Adding the two equations can hand over x + y without either letter being found
Read the question first
Take the pair 3x + 2y = 11 and 2x + 3y = 9, and suppose the question asks only for the value of x + y. You could find x and y and add them. But look at the coefficients first: the first equation has 3 and 2, and the second has 2 and 3. The numbers are swapped.
Add, then factor
Add the two equations. The x terms give 3x + 2x = 5x, and the y terms give 2y + 3y = 5y. The right sides give 11 + 9 = 20. So 5x + 5y = 20. Both letters now have the same coefficient, because 3 + 2 and 2 + 3 are both 5.
Take 5 out as a common factor: 5(x + y) = 20. Divide both sides by 5: x + y = 4. That answers the question, and neither x nor y has been found.
The long way gives the same total
Solving in full also works, but it takes longer. Multiply the first equation by 3 and the second by 2, so that both have 6y: 9x + 6y = 33 and 4x + 6y = 18. Subtract: 5x = 15, so x = 3. Then 3 × 3 + 2y = 11 gives 2y = 2, so y = 1. And x + y = 3 + 1 = 4, the same answer.
The lines cross where x = 3 and y = 1, and 3 + 1 = 4.
Subtracting gives x − y
Subtracting the equations works in the same way. (3x + 2y) − (2x + 3y) = 11 − 9 gives x − y = 2. So one addition gives x + y = 4, and one subtraction gives x − y = 2.
Those two short equations are easy to finish if x and y are wanted after all. Add them: 2x = 6, so x = 3. Then 3 + y = 4, so y = 1.
Look for this whenever one equation has the other's coefficients swapped. With 2x + y = 7 and x + 2y = 8, adding gives 3x + 3y = 15, so x + y = 5.
The usual mistakes
Stopping at 5x + 5y = 20 and giving 20 as the answer. That is five lots of x + y, so divide by 5 to get x + y itself.
Solving for x and y first when the question does not ask for them. It gives the right total, but by a longer road with more places to slip.
Equations and inequalities · Two Garden Center Receipts and the Cost of a Border in Pairs
QuestionA garden center sells rose bushes at one price and lavender plants at another. One receipt shows 7 rose bushes and 3 lavender plants for $101. Another shows 3 rose bushes and 7 lavender plants for $89. A gardener plans a border of 12 pairs, each pair one rose bush and one lavender plant. (a) Without finding either price, find the cost of one pair, and of the border. (b) The owner then asks for 5 of the 12 rose bushes to be replaced by lavender plants. Without finding either price, find how much the border costs now.
1.Let a rose bush cost x dollars and a lavender plant y dollars. The receipts give 7x + 3y = 101 (1) and 3x + 7y = 89 (2).
A rose bush costs x dollars and a lavender plant y dollars. Each receipt is one equation: 7x + 3y = 101 and 3x + 7y = 89. 2.Add the two equations: 10x + 10y = 190. Divide both sides by 10: x + y = 19.
Add the equations: 10x + 10y = 190. Divide both sides by 10: x + y = 19. 3.(a) One pair costs $19, and the border of 12 pairs costs 12 × 19 = $228.
(a) One pair costs $19, and 12 pairs cost 12 × 19 = $228. 4.Replacing a rose bush by a lavender plant lowers the cost by x − y. Subtract equation (2) from equation (1): 4x − 4y = 12, so x − y = 3. Each replacement saves $3.
Subtract (2) from (1): 4x − 4y = 12, so x − y = 3. Each rose bush replaced by a lavender plant saves $3. 5.(b) Five replacements save 5 × 3 = $15, so the border now costs 228 − 15 = $213. Check: x + y = 19 and x − y = 3 give x = 11 and y = 8, and 7 rose bushes and 17 lavender plants cost 77 + 136 = 213.
(b) Five replacements save 5 × 3 = $15, so the border now costs 228 − 15 = $213.
Answer: (a) $19 a pair, $228 for the border; (b) $213
Watch out for
- Solving for x and y by elimination before answering (a). It reaches the same total by a longer road. Because the receipts swap their numbers, adding them gives 10x + 10y at once, and the sum of the prices follows in one line.
- Working out (b) as 228 − 5 × 19, as if five whole pairs were removed. The rose bushes are replaced, not removed, so each one lowers the cost by x − y = $3.
Logic and counting · A Purse Worth 32 Dollars, and 38 with the Coin Counts Swapped
QuestionRachel had a purse containing only 20-cent coins and 50-cent coins with a total value of $32.00. If the number of 20-cent coins and 50-cent coins were swapped, the total value of the coins would increase by $6.00 to $38.00. (a) How many coins did Rachel have in her purse altogether? (b) How many 50-cent coins did Rachel have at first?
1.Put the purse and its swapped version side by side. Together they are worth $32 + $38 = $70, or 7000 cents.
The purse and its swapped version together: $70. 2.Every 20-cent coin in the purse is a 50-cent coin in the swapped version, and every 50-cent coin a 20-cent coin. So each coin of the purse pairs with one in the copy, 20 cents with 50 cents: 70 cents a pair.
Each coin is paired with its swap, 20¢ with 50¢: 70¢ a pair. 3.There are as many pairs as coins in the purse: 7000 ÷ 70 = 100.
(a) 7000 ÷ 70 = 100 coins. 4.(a) Rachel had 100 coins.
5.Swapping raised the value by $6, or 600 cents. It turned every 20-cent coin into a 50-cent coin and every 50-cent coin into a 20-cent coin, so the purse held more 20-cent coins.
Swapping raised the value by $6, so there were more 20-cent coins. 6.Each 20-cent coin that becomes a 50-cent coin gains 30 cents, and each 50-cent coin that becomes a 20-cent coin loses 30 cents.
Each extra 20-cent coin swapped gains 30¢: 600 ÷ 30 = 20 more of them. 7.The gains outweigh the losses by 600 cents, so there were 600 ÷ 30 = 20 more 20-cent coins than 50-cent coins.
Each extra 20-cent coin swapped gains 30¢: 600 ÷ 30 = 20 more of them. 8.(b) The 50-cent coins were (100 − 20) ÷ 2 = 40, and the 20-cent coins 60. Check: 60 × 20 + 40 × 50 = 1200 + 2000 = 3200 cents.
(b) (100 − 20) ÷ 2 = 40 fifty-cent coins.
Answer: (a) 100 coins; (b) 40 fifty-cent coins
Watch out for
- Giving the 60 to the 50-cent coins. The value went up when the counts were swapped, so the purse held fewer 50-cent coins: 40.
- Dividing the 600 cents by 70. The 70 cents is the value of a pair; each swapped coin changes the value by the difference, 30 cents.
Your turn
1.2x + 1y = 14 and 1x + 2y = 10. What is x + y?