Question 226·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Choose $k=3$ because it is positive and makes $3/k=1$. Type $k=3$, then $3/k\sim(x_1+2)/(2x_1-1)$. Here $x_1$ is the $x$ Desmos finds, and $\sim$ tells Desmos to fit the two sides. Under PARAMETERS, it shows $x_1=3$. This is a spot-check, not a formula for every $k$.

Since $k$ is positive, it is not $0$; $2x-1$ also cannot be $0$ because it is a denominator in the original equation. Multiply both sides by $k(2x-1)$ to cancel the fractions: $3(2x-1)=k(x+2)$.

Multiply each term inside its parentheses by the number outside. Distribute $3$ and $k$: $6x-3=kx+2k$.

Keep $k$ fixed and put the $x$-terms on one side. Subtract $kx$ from both sides: $6x-kx-3=2k$. Add $3$ to both sides: $6x-kx=2k+3$. The $-3$ becomes $+3$ because you add $3$ to remove it.

Both terms on the left contain $x$. Use the distributive property backward to factor it out: $x(6-k)=2k+3$. The factor is $6-k$ because multiplying back gives $6x-kx$, not $kx-6x$.

The problem says $k$ and $x$ are positive, so $2k+3>0$. Since $x(6-k)=2k+3$ and $x>0$, the factor $6-k$ must be positive too. In particular, $6-k$ is not $0$, so you can divide by it.

Divide both sides by $6-k$: $x=\frac{2k+3}{6-k}$. For the spot-check, add $(2k+3)/(6-k)$ in Desmos. It prints $3$ when $k=3$, matching the original equation. This formula expresses $x$ in terms of $k$. Choice D.