what is the solution to the system of equations shown in the graph?

Learning Outcomes

• Graph systems of equations
  • Graph a system of two linear equations
  • Graph a system of two linear inequalities
• Evaluate ordered pairs as solutions to systems
  • Make up one's mind whether an ordered pair is a solution to a system of linear equations
  • Decide whether an ordered pair is a solution to a organisation of linear inequalities
• Classify solutions to systems
  • Identify what type of solution a system will have based on its graph

The way 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 it is raining, and so along. If yous want to best describe its flow, you must have into account these other variables. A arrangement of linear equations can help with that.

A system of linear equations consists of two or more linear equations made upward of two or more than variables such that all equations in the system are considered simultaneously. You will find 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 detect, for example, a pattern of traffic flow that that is merely affected by weather. Accidents, time of day, and major sporting events are just a few of the other variables that can touch on the flow of traffic in a city. In this department, we will explore some bones principles for graphing and describing the intersection of two lines that make up a organization of equations.

Graph a system of linear equations

In this department, nosotros will wait at systems of linear equations and inequalities in 2 variables.  First, we will practice graphing 2 equations on the aforementioned set of axes, and and so we will explore the different considerations you need to make when graphing ii linear inequalities on the same set of axes. The same techniques are used to graph a system of linear equations as yous take used to graph single linear equations. We tin utilize tables of values, slope and y-intercept, or x– and y-intercepts to graph both lines on the same fix of axes.

For example, consider the post-obit system of linear equations in ii variables.

[latex]\brainstorm{array}{r}2x+y=-8\\ x-y=-ane\terminate{array}[/latex]

Let'south graph these using slope-intercept form on the aforementioned set of axes. Remember that slope-intercept course looks like [latex]y=mx+b[/latex],  so we will want to solve both equations for [latex]y[/latex].

First, solve for y in [latex]2x+y=-8[/latex]

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

Second, solve for y in [latex]10-y=-i[/latex]

[latex]\brainstorm{array}{r}x-y=-one\,\,\,\,\,\\ y=x+1\cease{array}[/latex]

The arrangement is now written as

[latex]\begin{array}{c}y=-2x - viii\\y=ten+ane\end{array}[/latex]

At present you tin can graph both equations using their slopes and intercepts on the same set of axes, as seen in the figure beneath. Note how the graphs share i indicate in common. This is their betoken of intersection, a betoken that lies on both of the lines.  In the next section we volition verify that this indicate is a solution to the system.

 In the post-obit example, y'all will be given a system to graph that consists of ii parallel lines.

Case

Graph the organisation [latex]\begin{array}{c}y=2x+1\\y=2x-iii\end{array}[/latex] using the slopes and y-intercepts of the lines.

In the adjacent instance, you lot will exist given a organisation whose equations look unlike, but after graphing, plow out to be the same line.

Example

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

Graphing a organization of linear equations consists of choosing which graphing method you want to use and drawing the graphs of both equations on the same set of axes. When you graph a organisation of linear inequalities on the same set of axes, in that location are a few more things you volition need to consider.

Graph a organisation of two inequalities

Recollect from the module on graphing that the graph of a unmarried linear inequality splits the coordinate plane into two regions. On one side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality [latex]y<2x+5[/latex].

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], as all of the points below the line will make the inequality truthful. If you lot doubt that, try substituting the ten and y coordinates of Points A and B into the inequality—you'll see that they piece of work. So, the shaded expanse shows all of the solutions for this inequality.

The purlieus line divides the coordinate aeroplane in half. In this case, information technology is shown equally 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.

Permit'due south graph another inequality: [latex]y>−x[/latex]. You can cheque a couple of points to decide which side of the purlieus line to shade. Checking points M and Due north yield true statements. So, we shade the area higher up the line. The line is dashed every bit points on the line are not true.

To create a system of inequalities, yous demand to graph 2 or more than inequalities together. Let's use [latex]y<2x+v[/latex] and [latex]y>−x[/latex] since we have already graphed each of them.

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

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

Examples

Graph the system [latex]\begin{array}{c}y\ge2x+1\\y\lt2x-3\end{array}[/latex]

In the next section, we will see that points can be solutions to systems of equations and inequalities.  Nosotros will verify algebraically whether a betoken is a solution to a linear equation or inequality.

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

The lines in the graph above are defined as

[latex]\begin{assortment}{r}2x+y=-eight\\ 10-y=-ane\end{array}[/latex].

They cross at what appears to be [latex]\left(-3,-2\right)[/latex].

Using algebra, we can verify that this shared point is really [latex]\left(-three,-ii\right)[/latex] and not [latex]\left(-ii.999,-one.999\right)[/latex]. By substituting the x– and y-values of the ordered pair into the equation of each line, you can test whether the point is on both lines. If the substitution results in a true argument, then you accept found a solution to the arrangement of equations!

Since the solution of the system must exist a solution to all the equations in the system, y'all will need to check the betoken in each equation. In the following example, we will substitute -3 for 10 and -two for y in each equation to exam whether information technology is really the solution.

Example

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

[latex]\begin{array}{r}2x+y=-8\\ ten-y=-1\end{array}[/latex]

Instance

Is (3, 9) a solution of the system

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

Call back About It

Is [latex](−ii,4)[/latex] a solution for the system

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

Before yous do any calculations, expect at the point given and the first equation in the organisation.  Can you predict the answer to the question without doing any algebra?

Call up that in order to be a solution to the system of equations, the values of the bespeak must be a solution for both equations. Once yous discover one equation for which the point is false, yous have determined that it is not a solution for the organization.

Nosotros tin employ the same method to determine whether a point is a solution to a organisation of linear inequalities.

Make up one's mind whether an ordered pair is a solution to a system of linear inequalities

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

In contrast, points M and A both prevarication outside the solution region (purple). While indicate K is a solution for the inequality [latex]y>−x[/latex] and signal A is a solution for the inequality [latex]y<2x+5[/latex], neither bespeak is a solution for the system. The following example shows how to test a bespeak to see whether it is a solution to a system of inequalities.

Case

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

Here is a graph of the arrangement in the example above. Notice that (2, 1) lies in the purple area, which is the overlapping expanse for the two inequalities.

Instance

Is the point (2, ane) a solution of the system [latex]ten+y>one[/latex] and [latex]3x+y<4[/latex]?

Here is a graph of this system. Observe that (ii, 1) is not in the imperial area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the 2d inequality (the blue region).

As shown above, finding the solutions of a system of inequalities can exist done past graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire procedure of defining the region of solutions on a graph for a system of two linear inequalities.  The general steps are outlined below:

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

Instance

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

Utilise a graph to allocate 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. At that place are an infinite number of solutions. Every bit nosotros saw in the final department, if you have a arrangement of linear equations that intersect at one signal, this point is a solution to the arrangement.  What happens if the lines never cantankerous, equally in the case of parallel lines?  How would you lot describe the solutions to that kind of system? In this section, we will explore the 3 possible outcomes for solutions to a system of linear equations.

Three possible outcomes for solutions to systems of equations

Recall that the solution for a organization of equations is the value or values that are truthful for all equations in the system. There are three possible outcomes for solutions to systems of linear equations.  The graphs of equations within a organization can tell y'all how many solutions exist for that system. Expect at the images beneath. Each shows two lines that make up a arrangement of equations.

One Solution	No Solutions	Infinite Solutions
If the graphs of the equations intersect, and so there is one solution that is true for both equations.	If the graphs of the equations do not intersect (for example, if they are parallel), then in that location are no solutions that are true for both equations.	If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

• I Solution: When a system of equations intersects at an ordered pair, the system has one solution.
• Infinite Solutions: Sometimes the two equations volition graph every bit the same line, in which case nosotros accept an infinite number of solutions.
• No Solution: When the lines that make up a system are parallel, there are no solutions because the two lines share no points in common.

Case

Using the graph of [latex]\brainstorm{array}{r}y=ten\\x+2y=6\end{array}[/latex], shown below, determine how many solutions the system has.

Example (Advanced)

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

Instance

How many solutions does the arrangement [latex]\begin{assortment}{r}y=2x+1\\−4x+2y=two\stop{array}[/latex] have?

In the next section, nosotros will learn some algebraic methods for finding solutions to systems of equations.  Call back that linear equations in one variable can have i solution, no solution, or many solutions and we tin verify this algebraically.  We will use the aforementioned ideas to classify solutions to systems in ii variables algebraically.

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

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