12. Find three consecutive even integers such that the sum of the smallest number and twice the middle number is 20 more than the largest number.

Let's denote the three consecutive even integers as \( x \), \( x+2 \), and \( x+4 \). Given that the sum of the smallest number and twice the middle number is 20 more than the largest number, we can set up the following equation: \[ x + 2(x+2) = (x+4) + 20 \] Now, let's solve this equation to find the value of \( x \). Solve the equation by following steps: - step0: Solve for \(x\): \(x+2\left(x+2\right)=\left(x+4\right)+20\) - step1: Simplify: \(x+2\left(x+2\right)=x+24\) - step2: Cancel equal terms: \(2\left(x+2\right)=24\) - step3: Divide both sides: \(\frac{2\left(x+2\right)}{2}=\frac{24}{2}\) - step4: Divide the numbers: \(x+2=12\) - step5: Move the constant to the right side: \(x=12-2\) - step6: Subtract the numbers: \(x=10\) The solution to the equation is \( x = 10 \). Therefore, the three consecutive even integers are 10, 12, and 14.