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Realizamos las restas algebraicas: 1) Restar \( -3 y^{2}+4 y-5 \) de \( -13 y^{2}-2 y+9 \)

Ask by Collins Salazar. in Bolivia
Mar 10,2025

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Answer

La resta es \( -10y^{2} - 6y + 14 \).

Solution

1. Planteamos la resta que se nos pide: \[ (-13y^2-2y+9) - (-3y^2+4y-5) \] 2. Distribuimos el signo negativo al segundo paréntesis: \[ -13y^2-2y+9 + 3y^2 - 4y + 5 \] 3. Combinamos términos semejantes: - Términos con \( y^2 \): \[ -13y^2 + 3y^2 = -10y^2 \] - Términos con \( y \): \[ -2y - 4y = -6y \] - Términos constantes: \[ 9 + 5 = 14 \] 4. Escribimos el resultado final: \[ -10y^2 - 6y + 14 \]

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The Deep Dive

Para restar \( -3 y^{2}+4 y-5 \) de \( -13 y^{2}-2 y+9 \), primero puedes cambiar el signo del primer polinomio y luego sumarlos. Esto se traduce a: \[ (-13 y^{2}-2 y+9) - (-3 y^{2}+4 y-5) = -13 y^{2}-2 y+9 + 3 y^{2} - 4 y + 5 \] Ahora combinamos términos semejantes: \[ (-13 y^{2} + 3 y^{2}) + (-2 y - 4 y) + (9 + 5) = -10 y^{2} - 6 y + 14 \] Por lo tanto, la respuesta final es \( -10 y^{2} - 6 y + 14 \).

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