Literal Equations and Formulas: Practice

Collect the $x$ terms on the left and the constants on the right.

$kx - 3x = 1 - 5 \;\Rightarrow\; (k - 3)x = -4$

The coefficient is $k - 3$ and the constant is $-4$.

There is no solution exactly when the coefficient is zero while the constant is not.

$k - 3 = 0 \;\Rightarrow\; k = 3, \qquad \text{constant } -4 \neq 0$

At $k = 3$ the equation is $0 \cdot x = -4$, so there is no solution.

Infinitely many solutions need the coefficient and the constant to vanish together. The constant is already $0$.

$k - 3 = 0 \;\Rightarrow\; k = 3$

At $k = 3$ the equation is $0 \cdot x = 0$, true for every $x$.

Subtract $b$ from both sides, then divide by $m$.

$y - b = mx \;\Rightarrow\; x = \frac{y - b}{m}$

Keep the whole numerator $y - b$ over $m$; dividing only $y$ gives the wrong $\frac{y}{m} - b$.

Subtract $2l$ from both sides, then divide the whole right side by $2$.

$P - 2l = 2w \;\Rightarrow\; w = \frac{P - 2l}{2}$

This equals $\frac{P}{2} - l$, not $\frac{P}{2} - 2l$.

Subtract $b$, then divide by $a$.

$ax = c - b \;\Rightarrow\; x = \frac{c - b}{a}$

Dividing before subtracting gives the wrong $\frac{c}{a} - b$.

You divided by the factor $a - c$, so that factor must be nonzero.

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

The condition is on the difference, not on $a$ and $c$ separately.

Collect the $x$ terms and the constants.

$kx - 2x = 6 + 4 \;\Rightarrow\; (k - 2)x = 10$

The equation reduces to $(k - 2)x = 10$. There is no solution when the coefficient is zero but the constant is not.

$k - 2 = 0 \;\Rightarrow\; k = 2, \qquad 0 \cdot x = 10$

So $k = 2$ gives no solution.

Every step multiplied by the constant $\frac{9}{5}$ or added $32$; nothing was divided by an expression that can vanish.

$F = \frac{9}{5}C + 32 \text{ has no excluded values}$

A condition appears only when you divide by something that can be zero, and that never happened here.

Subtract $7$ from both sides to reduce the equation.

$kx = 0$

When $k = 0$ this is $0 \cdot x = 0$, solved by every real number. For $k \neq 0$ the only solution is $x = 0$.

The left side is $0$ for every $x$, so it can never equal $6$.

$0 \cdot x = 6 \;\Rightarrow\; 0 = 6 \text{ (false)}$

The solution set is $\varnothing$: no solution.