Linear Equation Systems

Values That Satisfy Several Equations at Once

A total price can be produced by many different item combinations. Additional information, such as the total number of items, narrows those possibilities. A linear equation system writes each piece of information as an equation. Its solution must satisfy every equation at once.

Linear Equation System

A linear equation system is a collection of two or more linear equations that must be satisfied simultaneously. Each linear equation has the form:

a_1x_1 + a_2x_2 + \ldots + a_nx_n = b

Here $a_1, a_2, \ldots, a_n$ are coefficients, $x_1, x_2, \ldots, x_n$ are variables, and $b$ is a constant.

Two Variable Linear Equation Systems

A system with two variables, usually $x$ and $y$ , can contain any number of equations. A system with two equations has the form:

\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}

Example:

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ x + y = 10 & \ldots (2) \end{cases}

The solution to this system is the pair of values $(x,y)$ that satisfies both equations.

Three Variable Linear Equation Systems

A system may contain any number of equations. To determine unique values for $x,y,z$ , however, the equations must provide three independent constraints. A common square form is:

\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}

Example:

\begin{cases} x + 2y + 3z = 27 \\ x + y + z = 16 \\ x = 6 \end{cases}

Solving Linear Equation Systems

A linear equation system is solved by removing one variable at a time until a single value remains. Substitution and elimination each remove one variable, and both routes work on the same kind of system. Compare at the end when each one is the shorter route.

Substitution Method

In the substitution method, express one variable in terms of another, then insert that expression into the other equation. The following example shows each step.

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ x + y = 10 & \ldots (2) \end{cases}

Step 1: Express one variable from the simpler equation.

From equation (2): $x + y = 10$ , we express $x$ in terms of $y$ :

x = 10 - y \ldots (3)

Step 2: Substitute into the other equation.

Insert equation (3) into equation (1):

2(10 - y) + 3y = 21

Step 3: Solve the resulting equation.

20 - 2y + 3y = 21

Step 4: Substitute the calculated value to find the other variable.

Substitute the value $y = 1$ from equation (4) into equation (3):

x = 10 - y = 10 - 1 = 9

The solution is $x = 9$ and $y = 1$ .

Elimination Method

In the elimination method, add or subtract two equations to remove one variable. The following calculation solves the same example.

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ x + y = 10 & \ldots (2) \end{cases}

Step 1: Match the coefficients of one variable.

Multiply equation (2) by $2$ to match the coefficient of $x$ :

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ 2x + 2y = 20 & \ldots (3) \end{cases}

Step 2: Eliminate the variable by subtracting the equations.

Subtract equation (3) from equation (1):

(2x + 3y) - (2x + 2y) = 21 - 20

Step 3: Use the value of $y$ to find $x$ .

Substitute the value $y = 1$ from equation (4) into equation (2):

x + 1 = 10

The solution is $x = 9$ and $y = 1$ .

Verify:

Equation (1): $2(9) + 3(1) = 18 + 3 = 21$ ✓
Equation (2): $9 + 1 = 10$ ✓

Check Your Understanding

Find the pair that satisfies both equations:

\begin{cases} 3x+2y=16 \\ x-y=2 \end{cases}

Worked Solution

The second equation gives $x=y+2$ . Substitute this expression into the first equation:

3(y+2)+2y=16

Back-substitute into $x=y+2$ :

x=2+2=4

Both original equations confirm the result:

3(4)+2(2)=16

The solution is $(x,y)=(4,2)$ .

Linear Equation Systems in Context

Applied systems start from a word problem. The description has to be translated into equations before any method is applied, and the solution has to be checked against the original question.

Basketball Scoring

In five-on-five basketball, a successful free throw scores one point. A successful shot from the two-point area scores two points, and one from the three-point area scores three.

Wijaya scores 27 points from 16 successful shots, including six free throws. Missed shots are not counted in this problem. Let $a$ , $b$ , and $c$ be the numbers of successful one-, two-, and three-point shots respectively. Each variable is a nonnegative integer.

Translate the total points, successful shots, and successful free throws into three equations:

\begin{cases}a+2b+3c=27\\a+b+c=16\\a=6\end{cases}

Substituting $a = 6$ into the second equation:

6 + b + c = 16

Substitute $a = 6$ into the first equation:

6 + 2b + 3c = 27

From these two equations:

\begin{cases} 2b + 3c = 21 \\ b + c = 10 \end{cases}

Multiply the second equation by two, then subtract it from the first to eliminate the two-point shot count:

\begin{aligned}(2b+3c)-2(b+c)&=21-20\\c&=1\\b&=10-1=9\end{aligned}

Wijaya made $6$ free throws, $9$ $2$ -point shots, and $1$ $3$ -point shot.

Interpreting Solutions

For two equations that each represent a line, there are three possibilities. Intersecting lines have one solution, distinct parallel lines have no solution, and coincident lines have infinitely many solutions.

In three dimensions, a linear equation with at least one nonzero variable coefficient represents a plane. The solution set is the common intersection of all planes. An intersection of two planes need not satisfy a third plane. For example, $x=0$ and $y=0$ intersect along the z-axis, but no point on that axis satisfies $x+y=1$ .

For three or more planes, the common intersection can be a point, a line, a plane, or empty.

Visualizing Linear Equation Systems

The scenes below show three line configurations and how the number of common points decides the number of solutions.

Linear Equation System with One Solution

Two lines intersecting at a single point.

Linear Equation System with No Solution

Two parallel lines (no intersection).

Linear Equation System with Infinitely Many Solutions

The second equation is

2

times the first, so both graphs coincide and appear as one line. The line for the first equation is drawn wider so both equations remain visible along the same path.

Values That Satisfy Several Equations at Once

Linear Equation System

A linear equation system is a collection of two or more linear equations that must be satisfied simultaneously. Each linear equation has the form:

a_1x_1 + a_2x_2 + \ldots + a_nx_n = b

Here $a_1, a_2, \ldots, a_n$ are coefficients, $x_1, x_2, \ldots, x_n$ are variables, and $b$ is a constant.

Two Variable Linear Equation Systems

A system with two variables, usually $x$ and $y$ , can contain any number of equations. A system with two equations has the form:

\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}

Example:

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ x + y = 10 & \ldots (2) \end{cases}

The solution to this system is the pair of values $(x,y)$ that satisfies both equations.

Three Variable Linear Equation Systems

A system may contain any number of equations. To determine unique values for $x,y,z$ , however, the equations must provide three independent constraints. A common square form is:

\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}

Example:

\begin{cases} x + 2y + 3z = 27 \\ x + y + z = 16 \\ x = 6 \end{cases}

Solving Linear Equation Systems

Substitution Method

In the substitution method, express one variable in terms of another, then insert that expression into the other equation. The following example shows each step.

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ x + y = 10 & \ldots (2) \end{cases}

Step 1: Express one variable from the simpler equation.

From equation (2): $x + y = 10$ , we express $x$ in terms of $y$ :

x = 10 - y \ldots (3)

Step 2: Substitute into the other equation.

Insert equation (3) into equation (1):

2(10 - y) + 3y = 21

Step 3: Solve the resulting equation.

20 - 2y + 3y = 21

Step 4: Substitute the calculated value to find the other variable.

Substitute the value $y = 1$ from equation (4) into equation (3):

x = 10 - y = 10 - 1 = 9

The solution is $x = 9$ and $y = 1$ .

Elimination Method

In the elimination method, add or subtract two equations to remove one variable. The following calculation solves the same example.

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ x + y = 10 & \ldots (2) \end{cases}

Step 1: Match the coefficients of one variable.

Multiply equation (2) by $2$ to match the coefficient of $x$ :

\begin{cases} 2x + 3y = 21 & \ldots (1) \\ 2x + 2y = 20 & \ldots (3) \end{cases}

Step 2: Eliminate the variable by subtracting the equations.

Subtract equation (3) from equation (1):

(2x + 3y) - (2x + 2y) = 21 - 20

Step 3: Use the value of $y$ to find $x$ .

Substitute the value $y = 1$ from equation (4) into equation (2):

x + 1 = 10

The solution is $x = 9$ and $y = 1$ .

Verify:

Equation (1): $2(9) + 3(1) = 18 + 3 = 21$ ✓
Equation (2): $9 + 1 = 10$ ✓

Check Your Understanding

Find the pair that satisfies both equations:

\begin{cases} 3x+2y=16 \\ x-y=2 \end{cases}

Worked Solution

The second equation gives $x=y+2$ . Substitute this expression into the first equation:

3(y+2)+2y=16

Back-substitute into $x=y+2$ :

x=2+2=4

Both original equations confirm the result:

3(4)+2(2)=16

The solution is $(x,y)=(4,2)$ .

Linear Equation Systems in Context

Applied systems start from a word problem. The description has to be translated into equations before any method is applied, and the solution has to be checked against the original question.

Basketball Scoring

In five-on-five basketball, a successful free throw scores one point. A successful shot from the two-point area scores two points, and one from the three-point area scores three.

Translate the total points, successful shots, and successful free throws into three equations:

\begin{cases}a+2b+3c=27\\a+b+c=16\\a=6\end{cases}

Substituting $a = 6$ into the second equation:

6 + b + c = 16

Substitute $a = 6$ into the first equation:

6 + 2b + 3c = 27

From these two equations:

\begin{cases} 2b + 3c = 21 \\ b + c = 10 \end{cases}

Multiply the second equation by two, then subtract it from the first to eliminate the two-point shot count:

\begin{aligned}(2b+3c)-2(b+c)&=21-20\\c&=1\\b&=10-1=9\end{aligned}

Wijaya made $6$ free throws, $9$ $2$ -point shots, and $1$ $3$ -point shot.

Interpreting Solutions

For three or more planes, the common intersection can be a point, a line, a plane, or empty.

Visualizing Linear Equation Systems

The scenes below show three line configurations and how the number of common points decides the number of solutions.

Linear Equation System with One Solution

Two lines intersecting at a single point.

Linear Equation System with No Solution

Two parallel lines (no intersection).

Linear Equation System with Infinitely Many Solutions

The second equation is

2

times the first, so both graphs coincide and appear as one line. The line for the first equation is drawn wider so both equations remain visible along the same path.