What is the Solution to the System Graphed Below

What is the Solution to the System Graphed Below.

Learning Outcomes

  • Graph systems of equations
    • Graph a system of two linear equations
    • Graph a system of two linear inequalities
  • Evaluate ordered pairs every bit solutions to systems
    • Determine whether an ordered pair is a solution to a system of linear equations
    • Make up one’s mind whether an ordered pair is a solution to a system of linear inequalities
  • Allocate solutions to systems
    • Place what type of solution a system will have based on its graph

The mode a river flows depends on many variables including how big the river is, how much water it contains, what sorts of things are floating in the river, whether or not information technology is raining, so forth. If you want to best describe its menses, you must have into business relationship these other variables. A organisation of linear equations can assistance with that.

A
organisation of linear equations
consists of ii or more than linear equations made up of two or more variables such that all equations in the system are considered simultaneously. You volition observe systems of equations in every application of mathematics. They are a useful tool for discovering and describing how behaviors or processes are interrelated. It is rare to find, for case, a design of traffic flow that that is only afflicted by conditions. Accidents, fourth dimension of mean solar day, and major sporting events are merely a few of the other variables that tin bear on the flow of traffic in a city. In this section, we will explore some basic principles for graphing and describing the intersection of ii lines that brand up a system of equations.

Graph a organization of linear equations

In this section, nosotros will expect at systems of linear equations and inequalities in two variables.  Starting time, we will practice graphing two equations on the same set up of axes, and and then we will explore the different considerations yous need to make when graphing 2 linear inequalities on the same set of axes. The aforementioned techniques are used to graph a system of linear equations as you have used to graph single linear equations. Nosotros can use tables of values, slope and
y-intercept, or
x– and
y-intercepts to graph both lines on the same fix of axes.

For case, consider the following system of linear equations in two variables.

[latex]\begin{array}{r}2x+y=-eight\\ ten-y=-i\end{array}[/latex]

Let’due south graph these using slope-intercept form on the same fix of axes. Call up that slope-intercept grade looks like [latex]y=mx+b[/latex],  then we will desire to solve both equations for [latex]y[/latex].

First, solve for y in [latex]2x+y=-viii[/latex]

[latex]\begin{array}{c}2x+y=-8\\ y=-2x – eight\terminate{array}[/latex]

2nd, solve for y in [latex]x-y=-1[/latex]

[latex]\begin{array}{r}x-y=-1\,\,\,\,\,\\ y=10+1\end{assortment}[/latex]

The organization is now written as

[latex]\begin{assortment}{c}y=-2x – 8\\y=x+1\end{array}[/latex]

Now you tin graph both equations using their slopes and intercepts on the same gear up of axes, as seen in the figure below. Note how the graphs share ane betoken in common. This is their bespeak of intersection, a betoken that lies on both of the lines.  In the next section nosotros will verify that this bespeak is a solution to the system.

A graph of two lines running through the point negative 3, negative 2. The first line's equation is y equals minus 2x minus 8. The second line's equation is y equals x+1.

 In the following example, you will be given a organisation to graph that consists of two parallel lines.

Case

Graph the system [latex]\brainstorm{assortment}{c}y=2x+1\\y=2x-3\terminate{array}[/latex] using the slopes and y-intercepts of the lines.

In the next example, you will exist given a system whose equations look different, but after graphing, turn out to be the same line.

Example

Graph the arrangement [latex]\begin{array}{c}y=\frac{1}{ii}10+2\\2y-x=4\stop{array}[/latex] using the x – and y-intercepts.

Graphing a system of linear equations consists of choosing which graphing method you want to use and drawing the graphs of both equations on the aforementioned set of axes. When you graph a system of linear inequalities on the same set of axes, at that place are a few more things you will need to consider.

Graph a system of ii inequalities

Remember from the module on graphing that the graph of a single linear inequality splits the
coordinate aeroplane
into 2 regions. On one side prevarication all the solutions to the inequality. On the other side, at that place are no solutions. Consider the graph of the inequality [latex]y<2x+5[/latex].

An upward-sloping dotted line with the region below it shaded. The shaded region is labeled y is less than 2x+5. A is equal to (-1,1). B is equal to (3,1).

The dashed line is [latex]y=2x+5[/latex]. Every ordered pair in the shaded area below the line is a solution to [latex]y<2x+5[/latex], every bit all of the points below the line volition make the inequality truthful. If yous doubtfulness that, effort substituting the
ten
and
y
coordinates of Points A and B into the inequality—you lot’ll run into that they piece of work. So, the shaded area shows all of the solutions for this inequality.

The boundary line divides the coordinate plane in half. In this case, it is shown every bit a dashed line as the points on the line don’t satisfy the inequality. If the inequality had been [latex]y\leq2x+5[/latex], then the boundary line would have been solid.

Let’s graph another inequality: [latex]y>−10[/latex]. Y’all can check a couple of points to decide which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the expanse in a higher place the line. The line is dashed as points on the line are not true.

Downward-sloping dotted line with the region above it shaded. The shaded region is y is greater than negative x. Point M=(-2,3). Point N=(4,-1).

To create a system of inequalities, yous need to graph two or more inequalities together. Let’s utilise [latex]y<2x+5[/latex] and [latex]y>−x[/latex] since we have already graphed each of them.

The two previous graphs combined. A blue dotted line with the region above shaded and labeled y is greater than negative x. A red dotted line with the region below it shaded and labeled y is less than 2x+5. The region where the shaded areas overlap is labeled y is greater than negative x and y is less than 2x+5. The point M equals (-2,3) and is in the blue shaded region. The point A equals (-1,-1) and is in the red shaded region. The point B equals (3,1) and is in the purple overlapping region. The point N equals (4,-1) and is also in the purple overlapping region.

The purple expanse shows where the solutions of the two inequalities overlap. This expanse is the solution to the
organisation of inequalities. Any point within this regal region will exist true for both [latex]y>−x[/latex] and [latex]y<2x+5[/latex].

In the next case, you are given a arrangement of two inequalities whose boundary lines are parallel to each other.

Examples

Graph the system [latex]\begin{array}{c}y\ge2x+i\\y\lt2x-3\end{array}[/latex]

In the next department, nosotros will encounter that points tin exist solutions to systems of equations and inequalities.  We will verify algebraically whether a point is a solution to a linear equation or inequality.

Decide whether an ordered pair is a solution for a system of linear equations

image009-2

The lines in the graph higher up are defined as

[latex]\brainstorm{assortment}{r}2x+y=-viii\\ x-y=-1\end{assortment}[/latex].

They cross at what appears to exist [latex]\left(-three,-2\right)[/latex].

Using algebra, we can verify that this shared point is actually [latex]\left(-3,-2\right)[/latex] and not [latex]\left(-2.999,-1.999\right)[/latex]. By substituting the
x– and
y-values of the ordered pair into the equation of each line, you lot tin can exam whether the point is on both lines. If the commutation results in a true statement, and so you lot have found a solution to the system of equations!

Since the solution of the system must be a solution to
all
the equations in the organisation, you will need to cheque the point in each equation. In the following instance, we will substitute -iii for
10
and -2 for
y
in each equation to exam whether it is actually the solution.

Example

Is [latex]\left(-3,-2\right)[/latex] a solution of the system

[latex]\begin{array}{r}2x+y=-8\\ ten-y=-i\stop{array}[/latex]

Example

Is (3, nine) a solution of the system

[latex]\begin{assortment}{r}y=3x\\2x–y=6\stop{assortment}[/latex]

Think About It

Is [latex](−2,4)[/latex] a solution for the organisation

[latex]\begin{array}{r}y=2x\\3x+2y=1\end{array}[/latex]

Earlier you do any calculations, wait at the point given and the beginning equation in the system.  Tin you predict the answer to the question without doing any algebra?

Remember that in order to be a solution to the system of equations, the values of the point must be a solution for both equations. One time you observe one equation for which the point is fake, you have determined that information technology is not a solution for the system.

We can use the aforementioned method to determine whether a point is a solution to a system of linear inequalities.

Determine whether an ordered pair is a solution to a system of linear inequalities

image014

On the graph above, you tin can see that the points B and Due north are solutions for the organization because their coordinates will make both inequalities true statements.

In dissimilarity, points M and A both prevarication outside the solution region (purple). While point M is a solution for the inequality [latex]y>−x[/latex] and point A is a solution for the inequality [latex]y<2x+5[/latex], neither betoken is a solution for the
system. The following example shows how to examination a point to come across whether it is a solution to a system of inequalities.

Case

Is the point (2, 1) a solution of the system [latex]x+y>1[/latex] and [latex]2x+y<8[/latex]?

Hither is a graph of the system in the example above. Notice that (ii, ane) lies in the purple expanse, which is the overlapping area for the two inequalities.

Two dotted lines, one red and one blue. The region below the blue dotted line is shaded and labeled 2x+y is less than 8. The region above the dotted red line is shaded and labeled x+y is greater than 1. The overlapping shaded region is purple and is labeled x+y is greater than 1 and 2x+y is less than 8. The point (2,1) is in the overlapping purple region.

Example

Is the point (ii, ane) a solution of the system [latex]x+y>1[/latex] and [latex]3x+y<4[/latex]?

Hither is a graph of this system. Observe that (2, ane) is not in the purple surface area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).

A downward-sloping bue dotted line with the region below shaded and labeled 3x+y is less than 4. A downward-sloping red dotted line with the region above it shaded and labeled x+y is greater than 1. An overlapping purple shaded region is labeled x+y is greater than 1 and 3x+y is less than 4. A point (2,1) is in the red shaded region, but not the blue or overlapping purple shaded region.

As shown to a higher place, finding the solutions of a system of inequalities tin can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that bear witness the entire process of defining the region of solutions on a graph for a system of ii linear inequalities.  The general steps are outlined beneath:

  • Graph each inequality equally a line and decide whether it will be solid or dashed
  • Determine which side of each boundary line represents solutions to the inequality by testing a point on each side
  • Shade the region that represents solutions for both inequalities

Case

Shade the region of the graph that represents solutions for both inequalities.  [latex]x+y\geq1[/latex] and [latex]y–x\geq5[/latex].

In this department nosotros take seen that solutions to systems of linear equations and inequalities tin exist ordered pairs. In the adjacent section, nosotros will piece of work with systems that have no solutions or infinitely many solutions.

Utilize a graph to classify solutions to systems

Recall that a linear equation graphs as a line, which indicates that all of the points on the line are solutions to that linear equation. In that location are an infinite number of solutions. As we saw in the terminal section, if you have a arrangement of linear equations that intersect at one betoken, this point is a solution to the organisation.  What happens if the lines never cross, as in the case of parallel lines?  How would yous depict the solutions to that kind of system? In this department, we volition explore the iii possible outcomes for solutions to a system of linear equations.

Three possible outcomes for solutions to systems of equations

Remember that the solution for a arrangement of equations is the value or values that are true for
all
equations in the system. In that location are three possible outcomes for solutions to systems of linear equations.  The graphs of equations inside a system can tell yous how many solutions exist for that system. Expect at the images below. Each shows two lines that make up a system of equations.

One Solution No Solutions Infinite Solutions
Two intersecting lines. Two parallel lines Two identical lines that overlap so that they appear to be one line
If the graphs of the equations intersect, then there is ane solution that is true for both equations. If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations. If the graphs of the equations are the same, and then at that place are an infinite number of solutions that are true for both equations.
  • One Solution: When a system of equations intersects at an ordered pair, the system has ane solution.
  • Infinite Solutions:
    Sometimes the two equations will graph as the same line, in which case we have an infinite number of solutions.
  • No Solution:
    When the lines that brand upwards a system are parallel, at that place are no solutions considering the two lines share no points in common.

Example

Using the graph of [latex]\begin{array}{r}y=10\\x+2y=half-dozen\end{assortment}[/latex], shown beneath, determine how many solutions the organisation has.

A line labeled x+2y=6 and a line labeled y=x.

Case (Advanced)

Using the graph of [latex]\begin{assortment}{r}y=3.5x+0.25\\14x–4y=-4.5\finish{assortment}[/latex], shown below, determine how many solutions the system has.

Two parallel lines. One line is y=3.5x+0.25. The other line is 14x-4y=-4.5.

Example

How many solutions does the system [latex]\brainstorm{array}{r}y=2x+1\\−4x+2y=2\cease{array}[/latex] accept?

In the next department, we volition learn some algebraic methods for finding solutions to systems of equations.  Recall that linear equations in i variable can have ane solution, no solution, or many solutions and nosotros can verify this algebraically.  We will employ the aforementioned ideas to classify solutions to systems in ii variables algebraically.

What is the Solution to the System Graphed Below

Source: https://courses.lumenlearning.com/beginalgebra/chapter/introduction-to-systems-of-linear-equations/