Solving Linear Equations by Balancing

Picture two scooter-hire apps. One charges a chunky fee just to start a ride, then a little for each minute you are moving. The other starts you off almost free, then charges far more per minute. For a quick dash to the shops, one app is cheaper. For a long ride across town, the other one is. Somewhere in between sits a journey where the two apps charge you exactly the same. Finding that crossover means writing the same unknown, the number of minutes, on both sides of an equals sign, then working out what it has to be. Turning a worded situation into that equation is its own skill, and the video on setting up and solving an equation from a context takes it on. Here we start from the equation and get the answer out.

So, the machinery underneath all of it. Every equation is a statement that two things are equal, and that statement is what you protect. Treat the equals sign as a promise that the left side and the right side are the same value. Change one side and not the other, and the promise breaks: your answer now belongs to a different equation. That leaves one safe move in the whole topic: whatever you do, you do to both sides in the same moment. Adding, subtracting, multiplying or dividing, as long as both sides get it. Here it is at its plainest. If x plus four equals eleven, you take four from both sides, and x equals seven. The four did not vanish because you wanted it gone; a subtraction landed on both sides at once.

Quick check before the main event. You are looking at three x equals twenty-one, and you want x on its own. Do you subtract three from both sides, or divide both sides by three?

Take your pick. I'll wait.

The answer is divide by three. Three x means three lots of x, so the three is attached by multiplication, and only dividing undoes multiplying. Subtracting three would leave three x minus three, with x no closer to being alone.

The move you make has to be the opposite of the operation that is actually there.

Now for the case this video is built around, where x turns up on both sides at once. Five x minus three equals two x plus twelve. There is an x on the left and an x on the right, so there is nothing to isolate yet, because x is living in two places. The goal is every x term on one side and every plain number on the other, and the only tool for that is the move you already have. The smaller x term is two x, so subtract two x from both sides. On the left, five x minus two x leaves three x, and the minus three is untouched. On the right, the two x cancels itself, so twelve is all that survives. The equation now reads three x minus three equals twelve, with every x on one side. Taking the smaller x term is a choice, not a law. It leaves the x term positive, which is easier to finish without slipping. Look at what actually happened to that two x. It arrived on the right as a plus. It left the right completely, and it landed on the left as a subtraction. That is where the shortcut comes from: cross the line, flip the sign. Cross the line, flip the sign. The line is the equals sign, and the flip is not a rule invented to catch you out; it records the subtraction you did to both sides.

Your turn to catch it. Starting from five x minus three equals two x plus twelve, you take two x from both sides. Which line is right? A, three x minus three equals twelve. B, seven x minus three equals twelve. C, three x minus three equals two x plus twelve.

Pick one. I'll wait.

The answer is A. B is the sign error itself: two x was added on the right, so it has to be subtracted on the left, not added. C changed the left side and left the right alone, so the sides no longer match.

Finish it. From three x minus three equals twelve, add three to both sides, and three x equals fifteen. Divide both sides by three, and x equals five. You can check that yourself without asking anyone. Put five back into the original: five fives minus three is twenty-two, and two fives plus twelve is twenty-two as well. Both sides agree, so five is the solution.

One more shape to handle, and it is the one with a fraction sitting in it. Two x over three, plus one, equals seven. The x term has been divided by three, and a divided x makes people nervous, so they rush the step that needs care. The tempting move is to multiply the fraction by three so it disappears, and leave the rest of the equation as it was. That breaks the promise on the spot, because one term has been tripled and its neighbours have not. Multiplying by three is completely allowed. It is allowed on both sides, on every single term, in the same moment. Two x over three, times three, gives two x, the one becomes three, and the seven becomes twenty-one. So the equation reads two x plus three equals twenty-one, with no fraction left anywhere. Take three from both sides to get two x equals eighteen, then divide both sides by two, and x equals nine. Check it the same way. Two nines is eighteen, eighteen over three is six, and six plus one is seven.

Time to hand the whole method over to you, start to finish. Solve four x plus five equals x plus twenty, writing down every operation as it lands on both sides, in order. Pause it there and work it out. I'll wait. The answer is x equals five. Take x from both sides and three x plus five equals twenty. Take five from both sides and three x equals fifteen. Divide both sides by three, and there is your five. If you got that, you have the method itself. The numbers change from question to question, but those moves do not.

Now for what examiners actually see when they mark these. Here is one line from an examiner report on an AQA Foundation paper, describing students moving a value to the opposite side of an equation. Many students however are still forgetting to change the operator when using values on the opposite side, so we saw lots of attempts at fourteen divided by five, times two, which often went wrong due to the decimal nature of the result. Notice which operations that is about, a divide and a multiply, not a plus and a minus. Cross the line, flip the sign covers all four, because whatever carried a value across, its opposite meets it on the other side. The substitution check costs you about ten seconds. Put your answer back into the original, work out each side separately, and see whether they match. If they do not, you find out before the examiner does. Keep one job separate. Shuffling letters around a general formula to make a new subject uses moves that look identical, but the goal is different, and the video on using and rearranging a formula takes that on.

Before the summary, one more picture worth carrying with you. Draw the two sides of an equation as two straight-line graphs, and the solution is the x value where the lines cross, which is how a graph hands you an approximate answer when the exact one is awkward.

Cross the line, flip the sign. That chant sits in the middle of everything here, so here is the whole method wrapped around it. An equation is a promise that both sides are equal, so every operation you choose lands on both sides at the same time. With x on both sides, subtract the smaller x term from both sides first, then move the numbers, then divide by whatever is left multiplying x. With a fraction, multiply every term on both sides by the bottom number, not just the fraction, then solve the plain equation you are left with. Then put your answer back into the original and check both sides agree, because that turns a hopeful answer into a checked one.

Next in the chapter: Solving Linear Equations with Brackets, the same balancing, with a bracket to clear out of the way first.

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