Systems of equations

A system of equations asks for values that satisfy several conditions at the same time. Almost every word problem with two or more unknowns reduces to a system: prices, speeds, mixtures, rates of work. This lesson covers the solving methods (substitution, elimination, introducing a new variable), the number of solutions of a linear system and problems with a parameter, non-linear systems, and how to set up a system from a word problem.

§1. Basic ideas

Several equations joined by a curly bracket form a system. A solution of the system is a pair of numbers (x; y) (or a triple (x; y; z)) that makes all the equations true at once. A pair that satisfies only one of the equations is not a solution.
Geometric meaning: each equation describes a curve in the plane (a straight line, a parabola, a circle), and the solutions of the system are the intersection points of these graphs. So the number of solutions depends on how many times the graphs meet.

Two systems are equivalent if they have the same set of solutions. Every solving method replaces a system with an equivalent but simpler one. Substituting the pair you found into both equations is the most reliable check.

§2. The substitution method

Algorithm:

1. From one equation, express one unknown in terms of the other.
2. Substitute it into the second equation — you get an equation in one unknown.
3. Solve it and put the value back into the expression from step 1.

It is convenient to express the unknown whose coefficient is 1 or −1 — then no fractions appear. The method is especially effective when one equation is linear and the other is not.

Worked example

Solve 2x + 3y = 13, 3x − y = 3.

1. From the second, y = 3x − 3.
2. 2x + 3(3x − 3) = 13 → 11x − 9 = 13 → x = 2.
3. y = 3·2 − 3 = 3. Answer: (2; 3).

§3. The elimination (addition) method

Multiply the equations by suitable numbers, then add or subtract them — one unknown disappears. Why is this allowed? If A = B and C = D, then A + C = B + D is also true; so the new equation holds for every solution of the system.
If the coefficients are opposite (2y and −2y), add straight away; if they are equal, subtract; otherwise multiply first. When multiplying, multiply both sides of the equation.

Worked example

Solve 3x + 2y = 7, 5x − 2y = 9.

1. The coefficients of y are opposite, so add: 8x = 16 → x = 2.
2. Substitute into the first: 3·2 + 2y = 7 → 2y = 1.
3. y = 1/2. Answer: (2; 1/2).

§4. How many solutions a linear system has

Two straight lines either intersect, are parallel or coincide. For a₁x + b₁y = c₁, a₂x + b₂y = c₂:

Reason: if a₁/a₂ = b₁/b₂, the gradients −a/b of the lines are equal, so the lines are parallel. If the constants are in the same ratio too, one equation is a multiple of the other — it is the same line.
This scheme is the main tool for problems with a parameter: questions like "for which a are there no / infinitely many solutions" are answered with exactly these conditions.

Worked example

For which a does the system x + 2y = 3, 2x + ay = 5 have no solution?

1. For no solution 1/2 = 2/a → a = 4.
2. Constants: c₁/c₂ = 3/5 ≠ 1/2 — the condition holds.
3. Answer: a = 4.

§5. Non-linear systems

• If one equation is linear, use substitution; usually a quadratic appears and there may be two solutions.
• Symmetric systems (unchanged when x and y are swapped): substitute u = x + y, v = xy.

• Difference of squares: x² − y² = (x − y)(x + y) — divide one equation by the other.
• A homogeneous equation (such as x² − 5xy + 6y² = 0) — divide by y² and find t = x/y.

In a symmetric system, every solution (a; b) comes with the solution (b; a) — remember to write both.

Worked example

Solve x² + y² = 25, x + y = 7.

1. xy = ((x + y)² − (x² + y²))/2 = (49 − 25)/2 = 12.
2. x and y are the roots of t² − 7t + 12 = 0: 3 and 4.
3. Answer: (3; 4) and (4; 3).

§6. Three unknowns and word problems

A linear system in three unknowns follows the same idea: add or subtract the equations in pairs to eliminate one unknown and reduce to a system in two unknowns (Gaussian elimination).
How to turn a word problem into a system:

1. Define the unknowns precisely (with units: "x is the price of a pen in so‘m").
2. Write each condition as a separate equation.
3. Solve the system and check the answer makes sense (prices and times cannot be negative).

Worked example

Solve x + y + z = 6, x − y + z = 2, 2x + y − z = 1.

1. Subtract the second from the first: 2y = 4 → y = 2.
2. Two equations remain: x + z = 4, 2x − z = −1.
3. Add them: 3x = 3 → x = 1, z = 3. Answer: (1; 2; 3).

Worked example

20 l of a 25 % solution must be made from 30 % and 10 % solutions. How many litres of each are needed?

1. Let x be the volume of the 30 % solution and y of the 10 % one: x + y = 20.
2. Amount of salt: 0.3x + 0.1y = 0.25 · 20 = 5.
3. Substitute y = 20 − x: 0.3x + 2 − 0.1x = 5 → 0.2x = 3 → x = 15, y = 5.

Check yourself

1. Solve x + y = 10, x − y = 4.

2. Solve x − y = 2, xy = 15.

3. Solve x² − y² = 16, x − y = 2.

4. 2 pens and 3 notebooks cost 13 000 so‘m; 3 pens and 2 notebooks cost 12 000 so‘m. How much is a notebook?

5. How many solutions does 2x − y = 1, 4x − 2y = 2 have?

6. Solve x + y = 5, xy = 6.

7. Solve 4x + 3y = 12, 3x + 4y = 2.

8. For which a does 3x + ay = 6, x + 2y = 2 have infinitely many solutions?

Common mistakes

• Finding one unknown and giving the answer without the other.
• Losing one of the pairs (x; y) and (y; x) in a symmetric system.
• Multiplying an equation by a number in the elimination method but not multiplying the right-hand side.
• Not checking the ratio c₁/c₂ when counting solutions.
• Not changing the signs of all the terms of the second equation when subtracting.
• Not checking a word-problem answer for sense (a negative price, a fractional number of people).