Linear Systems in Two and Three Variables, precalculus lesson

A lab needs $12$ liters of a 30% acid solution, and the shelf holds only a 20% solution and a 50% solution. There are two unknown amounts and two facts that link them, the total volume and the total acid. Neither fact alone fixes the amounts, but together they do. Finding the numbers that satisfy several equations at the same time is what solving a system means.

Key terms

Simultaneous equations

Definition A linear equation in the variables $x_1, \dots, x_n$ is an equation of the form

a_1 x_1 + a_2 x_2 + \dots + a_n x_n = b

with real constants $a_1, \dots, a_n$ and $b$ . A system of linear equations, also called a set of simultaneous equations, is a finite list of such equations in the same variables. A solution of the system is an ordered list of numbers $(x_1, \dots, x_n)$ that satisfies every equation at once, and the solution set is the set of all solutions.

In plain words You are looking for values that make all the equations true together. A point that fits only some of them is not a solution.

Example For the system $x + y = 4$ and $x - y = 2$ , the pair $(3, 1)$ is a solution, since $3 + 1 = 4$ and $3 - 1 = 2$ . The pair $(2, 2)$ satisfies the first equation but not the second, so it is not a solution of the system.

Intersections

Definition An equation $ax + by = c$ with $a$ and $b$ not both zero has a line as its graph in the plane. An equation $ax + by + cz = d$ with $a$ , $b$ and $c$ not all zero has a plane as its graph in space. The solutions of a linear system are exactly the points that lie on every one of these lines or planes, that is, the points of their common intersection.

In plain words Solving two equations in $x$ and $y$ means finding where their lines cross. With three variables, it means finding where the planes meet.

Example The lines $y = 2x - 1$ and $y = -x + 5$ meet where $2x - 1 = -x + 5$ , so $x = 2$ and then $y = 3$ . The point $(2, 3)$ lies on both lines, and it is the only such point.

Substitution

Definition To solve by substitution, solve one equation for one variable in terms of the others, and replace that variable by the resulting expression in every other equation. Solve the smaller system that results, then substitute back to find the variable you removed. Each of these steps keeps the solution set the same.

In plain words Use one equation to write a variable in terms of the others, then put that expression everywhere the variable appears. It is quickest when some variable already has coefficient $1$ or $-1$ .

Example Solve $3x + 2y = 16$ and $x - y = 2$ . The second equation gives $x = y + 2$ . Substituting into the first,

\begin{gathered} 3(y + 2) + 2y = 16, \\[0.4em] 5y = 10, \qquad y = 2. \end{gathered}

Then $x = 2 + 2 = 4$ . The pair $(4, 2)$ checks in both equations, since $3 \cdot 4 + 2 \cdot 2 = 16$ and $4 - 2 = 2$ .

Elimination

Definition Elimination replaces a system with an equivalent one, meaning one with the same solution set, using only three operations. You may swap two equations, multiply an equation by a nonzero constant, or add a multiple of one equation to another. Choose the multiples so that a variable cancels. With three variables, remove the same variable from two different pairs of equations, solve the resulting system in two variables, then substitute back.

In plain words Add or subtract equations so that one variable drops out. Each round removes a variable, until a single equation in one variable is left.

Example Solve the system

\begin{aligned} x + y + z &= 6 \\ 2x - y + z &= 3 \\ x + 2y - z &= 2. \end{aligned}

Adding the first and third equations removes $z$ and gives $2x + 3y = 8$ . Adding the second and third also removes $z$ and gives $3x + y = 5$ . Three times that last equation is $9x + 3y = 15$ , and subtracting $2x + 3y = 8$ leaves $7x = 7$ . So $x = 1$ , then $y = 5 - 3 = 2$ , and the first equation gives $z = 6 - 1 - 2 = 3$ . The solution is $(1, 2, 3)$ .

One, none or infinitely many solutions

Definition Every system of linear equations has exactly one solution, no solution, or infinitely many solutions. It never has exactly two, because if two different points solve the system, so does every point on the line through them. For two equations in two variables,

\begin{gathered} a_1 x + b_1 y = c_1, \\[0.4em] a_2 x + b_2 y = c_2, \end{gathered}

there is exactly one solution precisely when $a_1 b_2 - a_2 b_1 \ne 0$ . When each equation describes a line, meaning its coefficients of $x$ and $y$ are not both zero, this condition says that the two lines are neither parallel nor the same line.

In plain words Two lines cross once, never meet, or lie on top of each other. Planes can sit in more ways, but the number of common points is still one, zero or infinitely many.

Example Three systems that share the first equation show all three outcomes.

System	Lines	Solutions
$x + y = 2$ and $x - y = 0$	cross at $(1, 1)$	one
$x + y = 2$ and $x + y = 5$	parallel	none
$x + y = 2$ and $2x + 2y = 4$	the same line	infinitely many

In the first system $a_1 b_2 - a_2 b_1 = 1 \cdot (-1) - 1 \cdot 1 = -2$ , which is not zero. In the other two it equals $0$ .

Inconsistent systems

Definition A system is inconsistent when it has no solution, and consistent when it has at least one. Elimination exposes an inconsistent system by producing a false equation $0 = c$ with $c \ne 0$ .

In plain words The equations contradict each other. With two equations in two variables the lines are parallel. In three variables the planes have no point in common, even when every pair of them meets.

Example Consider the system

\begin{aligned} x + y + z &= 2 \\ x - y + 2z &= 1 \\ 2x + 3z &= 5. \end{aligned}

Adding the first two equations gives $2x + 3z = 3$ , while the third says $2x + 3z = 5$ . Subtracting one from the other gives $0 = 2$ , which is false, so the system has no solution. Each pair of these planes meets in a line, but the three lines are parallel and never share a point.

Dependent systems

Definition A system is dependent when it is consistent and has infinitely many solutions. When there are as many equations as variables, this happens exactly when the system is consistent and some equation is a sum of multiples of the others, so it adds no new condition. Elimination then reduces one of the equations to $0 = 0$ , at least one variable stays free, and the solutions are described with one or more parameters.

In plain words One equation repeats information the others already give. The lines coincide, or the planes share a whole line or even a whole plane, so the solutions form a family rather than a single point.

Example Change the last constant of the previous system to $3$ .

\begin{aligned} x + y + z &= 2 \\ x - y + 2z &= 1 \\ 2x + 3z &= 3. \end{aligned}

Now the third equation is the sum of the first two, and elimination reduces it to $0 = 0$ . Let $z = t$ for any real number $t$ . The sum of the first two equations becomes $2x + 3t = 3$ , and the first equation gives $y = 2 - x - t$ , so

\begin{gathered} x = \frac{3 - 3t}{2}, \qquad y = \frac{1 + t}{2}, \\[0.8em] z = t. \end{gathered}

Each real number $t$ gives one solution, and every solution arises this way. For $t = 1$ this is $(0, 1, 1)$ , and indeed $0 + 1 + 1 = 2$ , $0 - 1 + 2 = 1$ and $0 + 3 = 3$ .

Units

Definition In an applied system each variable stands for a quantity measured in some unit. Every term of an equation must have the same unit as the right side, so coefficients carry units such as liters of acid per liter of solution, and the solution comes out in the units of the variables. A term whose unit does not match the others points to a mistake in the setup.

In plain words Before solving, write down what each letter measures and in which unit. Then read each equation with its units to make sure it only adds like quantities.

Example Back to the acid solution. Let $x$ be the number of liters of the 20% solution and $y$ the number of liters of the 50% solution. Counting liters of solution and liters of acid gives

\begin{gathered} x + y = 12, \\[0.4em] 0.2x + 0.5y = 0.3 \cdot 12 = 3.6. \end{gathered}

The term $0.2x$ is liters of acid per liter times liters, which is liters of acid, so every term of the second equation is in liters of acid. Substituting $x = 12 - y$ gives $2.4 + 0.3y = 3.6$ , so $y = 4$ and $x = 8$ . Mix $8$ liters of the 20% solution with $4$ liters of the 50% solution.

Check yourself

Solve the system $2x + 3y = 12$ and $5x - 3y = 9$ by elimination.

Show the answer

Adding the equations removes $y$ and gives $7x = 21$ , so $x = 3$ . Then $2 \cdot 3 + 3y = 12$ gives $y = 2$ . The solution is $(3, 2)$ , and the second equation checks, since $5 \cdot 3 - 3 \cdot 2 = 9$ .

How many solutions does the system $x - 2y = 4$ and $-3x + 6y = 5$ have?

Show the answer

None. Three times the first equation is $3x - 6y = 12$ , and adding it to the second gives $0 = 17$ , which is false. Both lines have slope $\frac{1}{2}$ but different intercepts, so they are parallel and the system is inconsistent.

Solve the system $x + 2y + z = 7$ , $2x + y - z = 2$ and $3x - y + 2z = 5$ .

Show the answer

Adding the first two equations removes $z$ and gives $3x + 3y = 9$ , or $x + y = 3$ . Twice the second equation plus the third also removes $z$ and gives $7x + y = 9$ . Subtracting, $6x = 6$ , so $x = 1$ and $y = 2$ . The first equation then gives $z = 7 - 1 - 4 = 2$ . The solution is $(1, 2, 2)$ .

Linear Systems in Two and Three Variables

Key terms

Simultaneous equations

Intersections

Substitution

Elimination

One, none or infinitely many solutions

Inconsistent systems

Dependent systems

Units

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus