1. **Problem:** Six times a number is increased by 7. Then this sum is multiplied by 4, and the result is 10 larger than 30 times the opposite of the number. Find the number. 2. **Step 1: Define the variable.** Let the number be $x$. 3. **Step 2: Translate the problem into an equation.** - Six times a number increased by 7: $6x + 7$ - This sum multiplied by 4: $4(6x + 7)$ - 30 times the opposite of the number: $30(-x) = -30x$ - The result is 10 larger than this: $4(6x + 7) = -30x + 10$ 4. **Step 3: Expand and simplify the equation.** $$4(6x + 7) = -30x + 10$$ $$24x + 28 = -30x + 10$$ 5. **Step 4: Collect like terms.** Add $30x$ to both sides: $$24x + 30x + 28 = 10$$ $$54x + 28 = 10$$ 6. **Step 5: Isolate $x$.** Subtract 28 from both sides: $$54x + \cancel{28} - \cancel{28} = 10 - 28$$ $$54x = -18$$ 7. **Step 6: Solve for $x$.** Divide both sides by 54: $$x = \frac{-18}{54}$$ $$x = \frac{\cancel{-18}}{\cancel{54}} = -\frac{1}{3}$$ 8. **Answer:** The number is $-\frac{1}{3}$.