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a) \( |4 x-3|+\sqrt{144}=(-3+7)^{2} \) b) \( |9+x|-3 \cdot(-9+7)=(7-10)^{2} \) c) \( 8+\left|2 x+2^{3}\right|=(-6+9) \cdot(-5+8) \) d) \( -3 \cdot|x+2|-\sqrt{25}=8 \)

Ask by Clark Schneider. in Argentina
Dec 14,2024

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a) \( x = \frac{7}{4}, -\frac{1}{4} \) b) \( x = -6, -12 \) c) \( x = -\frac{7}{2}, -\frac{9}{2} \) d) No hay soluciones.

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To solve these equations, start by simplifying the non-variable parts. In equation (a), we compute \( \sqrt{144} = 12 \) and \( (-3 + 7)^2 = 16 \), leading to \( |4x - 3| + 12 = 16 \). Next, isolate the absolute value: \( |4x - 3| = 4 \). This gives two cases: \( 4x - 3 = 4 \) or \( 4x - 3 = -4 \). For part (b), simplify \( (-9 + 7) = -2 \), so \( -3 \cdot (-2) = 6 \), and \( (7 - 10)^2 = 9 \). Thus, the equation becomes \( |9 + x| - 6 = 9 \), leading to \( |9 + x| = 15 \) which also results in two cases. For (c), you find \( (-6 + 9) \cdot (-5 + 8) = 3 \cdot 3 = 9 \), therefore \( 8 + |2x + 8| = 9 \). This simplifies to \( |2x + 8| = 1 \), producing two cases. In (d), simplify \( -3 \cdot |x + 2| - 5 = 8 \), resulting in \( -3 |x + 2| = 13 \). Here, you will find that a solution does not exist because the absolute value cannot yield a negative multiple. Keep your algebra tidy, and always remember that absolute values can create more than one scenario! It's like those puzzle pieces; often you find multiple ways to fit things together.

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