Algebra: Quadratics, Simultaneous Equations and Graphs

The algebra is not the hard part

Ask somebody who has just lost marks on an algebra question what went wrong, and they will usually say they could not do it. Watch them more closely and something different is going on. They can factorise. They can substitute. They can run the formula. What they did was start before deciding, pick whichever method they had practised most recently, and get four lines into it before finding out it was the wrong one. Every problem in this part of the course announces what it wants, and it announces it in one specific feature that takes two seconds to check. A pair of simultaneous equations tells you whether to subtract or to substitute by whether a variable already has matching coefficients. A quadratic tells you whether to factorise by whether its numbers work out neatly, and if they do not, the formula always works. A graph question tells you whether you need exact values or only a shape. So the habit this module builds is small and it is the whole thing: LOOK FIRST, DECIDE, THEN START. How to carry out each method once chosen is covered in modules of its own, and this one hands off to them rather than repeating them.

Five words for choosing

Each of these names a method or a feature you decide on. Knowing the names is what lets you say WHY you chose.

Which method does each want?

Subtract or substitute?

You are given 5x plus 3y equals 41, and 5x plus y equals 27. Which method should you choose, and why?

• Elimination, because the x terms already match, so subtracting removes x in one step
• Substitution, because the second equation is simpler
• The quadratic formula, because there are two unknowns
• Draw both graphs and read where they cross

Two ways into a pair

Both methods always work. Choosing the one the problem is set up for saves lines, and lines are where mistakes happen.

Eliminate, then finish

Solve this pair: 5x plus 3y equals 41, and 5x plus y equals 27. Give the value of y.

Decide before you start

Five steps for any algebra problem in this topic. Put them in the order that stops you starting the wrong method.

• Read the problem and name what kind it is
• Look for the feature that decides the method
• Choose the method, and be able to say in one line why
• Carry out the method, showing enough working to be followed
• Put the answer back into the original problem to check it

When the formula is the right call

The quadratic formula always works. Which THREE of these are good reasons to reach for it rather than factorising?

• The numbers do not give whole-number brackets
• You have tried factor pairs and none of them work
• The question asks for answers to a number of decimal places, which suggests they are not whole
• It is always quicker than factorising
• It avoids having to understand what a root is

Look before you start

Almost every mark lost in this topic is lost in the first ten seconds, before a single line is written. The pattern is always the same. Somebody sees a pair of equations and begins rearranging one of them, because rearranging is what they did last lesson, and only after three lines do they notice that the y terms were already identical and one subtraction would have finished it. Or they set off multiplying out brackets on a quadratic that was already factorised, which is undoing the work somebody has done for them. Or they reach for the formula on something that splits into two neat brackets in a moment, and spend twice as long for the same answer with more chances to slip. None of that is a failure of algebra. It is a failure to look. So spend the two seconds. For a pair, ask whether a variable already has matching coefficients, and if it does, subtract. For a quadratic, ask whether the numbers give a clean pair of factors, and if they do not, use the formula without further searching. And for anything that looks unfamiliar, ask first whether it belongs to this topic at all, because a single equation with one unknown and nothing squared is neither a quadratic nor a pair and needs neither method. One more thing worth knowing for the exam: when a question asks for answers to a set number of decimal places, it is telling you the roots are not whole numbers, which is the same as telling you not to bother trying to factorise.

Complete the method sentences

Adding or subtracting two equations so that one variable disappears is _____. Putting one equation into the other is _____. Writing a quadratic as two brackets multiplied together is _____. A value that makes an expression equal zero, and where its graph crosses the horizontal axis, is _____.

One quadratic, read properly

Take x squared minus 9x plus 20 equals 0, and work through the decision rather than the arithmetic. First, what kind of problem is it? One unknown, a squared term, set equal to zero, so it is a quadratic and the question is which method. Second, the deciding feature: do the numbers give a clean pair of brackets? You need two numbers that multiply to twenty and add to minus nine. Four and five multiply to twenty, and minus four and minus five add to minus nine, so yes, they do. That answers the method question, and factorising is the route. The brackets are x minus four and x minus five, and the zero-product rule finishes it: if two things multiply to zero then one of them is zero, so x is four or x is five. Now notice what those two answers also are. If you drew the graph of x squared minus 9x plus 20, it would cross the horizontal axis at four and at five, because a root is exactly the place where the expression is worth nothing. That is why solving and sketching are two views of one thing. Finally, the check, which takes seconds: put five back in. Twenty-five minus forty-five plus twenty is zero. Correct.

Find the wasted move

Five lines from somebody solving a pair of equations where the y terms were already identical. Tap the ONE that shows the wasted decision.

• The two equations are 4x plus 2y equals 22 and x plus 2y equals 13.
• I rearranged the second equation to make x the subject.
• Then I substituted that expression into the first equation.
• I expanded the brackets and collected the terms.
• That gave x equals 3, and then y equals 5.

Justify the method

A method answer earns more when it names the feature it was chosen on. Choose the option in each gap.

The method run

Five quick decisions. Three lives.

Three questions, three decisions

You are working through an exam paper and meet three algebra questions in a row. You have enough time, but not enough to take the long route on all three. Work through the decisions.

• The first gives you two equations in which the x terms are already identical. What do you do?
• The second is a quadratic, and the question asks for answers to two decimal places. What does that tell you?
• The third has one unknown, no squared term, and brackets on both sides. What kind of problem is it?

Explain how you chose

A friend says they always use the quadratic formula because it always works, and always use substitution because they find elimination confusing. Explain why choosing on the problem rather than on habit is worth doing, using examples.

• Explain what feature tells you elimination will be quick
• Explain what feature makes substitution the natural choice
• Say when factorising beats the formula, and when it does not
• Explain what a question asking for decimal places is telling you
• Finish with the two-second habit that makes all of this work