No. 20 Systems of equations

A system of two equations in two unknowns. The answer is every pair (x, y).

How to solve it

Systems of linear equations

A system of two linear equations, such as $x + 2y = 7$ and $3x - y = 7$, puts two conditions on the same pair of numbers $(x, y)$. You need the pair that satisfies both. The usual tools are substitution and adding the equations.

Step by step

Find an equation where one variable is easy to express (coefficient $1$ or $-1$) and express it: $x + 2y = 7$ gives $x = 7 - 2y$.
Substitute it into the other equation: $3(7 - 2y) - y = 7$. One unknown is left.
Solve it: $21 - 7y = 7$, $y = 2$. Then find the other variable: $x = 7 - 4 = 3$.
If expressing is awkward, add the equations after multiplying them so that the coefficients of one variable become opposite.
Check the pair in both equations and write the answer: $(3, 2)$.

The coefficients of $x$ are opposite, so adding the equations removes $x$.

Then $y = 2$, and either equation gives $x = \frac{1}{2}$. Check: $2 \cdot \frac{1}{2} + 3 \cdot 2 = 7$.

Common mistakes

Multiplying only one side of an equation by a number; both sides must be multiplied.
Substituting the found number and not checking the pair in the other equation.
Writing the pair in the wrong order: $x$ comes first, then $y$.

Example

Nonlinear systems

In a nonlinear system one equation is linear and the other is not: $x + y = 1$, $xy = -12$. There are usually two solutions, each a pair of numbers. Substitution works: express a variable from the linear equation and put it into the other.

Step by step

Express one variable from the linear equation: $y = 1 - x$.
Substitute it into the other equation: $x(1 - x) = -12$.
Solve the quadratic equation you get: $x^2 - x - 12 = 0$, with roots $4$ and $-3$.
For each root find the other variable: $y = 1 - 4 = -3$ and $y = 1 - (-3) = 4$. That gives two pairs.
Check both pairs and write them in the answer: $(4, -3)$ and $(-3, 4)$.

The sum of the unknowns.

Their product. This is Vieta's theorem read backwards.

Example: $x + y = 1$, $xy = -12$ give $t^2 - t - 12 = 0$ with roots $4$ and $-3$.

Common mistakes

Finding only $x$ and forgetting to compute $y$.
Writing one pair, although the quadratic has two roots and so the system has two solutions.
Swapping $x$ and $y$ in a pair.