Solving Systems by Graphing

SOLVING SYSTEMS BY GRAPHING

INTRODUCTION

The objective for this lesson on Solving Systems by Graphing is, the student will solve systems of equations by graphing.

The skills students should have in order to help them in this lesson include, solving equations.

We will have three essential questions that will be guiding our lesson. Number one, explain the three possible outcomes when solving systems of linear equations. Number two, describe the graph of a system of linear equations when there is only one solution. And number three, how do the graphs differ when comparing a system of equations with no solution and a system of equations with infinite solutions?

We will begin by completing the warm-up solving equations to prepare for solving systems by graphing in this lesson.

SOLVE PROBLEM – INTRODUCTION

The SOLVE problem for this lesson is, Steve is considering changing gyms. He asked the owners for some data on the total costs of the gym membership and each provided a table of the values that show the total gym costs after a specific amount of months. Based on the tables provided, when will the total costs be the same for both gyms?

The table that is provided is seen here.

In Step S, we will Study the Problem. First we need to identify where the question is located within the problem and underline the question. The question for this problem is, based on the tables provided, when will the total cost be the same for both gyms?

Now that we have identified the question, we want to put this question in our own words in the form of a statement. This problem is asking me to find the number of months when the cost will be the same for both gyms.

During this lesson we will learn how to solve systems of linear equations by graphing. We will use this knowledge to complete this SOLVE problem at the end of the lesson.

SOLVING SYSTEMS BY GRAPHING – ONE SOLUTION

What are the two equations given for Question one? They are y equals negative x plus five and y equals two x minus four.

Let’s graph these two lines on the coordinate plane.

What form are the lines given in? They are given in slope-intercept form. Which is y equals mx plus b. Where m, which is the coefficient of x, is the slope and b the number being added to x is the y intercept.

Explain how we graph a linear equation when it’s given in slope-intercept form. We graph the y-intercept and then apply the slope.

So what is the y-intercept of the first equation? Y equals negative x plus five. The y-intercept is five. Plot the y-intercept at point (zero, five) on the graph.

What is the slope of the first equation? The coefficient of x in the equation is negative one. So the slope is negative one, which can also be written as negative one over one.

What should we do after we plot the y-intercept? We need to apply the slope by moving down one unit and to the right one unit to plot more points. Let’s plot several points now. We go down one unit and to the right one unit to plot each point. Now we will connect each point together to form a line.

Let’s look at the second equation. What is the y-intercept of the second equation? The y-intercept is negative four. Plot the y-intercept at the point zero, negative four on the graph. Let’s use a different color to represent this line on the graph.

What is the slope of the second equation? The slope is two or we can write this as two over one.

What should we do after we plot the y-intercept? Justify your thinking. We need to apply the slope by moving up two units and to the right one unit to plot more points. Let’s plot several points now. We need to go up two units and to the right one unit for each point. Once we have several points plotted, we can connect these points to form a line.

Let’s label the lines. Our blue line represents our first equation, which was y equals negative x plus five. And our red line represents our second equation, which is y equals two x minus four.

When we are solving an equation, what does it mean to find one solution? The solution is the value that when substituted back into the equation will make the statement true.

Using the equations we began with, let’s substitute some values on the lines into the equations to see if they are solutions.

For the equations y equals negative x plus five, is the point (zero, five) a solution? Yes. Explain your thinking. If we substitute zero into the equation for x and substitute five into the equation for y, the left side of the equation simplifies to five and the right side of the equation also simplifies to five.

Explain your thinking. The equation was y equals negative x plus five. At the point zero, five, x equals zero and y equals five. When we plug these values into the equation we find that the two sides of the equation are equal.

For the equation y equals two x minus four, is the point (zero, five) a solution? No. Explain your thinking. If we substitute zero into the equation for x and substitute five into the equation for y, the left side of the equation simplifies to five and the right side of the equation simplifies to negative four. Therefore a false statement is created and the point is not a solution.

Y equals two x minus four, when we plug in the points (zero, five) we are plugging in the value of zero for x, and the value of five for y. When we solve both sides of the equation we find that the two sides are not equal. Five is not equal to negative four, so therefore the point (zero, five) is not a solution to the second equation.

Is (zero, five) a solution for both equations? No, it is only a solution to the first equation y equals negative x plus five.

Let’s look at another point. Is the point (two, zero) a solution for y equals negative x plus five? Let’s see. If we plug in the value of zero for y, and two for x, and then find out what the left and the right side of the equation are equal to, we find that the left side of the equation equals zero and the right side of the equation equals three. Since we find that the two sides of the equation are not equal, we know that the point two, zero is not a solution for the equation y equals negative x plus five.

Is the point (two, zero) a solution for y equal two x minus four? Let’s see. We need to plug in the value of y as zero and the value of x as two. When we simplify both sides of the equation we find that both sides equal zero. Zero is equal to zero, so yes the point (two, zero) is a solution for the equation y equals two x minus four.

Is the point (two, zero) a solution for both equations? No, it is only a solution to the second equation.

Is the point (three, two) a solution for the equation y equals negative x plus five? Let’s see. We need to take the equation, and plug in the value of three for x, and the value two for y. When we simplify both sides of the equation, we find that the two sides are equal to each other. So yes, the point (three, two) is a solution for the equation y equals negative x plus five.

Is the point (three, two) a solution for the equation y equals two x minus four? Let’s see. We will plug in the value of three for x and two for y into the equation. When we simplify both sides of the equation we find that the two sides are equal, so yes the point (three, two) is a solution for the equation y equals two x minus four.

So is the point (three, two) a solution for both equations? Yes, it creates true statements for both equations.

What do you notice about these three points in regard to the lines we plotted? Record your observations.

Let’s talk about these now. What do you notice about the point (zero, five)? This point lies on the first line but not the second line. What did we find out about the point (zero, five) when we substituted it into the equations? It was a solution for the first line but not the second.

What do you notice about the point (two, zero)? This point lies on the second line but not the first.

What did we find out about the point (two, zero) when we substituted it into the equations? It was a solution for the second line but not the first.

And what do you notice about the point (three, two)? This point is found on both lines because it is the intersection point of both lines.

What did we find out about the point (three, two) when we substituted it into the equations? It was a solution for both of the equations.

So what conclusions can be drawn from this experiment? Justify your thinking. A solution, in terms of a graph, is a point that will lie on the graph of a line. The point (zero, five) was a solution to the first equation because the point fell on its line, just as the point (two, zero) was a solution to the second equation. If a point is going to be a solution to more than one equation, it must fall on both lines, in which case it will be the one, and only, intersection point.

Therefore, what is the intersection point of the two lines? It is the solution to a system of linear equations.

Plot a point where the two lines intersect on the coordinate plane. This point is plotted at the point (three, two). So what is the solution to this system of equations? The solution is the point (three, two). X equals three and y equals two.

SOLVING SYSTEMS BY GRAPHING - NO SOLUTION

We are going to continue solving systems by graphing. Take a look at this first equation, x plus y equals six.

Is the first equation in slope-intercept form? No, slope-intercept form is the form y equals mx plus b.

What can we do to put it into slope-intercept for, for easy graphing? Explain your thinking. We can subtract x from each side of the equation so that y is isolated. Let’s do that now. When we subtract x from each side of the equation, what is the equation in slope-intercept form? The equation is y equals negative x plus six.

Now let’s look at the second equations, x plus y equals negative one.

Is the second equation in slope-intercept form? No. What can we do to put it into slope-intercept form? Explain your thinking. We can subtract x from each side of the equation so that y is isolated. Let’s do that now. When we subtract x from both sides of the equation, what is the equation in slope-intercept form? The equation is y equals negative x minus one.

Now let’s look at Question two. Is the first equation for Question two in slope-intercept form? The equations is y equals three x plus two. Yes, this equation is in slope-intercept form.

Is the second equations for Question two in slope-intercept form? The equation is y minus three x equals negative one. No, this equation is not in slope-intercept form.

So what can we do to change it to slope-intercept form? We can add three x to each side of the equation to that y is isolated. Let’s do that now. When we add three x to both sides of the equation, what is the equation in slope-intercept form? The equation is y equals three x minus one.

Now take a moment to graph the lines. Y equals negative x plus six and y equals negative x plus one for Question one.

First let’s graph the equation y equals negative x plus six. The y-intercept is at the point (zero, six). And the slope is negative one. Let’s plot the line y equals negative x plus six on the graph. Be sure to label the line with the equation.

Now let’s plot the line for y equals negative x minus one on the graph. The slope of the line is negative one and the y-intercept is negative one. We have now graphed both lines on the coordinate grid.

What does the graph of the equations look like? It looks like parallel lines.

Now take a moment to graph the lines y equals three x plus two and y equals three x minus one for Question two.

First let’s graph the equation y equals three x plus two. This line has a slope of three and a y-intercept of two. Let’s graph the line now. Be sure to label the line with the equation.

Now let’s graph the second line, y equals three x minus one. This line has a slope of three and a y-intercept at negative one. Now what does the graph of the equations look like? It looks like parallel lines.

So what do you notice about the slope of the lines? The slope in each problem is the same value.

What do you notice about the y-intercept of the lines? The y-intercept of each line is different in each problem.

So what conclusion can you make from these equations and graphs? Justify your thinking. If the slopes of the lines are the same and the y-intercepts are different, the lines are parallel.

Parallel lines never intersect; therefore there will be no solution to this system.

SOLVING SYSTEMS BY GRAPHING – INFINITE SOLUTIONS

Let’s continue solving system by graphing.

Take a look at the first equation seen here. The equation is two x plus two y equals two.

Is the first equation in slope-intercept form? No, slope-intercept form is the form y equals mx plus b.

What can we do to put it into slope-intercept form, for easy graphing? Explain your thinking. We can divide each term in the equation by two, to begin isolating y. Let’s do that now. When we divide each of the terms in the equation by two, what is the simplified form of the equation? It is x plus y equals one.

Is the equation in slope-intercept form now? No, remember that slope-intercept form is the form y equals mx plus b.

So what can we do to put it into slope-intercept form? We can subtract x from each side of the equation so that y is isolated. Let’s do that now. When we subtract x from each side of the equation, what is the equation in slope-intercept form? It is y equals negative x plus one.

Now let’s look at the second equation, y equals negative x plus one. Is the second equation in slope-intercept form? Yes.

Let’s now look at the equations for Question two, these equations are four x plus two y equals six and two x plus y equals three.