Lesson 11.1: Systems of Linear Equations: Two Variables

New Concept

Explore systems of linear equations, where two lines intersect. We'll find their common solution $(x,y)$ using graphing, substitution, and addition, and learn to identify when systems have one, infinite, or no solutions.

What’s next

Next, you’ll master solving these systems through interactive examples and practice cards covering graphing, substitution, and the addition method.

Property

A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently.

There are three types of systems of linear equations in two variables, and three types of solutions.

• An independent system has exactly one solution pair $(x, y)$. The point where the two lines intersect is the only solution.
• An inconsistent system has no solution. The two lines are parallel and will never intersect.
• A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.

Examples

• Is the ordered pair $(2, 1)$ a solution to the system $x + y = 3$ and $2x - y = 3$? Substituting gives $2 + 1 = 3$ (True) and $2(2) - 1 = 3$ (True). Yes, it is the solution.
• Is the ordered pair $(1, 4)$ a solution to the system $y = 3x + 1$ and $y = -x + 5$? Substituting gives $4 = 3(1) + 1$ (True) and $4 = -(1) + 5$ (True). Yes, it is the solution.
• Is the ordered pair $(-1, 3)$ a solution to the system $y - x = 4$ and $2x + y = 0$? Substituting gives $3 - (-1) = 4$ (True) and $2(-1) + 3 = 1$ (False). No, it is not a solution.

Property

For a system of linear equations in two variables, we can find the solution by graphing the system of equations on the same set of axes. The point of intersection of the two lines is the solution to the system. This method allows you to visually identify the solution and the type of system (independent, inconsistent, or dependent).

Examples

• To solve the system $y = x + 1$ and $y = -2x + 7$, graph both lines. You'll see they intersect at the point $(2, 3)$, which is the solution.
• For the system $x + y = 6$ and $x - y = 2$, first write them as $y = -x + 6$ and $y = x - 2$. Graphing these lines shows they intersect at $(4, 2)$.
• Consider the system $y = 2x$ and $y = 4x - 4$. Graphing both lines on the same axes reveals their intersection point is $(2, 4)$, which is the solution.

Explanation

Solving by graphing is like drawing two paths on a map. The point where the paths cross is the treasure—the unique $(x, y)$ solution that works for both equations. If they never cross, there is no solution!

Property

The substitution method involves solving one of the equations for one variable and then substituting the result into the second equation to solve for the second variable.

Given a system of two equations in two variables, solve using the substitution method.

1. Solve one of the two equations for one of the variables in terms of the other.
2. Substitute the expression for this variable into the second equation, then solve for the remaining variable.
3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
4. Check the solution in both equations.

Examples

• For the system $y = 2x + 1$ and $3x + y = 11$, substitute $2x + 1$ for $y$ in the second equation: $3x + (2x + 1) = 11$, which gives $5x = 10$, so $x = 2$. Then $y = 2(2) + 1 = 5$. The solution is $(2, 5)$.
• Solve the system $x - y = 4$ and $2x + 3y = 18$. From the first equation, $x = y + 4$. Substitute into the second: $2(y + 4) + 3y = 18$, so $5y + 8 = 18$, giving $y = 2$. Then $x = 2 + 4 = 6$. The solution is $(6, 2)$.
• In the system $2a + b = 5$ and $a - 2b = 5$, solve the first for $b$: $b = 5 - 2a$. Substitute: $a - 2(5 - 2a) = 5$, so $5a - 10 = 5$, which means $a = 3$. Then $b = 5 - 2(3) = -1$. The solution is $(3, -1)$.

Property

The addition method involves adding two equations so that one variable is eliminated. This may require multiplying one or both equations by a constant to ensure the coefficients of one variable are opposites.

Given a system of equations, solve using the addition method.

1. Write both equations with variables on the left and constants on the right.
2. Line up corresponding variables. If needed, multiply one or both equations so one variable has opposite coefficients.
3. Add the equations to eliminate one variable and solve for the other.
4. Substitute the value back into an original equation to find the second variable.
5. Check the solution.

Examples

• For the system $x + y = 10$ and $x - y = 4$, adding the equations gives $2x = 14$, so $x = 7$. Substituting back, $7 + y = 10$, so $y = 3$. The solution is $(7, 3)$.
• Solve $2x + 3y = 8$ and $x + 2y = 5$. Multiply the second equation by $-2$ to get $-2x - 4y = -10$. Adding this to the first equation gives $-y = -2$, so $y = 2$. Then $x + 2(2) = 5$, so $x = 1$. The solution is $(1, 2)$.
• For $3a + 2b = 7$ and $2a - 5b = -8$, multiply the first by $5$ and the second by $2$ to get $15a + 10b = 35$ and $4a - 10b = -16$. Adding them yields $19a = 19$, so $a = 1$. Then $3(1) + 2b = 7$, so $2b=4$, and $b = 2$. The solution is $(1, 2)$.

Property

An inconsistent system consists of parallel lines that have the same slope but different $y$-intercepts. There is no solution, and attempting to solve it results in a false statement, such as $12 = 0$.

A dependent system consists of two equations that represent the same line. There are an infinite number of solutions. Solving a dependent system results in an identity, such as $0 = 0$. The general solution can be expressed in terms of one variable, like $\left(x, \frac{1}{3}x+\frac{2}{3}\right)$

Examples

• Solve the system $y = 2x + 5$ and $y = 2x - 1$. Substituting the first into the second gives $2x + 5 = 2x - 1$, which simplifies to $5 = -1$. This is false, so there is no solution. The system is inconsistent.
• Solve the system $x + y = 3$ and $2x + 2y = 6$. Multiply the first equation by $-2$ to get $-2x - 2y = -6$. Adding this to the second equation gives $0 = 0$. This is an identity, so there are infinite solutions. The system is dependent.
• Consider $x - 3y = 4$ and $-2x + 6y = 1$. Multiply the first equation by $2$ to get $2x - 6y = 8$. Adding this to the second equation gives $0 = 9$. This is a contradiction, so the system is inconsistent and has no solution.

Property

The cost function, $C(x)$, calculates the total costs of doing business. The revenue function, $R(x)$, calculates the total money earned from sales. The point where cost equals revenue, $C(x) = R(x)$, is the break-even point. The profit function is the revenue minus the cost: $P(x) = R(x) - C(x)$. A profit is made when $P(x) > 0$.

Examples

• A company has costs $C(x) = 15x + 2000$ and revenue $R(x) = 35x$. The break-even point is where $15x + 2000 = 35x$, which gives $20x = 2000$, so $x = 100$. They must sell 100 units to break even.
• Given cost $C(x) = 0.50x + 10000$ and revenue $R(x) = 1.50x$, find the profit function. $P(x) = R(x) - C(x) = 1.50x - (0.50x + 10000) = x - 10000$. The company profits after selling 10,000 units.
• A ticket seller has a cost function $C(x) = 10x + 5000$ and a revenue function $R(x) = 50x$. To find the break-even point, set $C(x) = R(x)$, so $10x + 5000 = 50x$. This simplifies to $40x = 5000$, so $x = 125$. They must sell 125 tickets.

Explanation

The break-even point is where a business is not losing money but also not making any profit yet. It is the number of items you must sell to cover all production costs. Sell more, and you are making a profit!