$A = P(1 + rt) \;\Rightarrow\; P = \frac{A}{1 + rt}, \quad 1 + rt \neq 0$

The factor pulled out is $1 + rt$, not $rt$, so the condition is $1 + rt \neq 0$.