Solving Simultaneous Linear Equations by Elimination

Two shopping trips, one small shop. Three pens and two rulers came to three pounds forty. One pen and two rulers came to two pounds. Neither receipt tells you what a single pen costs. Take the first receipt on its own and you are stuck. Three pens and two rulers for three pounds forty works if pens are seventy pence and rulers are sixty-five. It also works if pens are forty pence and rulers are one pound ten. Both add up perfectly. One receipt leaves endless possibilities open. Put the two receipts together and exactly one pair of prices survives - and there is a method that hunts that pair down every time, on any two equations like these.

First, that method, on a clean pair of equations. It starts somewhere you might not expect: with how you write them down.

The two equations are three x plus two y equals nineteen, and x plus y equals seven. Write them stacked, one directly under the other, with the x terms in one column and the y terms in the column beside them. Numbers go in front of letters - four c, never c four. That layout is not tidiness for its own sake. Once the columns line up, the two terms you are about to attack are sitting one directly above the other, and your next move is visible instead of imagined.

Three moves make a variable vanish. Line up. Match up. Knock out. Two more moves then finish the job: back-substitute, and check.

Why get rid of a letter at all? Because an equation with two unknowns in it has endless answers. x plus y equals seven is true for three and four, for five and two, and for ten and minus three. Lose one letter and what is left is an equation you can actually solve.

Line up is done, so match up is next. We want the two y terms to be the same size. Here are three moves. Multiply the first equation by two. Multiply the second equation by two. Or multiply the second equation by three. Only one of those matches the y terms.

Pick one. I'll wait.

The answer is the second one. Multiply every term of x plus y equals seven by two, and it becomes two x plus two y equals fourteen. Both equations now carry two y. And notice what that multiplying did not cost you. Two x plus two y equals fourteen says exactly what x plus y equals seven says. Doubling both sides of a true statement leaves it true.

Knock out. Three x plus two y equals nineteen on the top row, two x plus two y equals fourteen underneath it, and now subtract the bottom row from the top one, column by column. Three x take away two x leaves x. Two y take away two y leaves nothing at all. Nineteen take away fourteen leaves five. So x equals five.

Subtracting a whole equation is allowed because both equations are true. Two x plus two y and fourteen are two names for the same number. Take that number off the left-hand side of the first equation and its other name off the right, and the balance holds.

Back-substitute. Put x equals five into the simpler original equation, so five plus y equals seven, which gives y equals two.

Now check in both originals. Three fives plus two twos is fifteen plus four, which is nineteen. Five plus two is seven. So the answer is a pair, x equals five and y equals two, and you write both values down, never just the one you found first.

So how do you know whether to add the two equations or subtract them? That single decision is where a lot of correct plans fall over.

Read the signs on the two matched terms. Plus two y sitting above plus two y is the same sign, so you subtract, because two y take away two y is nothing. Plus two y above minus two y is different signs, so you add, because two y plus negative two y is also nothing. Same sign, subtract. Different signs, add. Matching only cares about size, not sign. Four y and minus four y count as matched, and you simply finish with an addition instead of a subtraction.

You also get to choose which letter goes. Multiply x plus y equals seven by three instead, and the x terms match, so x is the one that vanishes and y comes out as two. Pick whichever letter is cheaper to match, and if one coefficient divides neatly into the other, that is your cheap one.

The route is your choice, and the pair of answers comes out the same either way.

Your turn now, on a pair where the coefficients have already matched themselves.

Solve four x plus y equals eighteen together with two x plus y equals ten. Take it all the way to a pair of values, and check them in both equations before you hear mine.

Pause it there and work it out. I'll wait.

The answer is x equals four and y equals two. Both equations already carry a single y, so matching cost you nothing. Same sign, so subtract. Four x take away two x is two x, the y terms vanish, and eighteen take away ten is eight. Two x equals eight, so x equals four. Back-substitute into two x plus y equals ten. Eight plus y equals ten, so y equals two. Check the other equation, where four fours plus two is eighteen. Both of them hold.

If you got rid of y but then could not get anywhere with x, go back and look at whether those two y terms really were the same size before you subtracted. That is almost always the spot.

There is a shortcut on questions like these that feels completely reasonable and gives the wrong price.

Here is one line from an examiner report. It is about a question where two drinks, tea and coffee, had to be priced from two totals.

This question was not well answered by the large majority with a significant number of non-attempts. There were very few students using simultaneous equations. Common errors were to divide three pounds forty by three, with one pound thirteen as the price of tea, or seven pounds thirty by five, with one pound forty-six as the price of coffee.

Different shop, same slip - and our receipt happens to hold the same numbers. Three pounds forty divided by three is one pound thirteen for a pen. But that three pounds forty paid for two rulers as well, so dividing hands the rulers' money to the pens. A pen is nowhere near one pound thirteen.

The honest route takes two lines to set up. Call a pen p and a ruler r. Three p plus two r equals three pounds forty, and p plus two r equals two pounds.

Line up, and both rows already carry two r, so matching is free. Same sign, so subtract. Three p take away p is two p, the r terms vanish, and three pounds forty take away two pounds is one pound forty. Two p equals one pound forty, so a pen is seventy pence.

Back-substitute into p plus two r equals two pounds. Seventy pence plus two r equals two pounds, so two r is one pound thirty, and a ruler is sixty-five pence. Money answers carry two figures after the point, so seventy pence is written nought point seven zero, not nought point seven.

Check both rows. Two pounds ten of pens and one pound thirty of rulers makes three pounds forty, and one pen with two rulers makes two pounds exactly.

The reports keep landing on the same wall. This line is about a simultaneous equations question on a Foundation paper.

This simultaneous equations question was answered very poorly and there were a large number of non-attempts. Many students knew that they needed to eliminate one variable but the very large majority of students were unable to do so correctly.

Knowing the word eliminate was never what stopped those students; carrying it out on the page was.

One more report points straight at the page. It is about students who had built the two equations correctly and then stalled.

Many students were able to construct the simultaneous equations, but weren't able to progress from there, perhaps because of the way they had written them, often, c four plus d equals seven, and c ten plus d equals fifteen.

Say those last two out loud and you can hear the problem. The number is sitting behind the letter, so nothing stacks into a column, and the pair of terms you are meant to knock out never lines up. Write four c, stack the rows, and keep the columns straight.

On this topic the layout is not decoration; it is the thing that lets you take the next step at all.

Right - here is the whole method in one breath. Line up, match up, knock out, then back-substitute and check.

Line up means stacked rows, x above x and y above y, with the numbers in front of the letters. Match up means multiplying one equation, or both, until your target terms are the same size. Knock out means adding or subtracting, and the signs decide which: same sign, subtract; different signs, add. Then back-substitute your value into the simpler original equation to get the second letter, and check the pair in both equations before you commit to it.

All of that works because both of these are straight-line equations, with no squared terms anywhere in them.

Next in the chapter: Simultaneous Equations, One Linear and One Quadratic. That is where you go when one of the two equations has an x squared in it, because elimination cannot clear a squared term and substitution can.

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