Word Problems to Equations

Start every word problem by naming the unknown. Three consecutive integers add to seventy-two. Let the smallest be x. Consecutive means one after the other, so the next two are x plus one and x plus two. Adding them gives three x plus three, and that equals seventy-two. Subtract three: three x equals sixty-nine, so x is twenty-three. But the question asked for three numbers, so the answer is twenty-three, twenty-four and twenty-five. Check: they do add to seventy-two.

Age problems are easiest in a table. Mia is four times as old as her brother. Let the brother be b years old, so Mia is four b. In six years, both of them are six years older: b plus six and four b plus six. Then Mia will be twice as old as him, so four b plus six equals two lots of b plus six. Expand: four b plus six equals two b plus twelve. Subtract two b and six: two b equals six, so b is three. The brother is three and Mia is twelve. Check: in six years they will be nine and eighteen, and eighteen is twice nine.

For perimeter problems, write every side in terms of one unknown. A rectangle's length is five centimetres more than twice its width, and its perimeter is forty-six centimetres. Let the width be w, so the length is two w plus five. The perimeter is two lots of length plus width: two lots of w plus two w plus five, equals forty-six. Divide both sides by two: three w plus five equals twenty-three, so w is six. The rectangle is six centimetres by seventeen centimetres. Check: six plus seventeen is twenty-three, and doubled that is forty-six.

With two kinds of ticket, use one unknown for one kind and write the other kind from it. Thirty tickets were sold, adults at twelve dollars and children at seven, for two hundred and eighty dollars. Let a be the number of adult tickets, so the number of child tickets is thirty minus a. The costs add up: twelve a plus seven lots of thirty minus a equals two hundred and eighty. Expand: twelve a plus two hundred and ten minus seven a, which is five a plus two hundred and ten. So five a equals seventy and a is fourteen. Fourteen adults and sixteen children. Check: one hundred and sixty-eight plus one hundred and twelve is two hundred and eighty.

For journeys, use distance equals speed times time, and a table. A car drives for five hours, part of the way at sixty kilometres an hour and the rest at ninety, and covers three hundred and ninety kilometres. Let t be the hours at sixty, so the rest of the time is five minus t. The two distances add up: sixty t plus ninety lots of five minus t equals three hundred and ninety. Expand: sixty t plus four hundred and fifty minus ninety t, which is four hundred and fifty minus thirty t. So thirty t is sixty and t is two. Two hours at sixty is one hundred and twenty kilometres, and three hours at ninety is two hundred and seventy. Check: those add to three hundred and ninety. Quiz and exercises now.