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# Graphs of Linear Systems

## Graph lines to identify intersection points

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Practice Graphs of Linear Systems
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Solving Systems with One Solution Using Graphing

Two of coin one plus four of coin two equals a total of 70 cents. One of coin one plus five of coin two equals a total of 50 cents. What is the value of each coin?

### Guidance

In this lesson we will be using various techniques to graph the pairs of lines in systems of linear equations to identify the point of intersection or the solution to the system. It is important to use graph paper and a straightedge to graph the lines accurately. Also, you are encouraged to check your answer algebraically as described in the previous lesson.

#### Example A

Graph and solve the system:

yy=x+1=12x2

Solution:

Since both of these equations are written in slope intercept form, we can graph them easily by plotting the y\begin{align*}y-\end{align*}intercept point and using the slope to locate additional points on each line.

The equation, y=x+1\begin{align*}y=-x+1\end{align*}, graphed in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*}, has \begin{align*}y-\end{align*}intercept 1 and slope \begin{align*}- \frac{1}{1}\end{align*}.

The equation, \begin{align*}y=\frac{1}{2}x-2\end{align*}, graphed in \begin{align*}{\color{red}\mathbf{red}}\end{align*}, has \begin{align*}y-\end{align*}intercept -2 and slope \begin{align*}\frac{1}{2}\end{align*}.

Now that both lines have been graphed, the intersection is observed to be the point (2, -1).

Check this solution algebraically by substituting the point into both equations.

Equation 1: \begin{align*}y=-x+1\end{align*}, making the substitution gives: \begin{align*}(-1)=(-2)+1. \end{align*}

Equation 2: \begin{align*}y=\frac{1}{2}x-2\end{align*}, making the substitution gives: \begin{align*}-1=\frac{1}{2}(2)-2. \end{align*}

(2, -1) is the solution to the system.

#### Example B

Graph and solve the system:

Solution: This example is very similar to the first example. The only difference is that equation 1 is not in slope intercept form. We can either solve for \begin{align*}y\end{align*} to put it in slope intercept form or we can use the intercepts to graph the equation. To review using intercepts to graph lines, we will use the latter method.

Recall that the \begin{align*}x-\end{align*}intercept can be found by replacing \begin{align*}y\end{align*} with zero and solving for \begin{align*}x\end{align*}:

Similarly, the \begin{align*}y-\end{align*}intercept is found by replacing \begin{align*}x\end{align*} with zero and solving for \begin{align*}y\end{align*}:

We have two points, (2, 0) and (0, 3) to plot and graph this line. Equation 2 can be graphed using the \begin{align*}y-\end{align*}intercept and slope as shown in Example A.

Now that both lines are graphed we observe that their intersection is the point (4, -3).

Finally, check this solution by substituting it into each of the two equations.

Equation 1: \begin{align*}3x+2y =6; 3(4)+2(-3)=12-6=6 \end{align*}

Equation 2: \begin{align*}y = - \frac{1}{2}x-1; -3= - \frac{1}{2}(4)-1 \end{align*}

#### Example C

In this example we will use technology to solve the system:

This process may vary somewhat based on the technology you use. All directions here can be applied to the TI-83 or 84 (plus, silver, etc) calculators.

Solution: The first step is to graph these equations on the calculator. The first equation must be rearranged into slope intercept form to put in the calculator.

The graph obtained using the calculator should look like this:

The first equation, \begin{align*}y=\frac{2}{3}x-\frac{10}{3}\end{align*}, is graphed in \begin{align*}{\color{blue}\mathbf{blue}}\end{align*}. The second equation, \begin{align*}y=-\frac{2}{3}x+4\end{align*}, is graphed in \begin{align*}{\color{red}\mathbf{red}}\end{align*}.

The solution does not lie on the “grid” and is therefore difficult to observe visually. With technology we can calculate the intersection. If you have a TI-83 or 84, use the CALC menu, select INTERSECT. Then select each line by pressing ENTER on each one. The calculator will give you a “guess.” Press ENTER one more time and the calculator will then calculate the intersection of (5.5, .333...). We can also write this point as \begin{align*}\left(\frac{11}{2},\frac{1}{3}\right)\end{align*}. Check the solution algebraically.

Equation 1: \begin{align*}2x-3y=10; 2\left(\frac{11}{2}\right)-3\left(\frac{1}{3}\right)=11-1=10 \end{align*}

Equation 2: \begin{align*}y=- \frac{2}{3}x+4; - \frac{2}{3}\left(\frac{11}{2}\right)+4=-\frac{11}{3}+\frac{12}{3}=\frac{1}{3} \end{align*}

If you do not have a TI-83 or 84, the commands might be different. Check with your teacher.

Intro Problem Revisit The system of linear equations represented by this situation is:

If you plot both of these linear equations on the same graph, you find that the point of intersection is (25, 5). Therefore coin one has a value of 25 cents and coin two has a value of 5 cents.

### Guided Practice

Solve the following systems by graphing. Use technology for problem 3.

1.

2.

3.

1.

The first line is in slope intercept form and can be graphed accordingly.

The second line is a horizontal line through (0, 2).

The graph of the two equations is shown below. From this graph the solution appears to be (2, 2).

Checking this solution in each equation verifies that it is indeed correct.

Equation 1: \begin{align*}2=3(2)-4 \end{align*}

Equation 2: \begin{align*}2=2 \end{align*}

2.

Neither of these equations is in slope intercept form. The easiest way to graph them is to find their intercepts as shown in Example B.

Equation 1: Let \begin{align*}y=0\end{align*} to find the \begin{align*}x-\end{align*}intercept.

Now let \begin{align*}x=0\end{align*}, to find the \begin{align*}y-\end{align*}intercept.

Now we can use (-2, 0) and (0, 4) to graph the line as shown in the diagram. Using the same process, the intercepts for the second line can be found to be (-6, 0) and (0, -4).

Now the solution to the system can be observed to be (-3, -2). This solution can be verified algebraically as shown in the first problem.

3.

The first equation here must be rearranged to be \begin{align*}y=-5x+10\end{align*} before it can be entered into the calculator. The second equation can be entered as is.

Using the calculate menu on the calculator the solution is (3, -5).

Remember to verify this solution algebraically as a way to check your work.

### Explore More

Match the system of linear equations to its graph and state the solution.

Solve the following linear systems by graphing. Use graph paper and a straightedge to insure accuracy. You are encouraged to verify your answer algebraically.

1. .
1. .
1. .
1. .
1. .
1. .

Solve the following linear systems by graphing using technology. Solutions should be rounded to the nearest hundredth as necessary.

1. .
1. .
1. .

Use the following information to complete exercises 14-17.

Clara and her brother, Carl, are at the beach for vacation. They want to rent bikes to ride up and down the boardwalk. One rental shop, Bargain Bikes, advertises rates of $5 plus$1.50 per hour. A second shop, Frugal Wheels, advertises a rate of $6 plus$1.25 an hour.

1. How much does it cost to rent a bike for one hour from each shop? How about 10 hours?
2. Write equations to represent the cost of renting a bike from each shop. Let \begin{align*}x\end{align*} represent the number of hours and \begin{align*}y\end{align*} represent the total cost.
3. Solve your system to figure out when the cost is the same.
4. Clara and Carl want to rent the bikes for about 3 hours. Which shop should they use?

### Vocabulary Language: English

Consistent

Consistent

A system of equations is consistent if it has at least one solution.
Dependent

Dependent

A system of equations is dependent if every solution for one equation is a solution for the other(s).
Independent

Independent

A system of equations is independent if it has exactly one solution.
linear equation

linear equation

A linear equation is an equation between two variables that produces a straight line when graphed.
Linear Function

Linear Function

A linear function is a relation between two variables that produces a straight line when graphed.