Finding the Unknown: questions with answers

Substitute position 3 into the rule: 2 × 3 + 8 = 14 sticks. The fixed 8 is added once, outside the multiplication.

Subtracting 55 from both sides of the first equation gives $35p = 420$. The second equation says $35(f + 4) = 420$. So $p = f + 4$: $f$ counts only the cups beyond the first 4.

The requested expression is twice the entire left side: 2 × (14u − 4). Its value is therefore 2 × 46 = 92. There is no need to find the letter first; fractional solutions are allowed.

The expression asked for is twice the whole left side: $8m + 16 = 2(4m + 8)$. Its value is therefore $2 \times 20 = 40$. Both the letter term and the constant must be doubled.

For $u + 9 = u + 4$, subtract $u$ from both sides to obtain $9 = 4$. This is false, so the equation has no solution. The other equations have solutions $u = 0$, $u = 1$, and $u = -5$, respectively. A solution of zero is different from having no solution.

Substitute $p = 5$ into the original equation. The LHS is $5 \times 5 + 2 = 27$, and the RHS is $32$. Since $27 \ne 32$, the proposed value is not a solution. A check compares both sides; calculating only the LHS is insufficient. Repeating the same solving steps can repeat the same mistake.

Subtract $4$ from both sides to keep them equal: $7w + 4 - 4 = 46 - 4$. Simplifying gives $7w = 42$.

The changes differ: one side increases by 9 while the other decreases by 9. This creates a difference of twice 9, so the proposed line is not equivalent. A valid addition step adds 9 to both sides.

Subtract 7 from both sides to get c/6 = 8. Multiply both sides by 6 to obtain c = 48. If multiplying first, every term must be multiplied, including the constant.

Divide both sides by the nonzero coefficient -7. This gives w = 21/(-7) = -3. Keep the quotient in that order; subtracting or multiplying by the coefficient does not undo it.

Subtract 7m from both sides, then subtract (8): (3 - 7)m = -4 - 8. Divide both sides by (3 - 7), giving m = (-4 - 8)/(3 - 7) = 3. These equal operations give the rule.

On the LHS, $2t - 7t = -5t$. On the RHS, subtracting $7t$ removes the unknown term. This gives $-5t + 15 = -5$. The negative number multiplying $t$ is valid; subtracting the same term from both sides keeps the equality.

Multiply every term inside the bracket by $5$, giving $10m + 35 = 55$. Dividing both sides of this equation by $5$ gives $2m + 7 = 11$, just as dividing the original equation does. Subtract $7$ from both sides, then divide by $2$: $m = (11 - 7)/2 = 2$.

Collect like terms: (9 − 4)t = 50, so 5t = 50. Divide both sides by 5 to get t = 10. The number 5 multiplies t, so subtracting 5 from 50 to get 45 is incorrect. Also, 50 + 4t cannot be simplified to 54, because 4t is not the number 4.

The total mass is $4 \times 4 = 16$ kg. If the clip weighs 6 kg, the peg must weigh $16 - 6 = 10$ kg. This positive mass makes the two sides balance. Equal masses are not required, and we can test the proposed mass without weighing the peg.