Solve for \( x \) : \( \begin{array}{ll}\text { (1) } 7+9 x=7 x-3 & \text { (2) }-9+2 x=6-x \\ \text { (3) }-6(1-x)=10-2 x & \text { (4) } 9(x-3)=5(x-1)-14 \\ \text { (5) } \frac{2}{3}(x+2)-\frac{x+1}{5}=3 & \text { (6) } \frac{x-1}{2}-\frac{x+6}{3}=-2 \\ \text { (7) } x(x-2)+1=(x-2)^{2} & \text { (8) } \\ (x+2)(x-2)=(x+3)^{2}\end{array} \)
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Alright, let’s dive into these equations and find the value of \( x \) for each one! 1. For the equation \( 7 + 9x = 7x - 3 \): - Rearranging gives \( 9x - 7x = -3 - 7 \) -> \( 2x = -10 \) -> \( x = -5 \). 2. For \( -9 + 2x = 6 - x \): - Rearranging gives \( 2x + x = 6 + 9 \) -> \( 3x = 15 \) -> \( x = 5 \). 3. For \( -6(1 - x) = 10 - 2x \): - Distributing gives us \( -6 + 6x = 10 - 2x \) -> \( 6x + 2x = 10 + 6 \) -> \( 8x = 16 \) -> \( x = 2 \). 4. For \( 9(x - 3) = 5(x - 1) - 14 \): - Expanding yields \( 9x - 27 = 5x - 5 - 14 \) -> \( 9x - 27 = 5x - 19 \) -> \( 9x - 5x = -19 + 27 \) -> \( 4x = 8 \) -> \( x = 2 \). 5. For \( \frac{2}{3}(x + 2) - \frac{x + 1}{5} = 3 \): - Multiply through by 15 (LCM of 3 and 5) to eliminate fractions: \( 10(x + 2) - 3(x + 1) = 45 \). - This simplifies to \( 10x + 20 - 3x - 3 = 45 \) -> \( 7x + 17 = 45 \) -> \( 7x = 28 \) -> \( x = 4 \). 6. For \( \frac{x - 1}{2} - \frac{x + 6}{3} = -2 \): - Multiply through by 6 (LCM of 2 and 3): \( 3(x - 1) - 2(x + 6) = -12 \). - This simplifies to \( 3x - 3 - 2x - 12 = -12 \) -> \( x - 15 = -12 \) -> \( x = 3 \). 7. For \( x(x - 2) + 1 = (x - 2)^{2} \): - Expanding gives us \( x^2 - 2x + 1 = x^2 - 4x + 4 \) -> \( -2x + 1 = -4x + 4 \). - Rearranging gives \( 4x - 2x = 4 - 1 \) -> \( 2x = 3 \) -> \( x = \frac{3}{2} \). 8. For \( (x + 2)(x - 2) = (x + 3)^{2} \): - Expanding leads us to \( x^2 - 4 = x^2 + 6x + 9 \) -> \( -4 = 6x + 9 \). - Rearranging yields \( 6x = -13 \) -> \( x = -\frac{13}{6} \). To summarize: 1. \( x = -5 \) 2. \( x = 5 \) 3. \( x = 2 \) 4. \( x = 2 \) 5. \( x = 4 \) 6. \( x = 3 \) 7. \( x = \frac{3}{2} \) 8. \( x = -\frac{13}{6} \)
