Simultaneous Equations: One Linear, One Quadratic

Picture a circular pond in a field, five metres from the centre post to the water's edge all the way round. A straight fence cuts across that field and clips the edge of the pond in two places. To describe one of those places you need two numbers, never one: how far across, and how far up. Algebra can find both places exactly, from two equations and nothing else. But the method you would use for two straight fences stops dead the moment one of them curves.

First, the tell. Here is how to spot in one glance that a pair of equations needs a different tool from the usual one.

A quick note on where this sits. Pairing a line with a curve is Higher paper material. If you are sitting Foundation, the video on solving simultaneous linear equations by elimination is the one written for your paper.

Measure everything from the centre of the pond, in metres. The fence is y equals x plus one. The pond's edge is x squared plus y squared equals twenty-five. One equation is linear, plain x and y terms. The other carries squared terms, so it is no straight line.

With two straight lines you would eliminate. You multiply one equation until a column of terms matches, then add or subtract, and one variable disappears because like terms cancel.

That cancelling is the whole engine, and it only runs on like terms. Here you have y in one equation and y squared in the other. Those are not like terms, so nothing cancels and you are left holding both squares with no way in.

So you take the other route, called substitution. The linear equation is not just a fact about y. It is an instruction: it tells you what y is worth, written in x.

That gives you your three-word handle for this video: swap, solve, pair. Swap y out of the curved equation, solve the quadratic you are left with, then pair every answer with its partner.

Your turn already, and it is one line of thinking. Take y equals two x, together with x squared plus y squared equals twenty. Which first move is the right one? A: add the two equations together. B: replace y in the second equation with two x. C: multiply the first equation by two, then subtract. Take your pick. I'll wait. The answer is B. A and C are both elimination moves, and elimination needs like terms to cancel, which this pair has none of. B is the swap: y is worth two x, so two x can stand in its place.

Spot the squared term, and you know the method before you have written a single line down.

Now the swap in full, back on the fence and the pond.

The fence says y equals x plus one. So everywhere the pond equation says y, write x plus one instead. That turns x squared plus y squared equals twenty-five into x squared plus, in brackets, x plus one, all squared, equals twenty-five.

The brackets are doing real work there. The letter y stood for the whole of x plus one, and the whole of it is being squared, so the whole of it goes inside the brackets.

Expand the bracket. x plus one, all squared, is x squared plus two x plus one. So the equation reads x squared plus x squared plus two x plus one equals twenty-five.

Collect up and move the twenty-five across: two x squared plus two x minus twenty-four equals zero. Every term divides by two, so make life easier, x squared plus x minus twelve equals zero.

That is an ordinary quadratic now, and it factorises. x plus four, times x minus three, equals zero. So x is minus four, or x is three.

If you want that factorising step slowed right down, it lives in the video on solving a quadratic by factorising.

You have two x values now, and that is not the same as two answers, because each one is only half of an answer.

Here is where correct working still turns into a wrong answer, so this step gets its own slow walk through.

A pair like this normally has two solution pairs, not two loose numbers. Every x has its own y, and they travel together as one answer.

To find each y, substitute back into the linear equation, y equals x plus one, and never back into the curved one.

The reason is that the linear equation hands you exactly one y for each x you feed it. Feed an x into the curved equation instead and it can hand back two values, and you have no way of knowing which one belongs with the x you started from.

So x equals minus four gives y equals minus four plus one, which is minus three. And x equals three gives y equals three plus one, which is four.

Write them out as pairs. When x is minus four, y is minus three. When x is three, y is four. Two places, each with its own matched pair of numbers.

Check one against the pond. Minus four squared is sixteen, minus three squared is nine, and sixteen plus nine is twenty-five. It fits, and so does the other pair.

Each pair is one of the places the fence meets the water. Why the algebra and the picture agree is its own video, on graph intersections as simultaneous solutions.

Mismatch those pairs, and a full page of correct algebra still ends in a wrong answer.

Exam papers dress this up a second way, where the opening move looks different but the method underneath is the same.

This time both equations start with y. y equals two x minus three, and y equals x squared minus four x plus five.

At a solution, y has one single value. If y equals two x minus three, and that same y also equals x squared minus four x plus five, then those two expressions must equal each other.

So two x minus three equals x squared minus four x plus five. Bring every term to one side and it becomes x squared minus six x plus eight equals zero.

That factorises to x minus two, times x minus four, equals zero. So x is two, or x is four.

Now pair back through the linear equation, exactly as before. So one solution is x equals two with y equals one, and the other is x equals four with y equals five.

That is the same three moves underneath, just a different way in.

Your turn on a whole one now. Solve y equals x minus two together with y equals x squared minus four x plus two. Take it all the way to two full pairs. Pause it there and work it out. I'll wait. Here is the answer. Setting the two expressions equal gives x squared minus five x plus four equals zero, which factorises to x minus one, times x minus four. So x is one, or x is four. Pair back through y equals x minus two. So one solution is x equals one with y equals minus one, and the other is x equals four with y equals two.

If you finished with four numbers in a list rather than two pairs, that is the habit worth changing today.

Time for the exam side of this. It comes from a report written after a paper where students had to solve a linear equation and a non-linear one together. Although this is a familiar type of question it is one demanding accurate algebra and an understanding of the stages needed to solve non-linear simultaneous equations. The correct start is one to eliminate one variable, in this case using the linear equation as written. Some tried to treat the equations as if they were both linear and solve by elimination with no success. Read that middle sentence closely. Eliminate one variable, using the linear equation as written. Examiners use eliminate to mean getting a variable out of the way, and using the linear equation as written is exactly the swap. What failed was adding and subtracting the two equations as though both were straight lines. The same report carries a second line, about the students who did get through the algebra. Those who got this far used factorisation or the quadratic formula to generate the 2 values; normally the y values. A significant number of students treated the 2 values coming out of a quadratic in y as x values, an error which should have been spotted. In that question the quadratic came out in y, so the two values were y values, and students wrote them as x values. The pairing step protects you: put each solved value back into the linear equation, and the equation tells you which letter you are holding. So write your answer as labelled pairs, in words if it helps. When x is three, y is four. A labelled pair cannot be mismatched, by a marker or by you.

Swap, solve, pair. Three words, and here is what each one does, gathered into one place. Swap. The linear equation says what one letter is worth, so write that expression into the other equation in its place, in brackets if it is about to be squared. Solve. What is left is an ordinary quadratic in one letter, so factorise it and expect two values. Pair. Put each value back into the linear equation and write both full pairs. Elimination stays behind with the two straight lines, because y and y squared never cancel.

Next in the chapter: Graph Intersections as Simultaneous Solutions.

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