Introduction: Linear Equation Lab
Equations like 3x – 2 = 2x + 1 can be solved by graphing each side of the equation as a separate line, and then finding the intersection of the two lines. The x-coordinate of the intersection corresponds to the solution of the equation. For example, to solve 3x – 2 = 2x + 1, you would graph y = 3x – 2 and y = 2x + 1. The lines intersect at the point (3, 7), so the solution is x = 3.
In this Interactivity, you will learn how to graph each side of an equation, and how to identify which equations have one solution, no solutions, and infinitely many solutions.
Linear Equation Lab
Exploration Guide
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How can you solve 2x + 5 = x – 1 using a graph?
- Start by graphing the left side of the equation. First, make sure that "Use parentheses" is off. Set the shaded slider under "Left side" to 2, and the non-shaded slider to 5. (You can also change a slider's value by typing in a number to the right of the slider and pressing ENTER.) Notice that the left side of the top equation becomes 2x + 5.
- To graph the right side of the equation, set the shaded slider under "Right side" to 1 and the non-shaded slider to –1. This creates the expression 1x – 1, which simplifies to x – 1. Notice that the complete equation at the top of the Interactivity is 2x + 5 = x – 1.
- Look at the graph on the right. Do the two lines intersect? What is the x-coordinate of the intersection point? Click on "Show number line" to see the intersection point's x-coordinate projected onto a number line.
- Solve the equation 2x + 5 = x – 1 on paper. Is your answer the same as the x-coordinate of the intersection point? Click on "Show algebraic solution" to check your work.
- Try solving the equation 3x – 7= –x + 5 on paper. Then check your answer by graphing each side of the equation as a separate line in the Interactivity.
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To solve equations like 2x + 1 = x + 2 algebraically, you need to get the variables on one side of the equation using the Addition or Subtraction Properties of Equality. For example, you could subtract x from each side of 2x + 1 = x + 2 to obtain x + 1 = 2. In this step, you will use a graph to verify that subtracting x from each side of an equation does not change its solution.
- Graph both sides of the equation 2x + 1 = x + 2 using the Interactivity. Click on "Show number line" to see the solution of this equation.
- Next, graph both sides of x + 1 = 2. How does this graph compare to the graph of 2x + 1 = x + 2? Notice that although the graphs are different, the x-coordinate of the intersection point is the same. Therefore, the solution of the equations 2x + 1 = x + 2 and x + 1 = 2 are the same.
- To complete the solution, subtract 1 from both sides of x + 1 = 2 to obtain x = 1. Graph both sides of this equation. How does this graph compare to the previous graphs?
- If you add the same number to both sides of an equation, will the equation's solution ever change? How about if you subtract the same number from both sides? Try several different numbers in the Interactivity to see what happens.
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Identities are equations that are true for every value of a variable.
- Graph both sides of the equation 3x + 2 = 3x + 2 by setting both shaded sliders to 3 and both non-shaded sliders to 2. Since both sides of the equation are the same, the two lines on the graph are identical.
- Click on "Show number line." Since the lines are identical, their intersection is equal to their entire lengths. Therefore, all points on the number line are solutions to this equation.
- Click on "Show algebraic solution." Notice that the variable can be eliminated from the equation, resulting in 0 = 0. This statement is true for any value of x.
- Turn off "Show algebraic solution," and then graph –2x + 6 = –2(x – 3). (To graph the right side, click on "Use parentheses" under "Right side" and set the shaded slider to –2 and the non-shaded slider to –3.) What does the graph look like? Is this equation an identity?
- Graph each of the following equations, and identify which ones are identities: 2x – 8 = x – 4; –(x – 3) = –x + 3; 3x – 6 = 3(x – 2).
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If there is no value of a variable that makes an equation true, then the equation has no solution.
- Consider the equation x + 1 = x – 1. Is there a value for x that would make this equation true? For example, if we substitute x = 1 into the equation, we would obtain 1 + 1 = 1 – 1, or 2 = 0. Since 2 = 0 is not a true statement, x = 1 is not a solution of this equation. Try some other numbers to see if you can find a solution.
- Graph both sides of x + 1 = x – 1. What is the relationship between the two lines on the graph? How does it compare to the graph of an identity?
- Click on "Show number line." Do the lines intersect at any point?
- Click on "Show algebraic solution." As with an identity, the variable can be eliminated from the equation, but the result is 0 = –2. This statement is false for any value of x. Equations that are false for all values of the variable have no solution.
- Graph both sides of the following equations, and determine which have one solution, no solutions, or infinitely many solutions: 3x – 5 = 3(x + 6); –3(x – 1) = –3x + 3; –3(x – 3) = 3x + 3
