1. **Problem:** Find four consecutive even integers such that 4 times the sum of the first and fourth is 8 greater than 12 times the third. 2. **Step 1: Define variables.** Let the first even integer be $x$. Then the next three consecutive even integers are $x+2$, $x+4$, and $x+6$. 3. **Step 2: Write the equation from the problem statement.** $$4 \times (x + (x+6)) = 12 \times (x+4) + 8$$ 4. **Step 3: Simplify inside parentheses.** $$4 \times (2x + 6) = 12(x+4) + 8$$ 5. **Step 4: Distribute multiplication.** $$8x + 24 = 12x + 48 + 8$$ 6. **Step 5: Simplify right side.** $$8x + 24 = 12x + 56$$ 7. **Step 6: Rearrange to isolate $x$.** $$8x + 24 - 12x = 56$$ $$\cancel{8x} + 24 - \cancel{12x} = 56 - 4x$$ Actually, subtract $12x$ from both sides: $$8x - 12x + 24 = 56$$ $$-4x + 24 = 56$$ 8. **Step 7: Subtract 24 from both sides.** $$-4x = 56 - 24$$ $$-4x = 32$$ 9. **Step 8: Divide both sides by -4.** $$x = \frac{32}{-4} = -8$$ 10. **Step 9: Find the four integers.** $$x = -8, \quad x+2 = -6, \quad x+4 = -4, \quad x+6 = -2$$ 11. **Answer:** The four consecutive even integers are $-8$, $-6$, $-4$, and $-2$. **Verification:** Calculate left side: $$4 \times (-8 + (-2)) = 4 \times (-10) = -40$$ Calculate right side: $$12 \times (-4) + 8 = -48 + 8 = -40$$ Both sides equal, so solution is correct.