\( C=2 y-1 \) and \( D=y-y^{2} \), find an expression that eqpals \( 3 C+D \) in standard
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To find an expression that equals \( 3C + D \), we start by substituting the given equations: 1. \( C = 2y - 1 \) 2. \( D = y - y^2 \) Now, calculate \( 3C \): \[ 3C = 3(2y - 1) = 6y - 3 \] Next, we add \( D \) to this expression: \[ 3C + D = (6y - 3) + (y - y^2) \] Combining the like terms gives: \[ 3C + D = 6y - 3 + y - y^2 = (6y + y) - y^2 - 3 \] This simplifies to: \[ 3C + D = 7y - y^2 - 3 \] To express this in standard form, we rearrange it: \[ 3C + D = -y^2 + 7y - 3 \] Thus, the expression in standard form is: \[ 3C + D = -y^2 + 7y - 3 \]