Subtract the following binomials. (Both by vertical and horizontal arrangement) \( \begin{array}{ll}\text { a) } 2 x+3 y \text { from } 3 x+5 y & \text { b) } 5 x-4 y \text { from } 2 x+5 y \\ \text { c) } 8 x y+2 x \text { from } 5 x y-2 x & \text { d) } 4 x-5 z \text { from } 6 x-4 z \\ \text { e) } 15 x^{2} y^{3}-3 x y^{2} \text { from }-5 x^{2} y^{3}-3 x y^{2} & \text { d) } 5+2 m n \text { from } 6-4 \text { mn }\end{array} \)
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Let's solve the problems step by step, using both vertical and horizontal arrangements. **a)** **Vertical Arrangement:** \[ \begin{array}{rl} & 3x + 5y \\ - & (2x + 3y) \\ \hline & 3x + 5y \\ & -2x - 3y \\ \hline & (3x - 2x) + (5y - 3y) \\ & = 1x + 2y \\ \end{array} \] **Horizontal Arrangement:** \( (3x + 5y) - (2x + 3y) = (3x - 2x) + (5y - 3y) = x + 2y \) --- **b)** **Vertical Arrangement:** \[ \begin{array}{rl} & 2x + 5y \\ - & (5x - 4y) \\ \hline & 2x + 5y \\ & -5x + 4y \\ \hline & (2x - 5x) + (5y + 4y) \\ & = -3x + 9y \\ \end{array} \] **Horizontal Arrangement:** \( (2x + 5y) - (5x - 4y) = (2x - 5x) + (5y + 4y) = -3x + 9y \) --- **c)** **Vertical Arrangement:** \[ \begin{array}{rl} & 5xy - 2x \\ - & (8xy + 2x) \\ \hline & 5xy - 2x \\ & -8xy - 2x \\ \hline & (5xy - 8xy) + (-2x - 2x) \\ & = -3xy - 4x \\ \end{array} \] **Horizontal Arrangement:** \( (5xy - 2x) - (8xy + 2x) = (5xy - 8xy) + (-2x - 2x) = -3xy - 4x \) --- **d)** **Vertical Arrangement:** \[ \begin{array}{rl} & 6x - 4z \\ - & (4x - 5z) \\ \hline & 6x - 4z \\ & -4x + 5z \\ \hline & (6x - 4x) + (-4z + 5z) \\ & = 2x + 1z \\ \end{array} \] **Horizontal Arrangement:** \( (6x - 4z) - (4x - 5z) = (6x - 4x) + (-4z + 5z) = 2x + z \) --- **e)** **Vertical Arrangement:** \[ \begin{array}{rl} & -5x^2y^3 - 3xy^2 \\ - & (15x^2y^3 - 3xy^2) \\ \hline & -5x^2y^3 - 3xy^2 \\ & -15x^2y^3 + 3xy^2 \\ \hline & (-5x^2y^3 - 15x^2y^3) + (-3xy^2 + 3xy^2) \\ & = -20x^2y^3 + 0 \\ \end{array} \] **Horizontal Arrangement:** \( (-5x^2y^3 - 3xy^2) - (15x^2y^3 - 3xy^2) = (-5x^2y^3 - 15x^2y^3) + (-3xy^2 + 3xy^2) = -20x^2y^3 \) --- **f)** **Vertical Arrangement:** \[ \begin{array}{rl} & 6 - 4mn \\ - & (5 + 2mn) \\ \hline & 6 - 4mn \\ & -5 - 2mn \\ \hline & (6 - 5) + (-4mn - 2mn) \\ & = 1 - 6mn \\ \end
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