Two unknowns at once

Lesson 5 dodged something. The theatre problem had two unknown numbers in it, standard tickets and concessions, and the way round it was to call the concessions c and write the standard tickets as 3c, so that only one letter ever appeared.

That works when one quantity is described directly in terms of the other. Often it's not. You know two facts about two quantities and neither fact hands you one in terms of the other. Then you need both facts written down as equations, and a way of working with the pair.

What a pair of equations is claiming

Take these two.

x + y = 90

12x + 7y = 880

Each one on its own has an enormous number of solutions. The first is satisfied by x = 1, y = 89 and by x = 20, y = 70 and by x = 90, y = 0, and by infinitely many other pairs. The second is satisfied by a different infinite collection.

Asking for a solution of the pair, which is what people mean by a system, is asking for the one pair of numbers that satisfies both. That's the whole idea, and everything else in this lesson is machinery for finding it.

The picture makes it obvious. Each equation is a straight line, in the sense of lesson 4: the line is made of all the pairs that satisfy it. Two lines that aren't parallel cross at exactly one point, and that point is on both lines, so its two coordinates satisfy both equations.

You could find the crossing point by drawing carefully and reading it off, and for a rough answer that is a perfectly good method. The algebra is for when you want it exactly.

Method one: substitution

Substitution is the method from lesson 5 made general. Get one letter alone in one equation, then put what it equals into the other, which leaves an equation in a single unknown, which is lesson 3.

The situation. A theatre sells adult tickets at $12 and child tickets at $7. One evening it sells 90 tickets and takes $880. How many of each?

Name the letters as numbers. Let x be the number of adult tickets and y the number of child tickets.

Write both facts.

• Ninety tickets in all: x + y = 90
• Eight hundred and eighty dollars in all: 12x + 7y = 880

Get one letter alone. The first equation is the easy one. Take x off both sides: y = 90 - x.

Substitute into the other. Everywhere the second equation says y, write 90 - x instead.

12x + 7(90 - x) = 880

Now it's lesson 3. Expand the bracket: 12x + 630 - 7x = 880. Collect the x terms: 5x + 630 = 880. Take 630 off both sides: 5x = 250. Divide by five: x = 50.

Find the other letter. Put x = 50 back into whichever equation is simpler: y = 90 - 50, so y = 40.

Check in both equations, not one. First: 50 + 40 = 90. Second: 12 × 50 is 600, 7 × 40 is 280, and 600 + 280 = 880. Both hold, so the answer is fifty adult tickets and forty child tickets.

The step people skip is the last one, and it's skipped because after all that work the answer feels earned. It is not earned until it survives both equations.

Method two: elimination

Sometimes neither equation gives a letter up easily, and substitution drags fractions in. Elimination goes at it from the other side: combine the two equations so that one letter disappears.

Start with a pair where it works with no preparation at all.

4x + y = 14

2x - y = 4

Look at the +y in the first and the -y in the second, and ask what would happen if you added the two equations, left sides together and right sides together.

Then substitute back into either equation: 4 × 3 + y = 14 gives 12 + y = 14, so y = 2. Check in both: 12 + 2 = 14, and 6 - 2 = 4. Both hold.

Why is adding two equations allowed? Because of the balance from lesson 2. The second equation says that 2x - y and 4 are two names for the same number. Adding that number to the left side of the first equation and adding 4 to its right side is doing the same thing to both sides, which is the one move that keeps an equation true. Elimination isn't a new rule; it's the old rule with a slightly cleverer choice of what to add.

When the coefficients do not match

Most pairs aren't so obliging. Here the letters don't cancel as they stand.

The situation. Two coffees and three teas cost $11.50. Four coffees and one tea cost $13.50. What does each drink cost?

Let c be the price of a coffee in dollars and t the price of a tea.

2c + 3t = 11.50

4c + t = 13.50

Adding these gives 6c + 4t = 25, which eliminates nothing. Subtracting gives -2c + 2t = -2, which is a simpler equation but still has both letters.

The move. Multiply one whole equation by a number so that one letter matches in both. Multiplying the first equation by 2 gives:

4c + 6t = 23

Every term doubled, including the right side, because multiplying both sides of an equation by the same number keeps it true. Now the c terms match, so subtract the second equation from this one.

4c + 6t = 23

4c + t = 13.50

Subtracting: the c terms give zero, the t terms give 5t, and the right sides give 9.50. So 5t = 9.50, and t = 1.90.

Substitute back into the second original equation: 4c + 1.90 = 13.50, so 4c = 11.60 and c = 2.90.

Check in both originals. First: 2 × 2.90 is 5.80, 3 × 1.90 is 5.70, and they add to 11.50. Second: 4 × 2.90 is 11.60, plus 1.90 is 13.50. A coffee is $2.90 and a tea is $1.90.

The step that feels arbitrary is multiplying an equation by 2 for no reason visible in the problem. The reason is entirely tactical: you are choosing a form of the same true statement that happens to cancel neatly against the other one. Nothing about the situation changed. Two coffees and three teas still cost $11.50; four coffees and six teas costing $23 is the same fact said louder.

Which method to use

Neither is better in general, and both give the same answer, since both are just ways of getting down to one unknown.

Substitution is easier when a letter is already alone, or has a coefficient of 1 so that getting it alone costs nothing. y = 2x - 1 is begging to be substituted.

Elimination is easier when neither letter is alone and the coefficients are close to matching. The coffee problem done by substitution means writing t = 13.50 - 4c and pushing that through, which works and is messier.

If you cannot decide, pick one. The wasted effort of choosing the clumsier method is smaller than the wasted effort of deliberating.

When there's no crossing point

Two lines that aren't parallel cross exactly once. Two that are parallel never cross, and one line drawn twice crosses everywhere. Both of those show up in the algebra as something odd.

No solution. Take y = 2x + 1 and y = 2x + 5. Try substituting the first into the second before you open the answer, and see what you're left with.

Infinitely many solutions. Take 2x + y = 6 and 4x + 2y = 12. Multiply the first by 2 and you get the second exactly. Eliminating gives 0 = 0, which is true and tells you nothing about x. It's one line written in two ways, and every point on it is a solution.

So when the letters vanish, read what is left. A false statement like 0 = 7 means the lines are parallel and there is no answer. A true statement like 0 = 0 means the two equations are the same line. Neither is a mistake, and both are worth recognising rather than staring at.

What people get wrong

Forgetting that the answer is a pair. "The solution is 3" isn't an answer to a system. The solution is x = 3 and y = 2, and a question about tickets or drinks wants both numbers with their units.

Checking in one equation only. Half the arithmetic slips available in this lesson produce a pair that satisfies one equation and fails the other. Checking one isn't much better than checking none.

Multiplying only part of an equation. Doubling 2c + 3t = 11.50 means doubling the 11.50 as well. Leaving the right side alone breaks the claim, and the answer that follows will fail the check.

Losing a minus sign when subtracting equations. Subtracting 4c + t = 13.50 means subtracting the t and the 13.50 too. This is lesson 3's misconception in a new place, and the fix is the same: write the subtraction out rather than doing it in your head.

Treating 0 = 0 as an error. It means the two facts you were given were the same fact, which is a real thing that happens when a problem is written carelessly or when two measurements say the same thing.

Connections

Lesson 2's balance is what makes both methods legal. Adding one equation to another is adding the same thing to both sides, and multiplying an equation through is multiplying both sides by the same number.

Lesson 3 is the engine. Every system here was reduced to a single equation in one unknown and then solved exactly as that lesson taught.

Lesson 4 supplies the picture, and the printer question in that lesson was already a system: two cost lines, and the question of where they cross.

Lesson 5's naming and numerical checking is what turns a situation with two unknowns into a pair of equations at all, which is steps one and two of every word problem here.

Where this course stops, and what comes next

That's the course. You can say what a letter means, keep an equation true while you change it, solve one for its unknown, read and write a straight line, turn a sentence into an equation, and handle two unknowns at once. That's the working core of elementary algebra.

What this course deliberately leaves out is quadratics, factoring, exponents, and inequalities. Those need more space than a foundation course has. Two free places to get them, both good:

• OpenStax, Elementary Algebra 2e, chapters 6 to 10, which continues where this leaves off, with full worked solutions. Inequalities are the exception: they sit back in chapter 2, section 2.7, rather than in that later block.
• Khan Academy's Algebra 1, for practice with immediate feedback, which is the thing a written course cannot give you.

Go deeper

• OpenStax, Elementary Algebra 2e, chapter 5, free online, on systems of linear equations, with a longer treatment of graphing them.
• Khan Academy, systems of equations, for practice with both methods.