Simultaneous Equations Explained: Supply and Demand Equilibrium and the Effect of a Tax

In 30 seconds

Simultaneous equations are two (or more) equations that must be true at the same time. Their solution is the point where their graphs cross.
A market equilibrium is exactly that: the price where the quantity buyers want equals the quantity sellers offer.
Substitution: make one variable the subject and put it into the other equation. Elimination: add or subtract the equations to cancel one variable.
With one linear and one quadratic equation, substitute the linear one into the quadratic, then solve the quadratic. You may get two, one or no solutions.
Always check the answer in both equations, and reject solutions that make no sense in context (such as a negative price).

The case: where does the price of a coffee settle?

Picture the market for takeaway coffee in a small city centre on a weekday. Coffee shops will sell more cups if the price is higher, because it is worth opening an extra till or staying open later. Customers will buy fewer cups as the price rises. Suppose market research gives these daily schedules (the numbers are illustrative):

Demand: $ Q = 900-60P $
Supply: $ Q = 150P-360 $

Here $ P $ is the price of a cup in dollars and $ Q $ is the number of cups per day. The council is also considering a tax of $1.40 on every cup sold, paid by the shops. Two questions follow. What price and quantity will the market settle at now? And after the tax, what will customers pay, what will shops keep, and how much will the council raise?

The concept: two conditions, one point

Each schedule on its own has endless pairs of $ (P, Q) $ that fit it. At $5, buyers want 600 cups; at $8, sellers offer 840 cups. Neither pair is an equilibrium, because buyers and sellers disagree. At equilibrium the same price must give the same quantity on both schedules. So we need the one pair $ (P, Q) $ that satisfies both equations at once. Drawn on a graph, that is where the two lines cross.

Two straight lines can cross once, never (if they are parallel) or everywhere (if they are the same line). So a pair of linear equations has one solution, no solution or infinitely many^[5]. Real demand and supply lines slope in opposite directions, so they always cross exactly once.

Key termSimultaneous equations: a set of equations in the same unknowns that must all hold at the same time. A solution is a set of values that makes every equation true; graphically, it is a point that lies on every graph.

The formula

For a linear market with demand $ Q = a-bP $ and supply $ Q = -c + dP $, where $ a, b, c, d > 0 $, setting demand equal to supply and solving gives

\[ P^* = \frac{a + c}{b + d} \] \[ Q^* = a-bP^* \]

Here $ a $ is the quantity demanded at a zero price, $ b $ is how many fewer units are bought for each $1 rise in price, $ c $ and $ d $ describe supply in the same way ($ d $ is the extra units offered per $1), and $ P^* $, $ Q^* $ are the equilibrium price and quantity^[1]. You do not need to memorise this. It is simply what substitution gives every time, and it is a useful check.

Chapter 2 of Mathematics Made Simple covers simultaneous equations, linear and quadratic, step by step.Get the textbook · $12

Why this method fits the case

An equilibrium is defined by two conditions holding together: buyers are on their demand schedule, and sellers are on their supply schedule. That is precisely a pair of simultaneous equations. Solving one equation on its own tells you nothing, because it has infinitely many solutions. Only together do they pin down a single price.

Substitution is the natural choice here, because both equations already have $ Q $ as the subject. Elimination is often quicker when the equations are written in the form $ ax + by = c $ with matching coefficients. When there are three or more unknowns, as in a model of several linked markets, the same ideas extend, but matrices or Cramer’s rule become the efficient tool. And when the schedules are curved, the algebra becomes a quadratic or worse, so you must allow for more than one crossing point.

Setting it up

Unknowns: $ P $ (dollars per cup) and $ Q $ (cups per day).
Equations: $ Q = 900-60P $ and $ Q = 150P-360 $.
Domain: $ P > 0 $ and $ Q \ge 0 $. Supply is positive only when $ P > 2.40 $, and demand is positive only when $ P < 15 $, so the answer must lie between those prices.
The tax: a $1.40 tax on sellers means a shop that charges $ P $ keeps only $ P-1.40 $. Supply depends on what shops keep, so the new supply is $ Q = 150(P-1.40)-360 $.

Solving it

Worked example 1 (substitution)

Find the equilibrium price and quantity before the tax.

Step 1Both equations give $ Q $, so set them equal: $ 900-60P = 150P-360 $.

Step 2Collect terms: $ 900 + 360 = 150P + 60P $, so $ 1260 = 210P $.

Step 3Solve: $ P = \dfrac{1260}{210} = 6 $.

Step 4Substitute back: $ Q = 900-60(6) = 900-360 = 540 $.

Step 5Check in the other equation: $ 150(6)-360 = 900-360 = 540 $. ✓

Step 6Check with the formula: $ P^* = \dfrac{900 + 360}{60 + 150} = \dfrac{1260}{210} = 6 $. ✓

The market settles at $6 a cup, with 540 cups sold each day.

Worked example 2 (elimination, with the tax)

Find the new equilibrium after a $1.40 tax per cup, the price buyers pay, the price shops keep, and the tax revenue.

Step 1New supply: $ Q = 150(P-1.40)-360 $, which is $ 150P-210-360 $, so $ Q = 150P-570 $.

Step 2Write both in the form $ Q + kP = \text{constant} $: demand $ Q + 60P = 900 $, supply $ Q-150P = -570 $.

Step 3Subtract the second from the first to eliminate $ Q $: $ 60P-(-150P) = 900-(-570) $.

Step 4So $ 210P = 1470 $, giving $ P = 7 $.

Step 5Then $ Q = 900-60(7) = 900-420 = 480 $. Check: $ 150(7)-570 = 1050-570 = 480 $. ✓

Step 6Buyers pay $7. Shops keep $ 7-1.40 = 5.60 $ dollars.

Step 7Tax revenue: $ 1.40 \times 480 = 672 $ dollars per day.

The tax raises the price to $7, cuts sales to 480 cups and raises $672 a day for the council.

Worked example 3 (one linear, one quadratic)

In a specialist market, supply grows faster at higher prices: $ Q = P^2 + 4P $. Demand is $ Q = 96-6P $. Find the equilibrium.

Step 1Substitute the linear equation into the quadratic one: $ P^2 + 4P = 96-6P $.

Step 2Rearrange to equal zero: $ P^2 + 10P-96 = 0 $.

Step 3Factorise: $ (P + 16)(P-6) = 0 $, so $ P = 6 $ or $ P = -16 $.

Step 4A price cannot be negative, so reject $ P = -16 $.

Step 5$ Q = 96-6(6) = 60 $. Check: $ 6^2 + 4(6) = 36 + 24 = 60 $. ✓

The equilibrium is a price of $6 and a quantity of 60. In a pure maths question, both solutions would count: the graphs also meet at $ (-16, 192) $.

Practise equilibrium, elimination and line–curve problems with 1,565 questions and worked answers.Get 1,565 questions · $9

Coffee market: demand, supply and a $1.40 tax price $ per cup

demand: Q = 900 − 60Psupply: Q = 150P − 360supply with tax

Figure 1. Each equilibrium is the intersection of two lines, so it is the solution of a pair of simultaneous equations. The tax lifts supply by $1.40; buyers now pay $7, sellers keep $5.60, and the shaded rectangle is the government’s revenue.

What the answer tells us

Before the tax, 540 cups change hands at $6. After it, the price buyers pay rises by $1.00, while the amount shops keep falls by only $0.40. The tax is paid by the shops on paper, but most of it is passed on to customers. Who really bears a tax depends on how strongly each side responds to price, not on who sends the money to the council^[2].

The numbers explain why. Supply responds by 150 cups per $1, demand by only 60. When the tax pushes the price up, buyers keep buying (they are less responsive), so they carry the larger share: $ \dfrac{150}{60 + 150} = \dfrac57 $ of $1.40, which is exactly $1.00. Shops carry the other $ \dfrac27 $, which is $0.40.

The answers pass a reasonableness check. The new price ($7) is between the old price ($6) and the old price plus the full tax ($7.40). Sales fall, as they should. And the units are right: dollars per cup times cups per day gives dollars per day.

For sensitivity, try a bigger tax. At $2.10 a cup, the same steps give $ P = 7.50 $, $ Q = 450 $ and revenue of $945. Revenue rises, but less than in proportion, because the tax base (cups sold) shrinks as the tax grows. The lost sales, 90 cups at the $1.40 tax, are the source of what economists call the deadweight loss, here $ \tfrac12 \times 1.40 \times 60 = 42 $ dollars a day.

One linear and one quadratic: Q = 96 − 6P and Q = P² + 4P

demand Q = 96 − 6Psupply Q = P² + 4P

Figure 2. With a curved supply schedule, the two graphs still cross only once at a positive price, P = $6. The other algebraic root, P = −16, is outside the domain and is rejected.

Common mistakeShifting the wrong way. A tax on sellers means they keep $ P-t $, so supply becomes $ Q = 150(P-1.40)-360 $. Writing $ 150(P + 1.40) $ moves supply the wrong way and makes the price fall, which should set off alarm bells.

Common mistakeSign errors in elimination. When you subtract $ Q-150P = -570 $ from $ Q + 60P = 900 $, both minus signs flip: $ 60P + 150P = 900 + 570 $. Write the subtraction out in full before simplifying.

Exam tipCambridge 9709 Pure Mathematics 1 (topic 1.1) asks you to “solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic”, often as the intersection of a line and a curve^[3]. IB Mathematics: analysis and approaches HL (AHL 1.16) extends this to systems of up to three linear equations, including the no-solution and infinite-solution cases^[4]. For AP Calculus, this is assumed algebra: you will use it to find where curves meet before computing the area between them.

To build a full algebra toolkit before calculus, the Mathematics Made Simple textbook starts from the basics and includes Appendix B, Mathematics for economists. For drill, the Mathematics Made Simple practice questions book has every answer worked in full.

Get both Mathematics Made Simple books for $16 and take your algebra from equilibrium to exam ready.Both books · $16

Practice questions

EasyQ1. Solve $ 3x + 2y = 16 $ and $ x-2y = 0 $. Show worked answer

Add the equations to eliminate $ y $: $ 4x = 16 $, so $ x = 4 $.
Then $ 4-2y = 0 $, so $ y = 2 $.
Check: $ 3(4) + 2(2) = 16 $. ✓

EasyQ2. Demand is $ Q = 200-4P $ and supply is $ Q = 6P-50 $. Find the equilibrium price and quantity. Show worked answer

$ 200-4P = 6P-50 $, so $ 250 = 10P $ and $ P = 25 $.
$ Q = 200-4(25) = 100 $. Check: $ 6(25)-50 = 100 $. ✓

MediumQ3. A cinema sells adult tickets at $12 and child tickets at $7. One evening it sells 150 tickets for a total of $1,500. How many of each did it sell? Show worked answer

Let $ a $ adults and $ c $ children: $ a + c = 150 $ and $ 12a + 7c = 1500 $.
Substitute $ c = 150-a $: $ 12a + 7(150-a) = 1500 $.
$ 12a + 1050-7a = 1500 $, so $ 5a = 450 $ and $ a = 90 $.
So $ c = 60 $. Check: $ 12(90) + 7(60) = 1080 + 420 = 1500 $. ✓

MediumQ4. Find the points where the line $ y = 2x-2 $ meets the curve $ y = x^2-3x + 4 $. Show worked answer

Substitute: $ x^2-3x + 4 = 2x-2 $, so $ x^2-5x + 6 = 0 $.
Factorise: $ (x-2)(x-3) = 0 $, so $ x = 2 $ or $ x = 3 $.
$ y = 2(2)-2 = 2 $ and $ y = 2(3)-2 = 4 $.
The points are $ (2, 2) $ and $ (3, 4) $.

ExamQ5. In a market, inverse demand is $ P = 40-0.5Q $ and inverse supply is $ P = 10 + Q $. (a) Find the equilibrium price and quantity. [3] (b) The government pays producers a subsidy of $6 per unit, so inverse supply becomes $ P = 4 + Q $. Find the new quantity, the price consumers pay and the price producers receive. [3] (c) Find the total cost of the subsidy, and how the $6 per unit is shared between consumers and producers. [4] Show worked answer

(a) $ 40-0.5Q = 10 + Q $ [1], so $ 30 = 1.5Q $ and $ Q = 20 $ [1]. $ P = 10 + 20 = 30 $ [1].
(b) $ 40-0.5Q = 4 + Q $, so $ 36 = 1.5Q $ and $ Q = 24 $ [1]. Consumers pay $ P = 40-0.5(24) = 28 $ [1]. Producers receive $ 28 + 6 = 34 $ [1].
(c) Cost: $ 6 \times 24 = 144 $ dollars [1]. Consumers pay $ 30-28 = 2 $ dollars less [1]; producers receive $ 34-30 = 4 $ dollars more [1]. So consumers gain $2 and producers $4 of each $6 [1].

Frequently asked questions

Should I use substitution or elimination?

Use substitution when one equation already has a variable as its subject, as with demand and supply in the form $ Q = \dots $. Use elimination when both are in the form $ ax + by = c $ and a coefficient matches, or can be made to match by multiplying. For a linear and a quadratic equation, always substitute.

What if the equations have no solution?

Two linear equations with the same gradient but different intercepts are parallel lines, which never meet. When you solve them, the variables cancel and leave a false statement such as $ 0 = 5 $. If they cancel to $ 0 = 0 $, the equations describe the same line and there are infinitely many solutions.

How many solutions can a linear and a quadratic equation have?

Two, one or none. After substituting, you get a quadratic, and its discriminant $ b^2-4ac $ tells you which: positive gives two crossing points, zero means the line is a tangent, and negative means they do not meet.

Why reject a negative solution?

Algebra does not know that $ P $ is a price. A negative price or quantity is a valid answer to the equations but not to the economic question, so you state it and then reject it with a reason.

References

Chiang, A.C. and Wainwright, K. (2005) Fundamental Methods of Mathematical Economics, 4th edn, Chapter 3 “Equilibrium Analysis in Economics”. New York: McGraw-Hill.
Mankiw, N.G. (2021) Principles of Economics, 9th edn, Chapter 6 “Supply, Demand, and Government Policies”. Boston, MA: Cengage.
Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027, Pure Mathematics 1, topic 1.1 Quadratics. Cambridge: Cambridge University Press & Assessment.
International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide (first assessment 2021), AHL 1.16. Cardiff: IBO.
Abramson, J. (2021) College Algebra 2e, Section 7.1 “Systems of Linear Equations: Two Variables”. Houston, TX: OpenStax. https://openstax.org/details/books/college-algebra-2e