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Solve for x. Assume the equation has a solution for x. {:[d*(-3+x)=kx+9],[x=◻]:}

Solve for x x .\newlineAssume the equation has a solution for x x .\newlined⋅(−3+x)=kx+9x=□ \begin{array}{l} d \cdot(-3+x)=k x+9 \\ x=\square \end{array}

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Full solution

Q. Solve for x x .\newlineAssume the equation has a solution for x x .\newlined⋅(−3+x)=kx+9x=□ \begin{array}{l} d \cdot(-3+x)=k x+9 \\ x=\square \end{array}
  1. Simplify equation: First, let's simplify the first equation in the system: d∗(−3+x)=kx+9d*(-3+x) = kx + 9. We distribute dd across the terms inside the parentheses: −3d+dx=kx+9-3d + dx = kx + 9.
  2. Isolate xx terms: Next, we want to isolate xx terms on one side of the equation. To do this, we can add 3d3d to both sides: dx=kx+9+3ddx = kx + 9 + 3d.
  3. Subtract kxkx: Now, we need to get all the xx terms on one side by subtracting kxkx from both sides: dx−kx=9+3ddx - kx = 9 + 3d.
  4. Factor out xx: We can factor out xx from the left side of the equation: x(d−k)=9+3dx(d - k) = 9 + 3d.
  5. Solve for x: To solve for x, we divide both sides of the equation by (d−k)(d - k), assuming d−k≠0d - k \neq 0: x=9+3dd−kx = \frac{9 + 3d}{d - k}.
  6. Final solution: Since the second part of the system of equations indicates that xx equals a blank square, we can assume that the blank square is the solution we found: x=9+3dd−kx = \frac{9 + 3d}{d - k}.

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