Graphing Linear Equations in Two Variables


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1 Math 123 Section Graphing Linear Equations Using Intercepts  Page 1 Graphing Linear Equations in Two Variables I. Graphing Lines A. The graph of a line is just the set of solution points of the linear equation. B. Graphing a line 1. Pick three values for x. 2. Substitute each value into the equation (separately of course) and solve the equation for y. 3. Plot the points. 4. Draw the line. C. Examples Graph each of the following. 1. x = 6 First, we will make a table. But note that for this line, we have no choice for the values we choose for x, namely, x = 6. However, the yvalues can be anything Now we can plot the points and draw the line. What type of line do we have? Note that the equation for any vertical line is x = Some Number. 2. Now you try one: y = 3 Note that the equation for any horizontal line is y = Some Number.
2 Math 123 Section Graphing Linear Equations Using Intercepts  Page 2 D. Graphing using the x and yintercepts 1. The xintercept of a graph is the point where the graph crosses the xaxis. 2. The yintercept of a graph is the point where the graph crosses the yaxis. 3. To find the intercepts: a. xintercept Set y = 0 and solve for x. b. yintercept Set x = 0 and solve for y. E. Graph by finding the x and yintercepts. 1. 2x + 4y = 8 We first make a table and set x = 0 and find y. 2(0) + 4y = 8 Solve this for y. y = 2 Now we set y = 0 and solve for x. 2x + 4(0) = 8  Solve this for x. x = 4 So the table is: We can now plot the points and draw the graph Now you try one: 2x 3y = 6
3 Math 123 Section Graphing Linear Equations Using Intercepts  Page x + 25y = 50 Set x = 0 and solve for y. 30(0) + 25y = 50 y = 2 Set y = 0 and solve for x. 30x + 25(0) = 50 30x = 50 The answer is going to be a fraction! We don t like that, so choose some other value for x and solve for y, one that won t give us a fraction! Try setting x = 5. 30(5) + 25y = 50 OR y = 50 Subtract 150 from each side to get: 25y = 100 Divide by 25 to get: y = 4 So our table will be: We can now plot the points and draw the line. II. Determining the Equation from the Graph. A. If it is a vertical line, remember that the equation for any vertical line is x = Some Number. B. If it is a horizontal line, remember that the equation for any horizontal line is y = Some Number. C. Examples Write an equation for each graph. 1. Page 232, #42 We see that the line is a horizontal line, so the equation is y = Some Number. What does y =? Answer: y = 5
4 Math 123 Section Graphing Linear Equations Using Intercepts  Page 4 2. Page 232, #46 We see that the line is a vertical line, so the equation is x = Some Number. What does x =? Answer: x = 0 3. Now you try one: Page 232, #43 Answer: x = 3 III. Tricky graphs. A. The book tries to get cute, but you guys will be able to see through that and easily come up with the graphs. B. Examples Graph each equation. 1. y 2.5 = 0 We first have to add 2.5 to both sides to isolate the variable. y = 2.5 This is a horizontal line. 2. Now you try one: 12 3x = 0
5 Math 123 Section Graphing Linear Equations Using Intercepts  Page 5 IV. Applications A. Definitions 1. The yintercept is the initial value of an application. 2. A graph is increasing when it rises as it goes from left to right. 3. A graph is decreasing when if drops as it goes from left to right. 4. A graph is constant if it is horizontal as it goes from left to right. 5. Note that a graph can accomplish all three of these (increase, decrease, constant), but not all at the same time. B. Example A new car worth $24,000 is depreciating in value by $3000 per year. The mathematical model: y = 3000x + 24,000 describes the cars value, y, in dollars, after ears. (Page 233, #76) 1. Find the xintercept. Describe what this means in terms of the car's value. To find the xintercept, we set y = 0 and solve the equation for x. 0 = 3000x + 24,000 Subtract 24,000 from both sides. 24,000 = 3000x Divide both sides by = x Answer the question. Answer: The xintercept is 8. This means that after 8 years, the car is worthless. 2. Find the yintercept. Describe what this means in terms of the car's value. To find the yintercept, we set x = 0 and solve the equation for y. y = 3000(0) + 24,000 Multiply 3000 & 0. y = ,000 Add. y = 24,000 Answer the question. Answer: The yintercept is 24,000. This car is worth brand new. 3. Use the intercepts to graph the linear equation. Because x and y must be nonnegative (why?), limit your graph to quadrant 1 and its boundaries. y 24,000 Value 8 x 4. Use your graph to estimate the car's value after five years. Number of Years Clearly, with my graph that is impossible! So let's use the formula. Remember that x represents the number of years. So we substitute 5 in for x in the formula.
6 Math 123 Section Graphing Linear Equations Using Intercepts  Page 6 y = 3000(5) + 24,000 Multiply 3000 and 5. y = 15, ,000 Add. y = 9,000 Answer the question. Answer: After 5 years, the car is worth $9,000.
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