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Systems of Linear Equations: Two Variables |
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A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions.
In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations{: data-type="term"} consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.
In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.
The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.
In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system{: data-type="term"} of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.
Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different *y-*intercepts. There are no points common to both lines; hence, there is no solution to the system.
-
An independent system{: data-type="term"} has exactly one solution pair ( x,y ).
The point where the two lines intersect is the only solution.
-
An inconsistent system{: data-type="term"} has no solution. Notice that the two lines are parallel and will never intersect.
-
A dependent system{: data-type="term"} 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.
[link] compares graphical representations of each type of system.
- Substitute the ordered pair into each equation in the system.
- Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution. {: type="1"}
is a solution to the given system of equations.
into both equations.
satisfies both equations, so it is the solution to the system.
is a solution to the following system.
There are multiple methods of solving systems of linear equations. For a system of linear equations{: data-type="term" .no-emphasis} in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.
The lines appear to intersect at the point ( −3,−2 ).
We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.
so the system is independent.
Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.
Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations{: data-type="term" .no-emphasis} that are more precise than graphing. One such method is solving a system of equations by the substitution method{: data-type="term"}, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.
- Solve one of the two equations for one of the variables in terms of the other.
- Substitute the expression for this variable into the second equation, then solve for the remaining variable.
- 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.
- Check the solution in both equations. {: type="1"}
for y
in the second equation.
into the first equation and solve for y.
Check the solution by substituting ( 8,3 )
into both equations.
Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.
A third method of solving systems of linear equations{: data-type="term" .no-emphasis} is the addition method{: data-type="term"}. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.
- Write both equations with x- and y-variables on the left side of the equal sign and constants on the right.
- Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
- Solve the resulting equation for the remaining variable.
- Substitute that value into one of the original equations and solve for the second variable.
- Check the solution by substituting the values into the other equation. {: type="1"}
in the second equation, –1, is the opposite of the coefficient of x
in the first equation, 1. We can add the two equations to eliminate x
without needing to multiply by a constant.
we can solve the resulting equation for y.
into one of the original equations and solve for x.
Check the solution in the first equation.
in it and the second equation has x.
So if we multiply the second equation by −3,
the x-terms will add to zero.
into one of the original equations and solve for x.
See [link]. Check the solution in the original second equation.
and the other has 5x.
The least common multiple is 10x
so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate x
by multiplying the first equation by −5
and the second equation by 2.
into the original first equation.
Check it in the other equation.
so that we can eliminate the x-variable.
into the first equation.
Check it in the other equation.
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system{: data-type="term" .no-emphasis} consists of parallel lines that have the same slope but different y
-intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12=0.
the most obvious step is to use substitution.
Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
Recall that a dependent system{: data-type="term" .no-emphasis} of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0=0.
If we multiply both sides of the first equation by −3,
then we will be able to eliminate the x
-variable.
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function{: data-type="term"} is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R=xp,
where x=
quantity and p=
price. The revenue function is shown in orange in [link].
The cost function{: data-type="term"} is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in [link]. The x
-axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point{: data-type="term"}. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function{: data-type="term"} is the revenue function minus the cost function, written as P(x)=R(x)−C(x).
Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
and the revenue function R(x)=1.55x,
find the break-even point and the profit function.
to replace function notation.
from the first equation into the second equation and solve for x.
into either the cost function or the revenue function.
The profit function is found using the formula P(x)=R(x)−C(x).
We see from the graph in [link] that the profit function has a negative value until x=50,000,
when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.
for children and $50.00
for adults. On a certain day, attendance at the circus is 2,000
and the total gate revenue is $70,000.
How many children and how many adults bought tickets?
The total number of people is 2,000.
We can use this to write an equation for the number of people at the circus that day.
by the number of children, 25c.
The revenue from all adults can be found by multiplying $50.00
by the number of adults, 50a.
The total revenue is $70,000.
We can use this to write an equation for the revenue.
or a.
We will solve for a.
in the second equation for a
and solve for c.
into the first equation to solve for a.
children and 800
adults bought tickets to the circus that day.
for children and $12.00
for adults. If 1,650
meal tickets were bought for a total of $14,200,
how many children and how many adults bought meal tickets?
- A system of linear equations consists of two or more 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. See [link].
- Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
- One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See [link].
- Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See [link].
- A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See [link].
- It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See [link], [link], and [link].
- Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See [link].
- The solution to a system of dependent equations will always be true because both equations describe the same line. See [link].
- Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See [link] and [link].
or y
), graphically, or by addition.
For the following exercises, determine whether the given ordered pair is a solution to the system of equations.
and (4,0)
and (−6,1)
and (2,3)
and (−1,1)
and (3,5)
For the following exercises, solve each system by substitution.
For the following exercises, solve each system by addition.
For the following exercises, solve each system by any method.
For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions.
For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth.
For the following exercises, solve each system in terms of A,B,C,D,E,
and F
where A–F
are nonzero numbers. Note that A≠B
and AE≠BD.
For the following exercises, solve for the desired quantity.
and a revenue function R=20x.
Find the break-even point.
and a revenue function R(x)=5x.
When does the company start to turn a profit?
and a revenue function R(x)=200x.
What is the break-even point?
where x
is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point?
If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function.
For the following exercises, use a system of linear equations with two variables and two equations to solve.
interest, and the investor earns $737.50 annual interest, how much was invested in each account?
addition method : an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable ^
break-even point : the point at which a cost function intersects a revenue function; where profit is zero ^
consistent system : a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system ^
cost function : the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs ^
dependent system : a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system ^
inconsistent system : a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common ^
independent system : a system of linear equations with exactly one solution pair ( x,y ) ^
profit function : the profit function is written as P(x)=R(x)−C(x),
revenue minus cost ^
revenue function : the function that is used to calculate revenue, simply written as R=xp,
where x=
quantity and p=
price ^
substitution method : an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable ^
system of linear equations : a set of two or more equations in two or more variables that must be considered simultaneously.