The balance
Tonight's idea
An equation like \(3x + 7 = 22\) says: three mystery weights and a 7-gram weight on the left pan balance 22 grams on the right. You may do anything you like to the scales as long as you do it to both pans: remove 7 grams from each side, and \(3x = 15\). Split each side into three equal piles, and \(x = 5\). Every step keeps the balance, so the answer at the end is the answer at the start.
Undo what was done to \(x\), in reverse order, doing the same thing to both sides. Addition is undone by subtraction, multiplication by division. Check by putting the answer back in: \(3 \times 5 + 7 = 22\).
An inequality like \(2x + 3 < 11\) is a tilted scale: the left side is lighter. The same moves work, with one trap: multiplying or dividing both sides by a negative number flips the tilt. \(2 < 5\) is true, but multiply both by \(-1\) and \(-2 > -5\).
Watch one
The trick: think of a number. Double it. Add 10. Halve it. Subtract the number you first thought of. I know your answer is 5.
Call the number \(n\). Double: \(2n\). Add 10: \(2n + 10\). Halve: \(n + 5\). Subtract \(n\): \(5\). The \(n\) was always going to cancel.
Solve \(18 - 3x = 3\). Take 18 from both sides: \(-3x = -15\). Divide by \(-3\): \(x = 5\). Check: \(18 - 15 = 3\).
Your turn
- Solve, and check each answer: \(x - 9 = -4\); \(\;4y = 30\); \(\;\dfrac{m}{3} = -6\); \(\;2t + 5 = 17\).
- Two-step: \(5(x - 2) = 35\); \(\;\dfrac{x}{4} + 7 = 10\); \(\;3(x + 4) = 2(x + 9)\).
- Solve the inequalities and draw each answer on a number line (open circle for \(\lt\), filled circle for \(\le\)): \(2x + 3 < 11\); \(\;-3x \ge 12\). Explain the flip in the second one.
- A number is tripled and then 8 is subtracted. The result is 25. Find the number. Then: a rectangle has perimeter 46 and its length is 5 more than its width. Find both.
- Explain this trick with algebra: think of a number, add 3, double it, subtract 4, halve it, subtract the number you started with. What does everyone get?
- Find all whole numbers \(n\) with \(3n - 4 < 2n + 1\).
Stretch: two numbers, and your own trick
(a) Two numbers add to 20 and differ by 6. Find them two ways: first by calling the smaller one \(x\) and writing an equation; second by thinking about their average and how far each sits from it. Then give a general rule for numbers with sum \(S\) and difference \(D\).
(b) Design a think-of-a-number trick with at least four steps whose answer is always 7, and prove it works with algebra.
Answers
- \(x = 5\); \(y = 7.5\); \(m = -18\); \(t = 6\).
- \(x - 2 = 7\), so \(x = 9\). \(\tfrac{x}{4} = 3\), so \(x = 12\). \(3x + 12 = 2x + 18\), subtract \(2x\) from both sides, \(x + 12 = 18\), \(x = 6\).
- \(2x < 8\), \(x < 4\) (open circle at 4, shade left). \(-3x \ge 12\): dividing by \(-3\) flips the sign, \(x \le -4\) (filled circle at \(-4\), shade left). Check with \(x = -5\): \(-3 \times -5 = 15 \ge 12\), true; with \(x = 0\): \(0 \ge 12\), false.
- \(3n - 8 = 25\), \(n = 11\). Width \(w\), length \(w + 5\): \(2(w + w + 5) = 46\), \(4w + 10 = 46\), \(w = 9\), length 14.
- \(n \to n + 3 \to 2n + 6 \to 2n + 2 \to n + 1 \to 1\). Everyone gets 1.
- \(n < 5\): the whole numbers 0, 1, 2, 3, 4.
Stretch. (a) \(x + (x + 6) = 20\), so \(2x = 14\), \(x = 7\): the numbers are 7 and 13. Or: their average is 10 and they sit 3 either side of it, giving 7 and 13. In general the numbers are \(\tfrac{S + D}{2}\) and \(\tfrac{S - D}{2}\). (b) One design: think of \(n\), multiply by 4, add 28, divide by 4, subtract \(n\). Algebra: \(4n \to 4n + 28 \to n + 7 \to 7\). Any trick that ends by subtracting the original number, after steps that have left exactly one copy of \(n\), will work.