Lesson 2: Linear Systems

New Concept

A linear system uses two or more equations to solve a problem with multiple unknowns. The goal is to find a single ordered pair, like $(x, y)$, that satisfies all equations—the unique point where their graphs intersect.

What’s next

Next, you'll see how graphing reveals the solution. Get ready for interactive examples and practice cards to pinpoint where the lines cross.

Property

A pair of linear equations like $3t + 6r = 48$ and $5t + 2r = 32$ is an example of a linear system of two equations in two unknowns (or a $2 \times 2$ linear system). A solution to the system is an ordered pair of numbers, $(t, r)$, that satisfies both equations in the system.

Examples

• To check if $(5, -2)$ is a solution to the system $3x - 2y = 19$ and $4x + 5y = 10$, we substitute the values. $3(5) - 2(-2) = 19$ is true, and $4(5) + 5(-2) = 10$ is also true, so it is the solution.

• Is $(3, 5)$ the solution to the system $a + b = 8$ and $2a - b = 1$? We check both equations. $3 + 5 = 8$ is true, and $2(3) - 5 = 1$ is also true. Yes, it is the solution.

Property

Every point on the graph of an equation represents a solution to that equation. A solution to a system of two equations must be a point that lies on both graphs. Therefore, a solution to the system is a point where the two graphs intersect.

Examples

• For the system $y = x + 2$ and $y = -x + 4$, graphing the lines shows they intersect at the point $(1, 3)$. Therefore, the solution to the system is $x=1, y=3$.

• To solve the system $y = 1.7x + 0.4$ and $y = 4.1x + 5.2$, we can graph both lines. The intersection point occurs at $(-2, -3)$, which is the solution to the system.

Property

There are three types of $2 \times 2$ linear system:

1. Consistent and independent system. The graphs of the two lines intersect in exactly one point. The system has exactly one solution.
2. Inconsistent system. The graphs of the equations are parallel lines and hence do not intersect. An inconsistent system has no solutions.
3. Dependent system. All the solutions of one equation are also solutions to the second equation. The graphs of the two equations are the same line. A dependent system has infinitely many solutions.

Examples

• The system $y = 2x + 3$ and $y = 2x - 1$ is inconsistent. The lines have the same slope ($2$) but different y-intercepts, so they are parallel and never intersect. There is no solution.

• The system $x+y=5$ and $2x+2y=10$ is dependent. The second equation is just the first multiplied by 2. Their graphs are the same line, resulting in infinitely many solutions.

Property

The demand equation gives the number of units of the product that consumers will buy at a given price. The supply equation gives the number of units that the producer will supply at that price. The price at which the supply and demand are equal is called the equilibrium price.

Examples

• A mill's supply is $y = 400x$ and demand is $y = 6000 - 100x$, where $x$ is the price. At equilibrium, $400x = 6000 - 100x$. Solving for $x$ gives an equilibrium price of 12 dollars, where 4800 yards are sold.

• A farmer's supply of pumpkins is $S = 20p - 50$ and demand is $D = 100 - 10p$, where $p$ is price. Equilibrium occurs when $S=D$, so $20p - 50 = 100 - 10p$. The equilibrium price is 5 dollars.

Property

A business venture calculates its profit by subtracting its costs from its revenue. The formula is Profit = Revenue - Cost. If the company's revenue exactly equals its costs (so that their profit is zero), we say that the business venture will break even.

Examples

• A company's cost is $C = 200 + 4x$ and its revenue is $R = 12x$. To break even, $C=R$, so $200 + 4x = 12x$. Solving for $x$ gives $x=25$. The company must sell 25 pendants to break even.

• A bakery has daily costs of $C = 90 + 0.60x$ and revenue of $R = 1.50x$. The break-even point is when $90 + 0.60x = 1.50x$. Solving this shows the bakery must sell 100 loaves to break even.