Pesume yfabsieriue: 1) \( (x-1)^{2}-(x+3)(x-3)=12 \) 2) \( (2 y-3)(3 y+1)+2(y-5)(y+5)=2(1-2 y)^{2}+6 y \)
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To solve the first equation \( (x-1)^{2}-(x+3)(x-3)=12 \), start by expanding the terms: 1) Expand \( (x-1)^{2} = x^{2} - 2x + 1 \). 2) Expand \( (x+3)(x-3) = x^{2} - 9 \). Substituting these into the equation gives: \[ (x^{2} - 2x + 1) - (x^{2} - 9) = 12 \] This simplifies to: \[ -2x + 1 + 9 = 12 \] Combine like terms: \[ -2x + 10 = 12 \] Now solve for \( x \): \[ -2x = 12 - 10 \] \[ -2x = 2 \implies x = -1 \] Now for the second equation \( (2 y-3)(3 y+1)+2(y-5)(y+5)=2(1-2 y)^{2}+6 y \): First, expand both sides: Left-hand side: 1) Expand \( (2y-3)(3y+1) = 6y^{2} + 2y - 9y - 3 = 6y^{2} - 7y - 3 \). 2) Expand \( 2(y-5)(y+5) = 2(y^{2} - 25) = 2y^{2} - 50 \). So, \[ 6y^{2} - 7y - 3 + 2y^{2} - 50 = 8y^{2} - 7y - 53 \] Right-hand side: Expand \( 2(1 - 2y)^{2} = 2(4y^{2} - 4y + 1) = 8y^{2} - 8y + 2 \). Thus, we have: \[ 8y^{2} - 7y - 53 = 8y^{2} - 8y + 2 \] Cancel \( 8y^{2} \) from both sides, yielding: \[ -7y - 53 = -8y + 2 \] Rearranging gives: \[ y - 53 = 2 \implies y = 55 \] Final answers: 1) \( x = -1 \) 2) \( y = 55 \)