rach equation in standard form using intergers \[ +5=-(x+3) \] \[ \text { 26.) } y-1=-\frac{1}{2}(x-4) \]

Rewrite the equation by following steps: - step0: Rewrite in standard form: \(5=-\left(x+3\right)\) - step1: Calculate: \(5=-x-3\) - step2: Rewrite the expression: \(-x-3=5\) - step3: Move the constant to the right side: \(-x=8\) - step4: Multiply both sides: \(x=-8\) Rewrite the equation \( y-1=-\frac{1}{2}(x-4) \). Rewrite the equation by following steps: - step0: Rewrite in standard form: \(y-1=-\frac{1}{2}\left(x-4\right)\) - step1: Calculate: \(y-1=-\frac{1}{2}x+2\) - step2: Multiply both sides of the equation by LCD: \(2y-2=-x+4\) - step3: Move the variable to the left side: \(2y-2+x=4\) - step4: Move the constant to the right side: \(2y+x=6\) - step5: Reorder the terms: \(x+2y=6\) Let's convert each equation into standard form using integers. 1. For the equation \( +5 = -(x + 3) \): - Start by simplifying the equation: \[ +5 = -x - 3 \] - Rearranging gives: \[ x + 5 + 3 = 0 \quad \Rightarrow \quad x + 8 = 0 \] - Thus, the standard form is: \[ x = -8 \] 2. For the equation \( y - 1 = -\frac{1}{2}(x - 4) \): - Start by distributing the \(-\frac{1}{2}\): \[ y - 1 = -\frac{1}{2}x + 2 \] - Rearranging gives: \[ \frac{1}{2}x + y - 1 - 2 = 0 \quad \Rightarrow \quad \frac{1}{2}x + y - 3 = 0 \] - To eliminate the fraction, multiply the entire equation by 2: \[ x + 2y = 6 \] - Thus, the standard form is: \[ x + 2y = 6 \] In summary, the equations in standard form are: 1. \( x = -8 \) 2. \( x + 2y = 6 \)