Example Question - graphing linear equations
Here are examples of questions we've helped users solve.
Plotting a Linear Function
Đề bài cho hàm số y = 4x - 3 và yêu cầu: a) Vẽ đồ thị của hàm số trên. Để vẽ đồ thị hàm số y = 4x - 3, bạn cần lập bảng giá trị cho x và tìm các giá trị tương ứng của y. Chọn một số giá trị của x, ví dụ x=0, x=1, và tìm giá trị y tương ứng: - Khi x = 0: y = 4(0) - 3 = -3 - Khi x = 1: y = 4(1) - 3 = 1 Với hai điểm này, bạn có thể vẽ một đường thẳng, vì hàm số cho là một hàm số bậc nhất, đồ thị sẽ là một đường thẳng. b) Tìm điểm thuộc đồ thị hàm số trên có hoành độ bằng 1/4 Để tìm điểm thuộc đồ thị với x = 1/4, bạn chỉ cần thay x = 1/4 vào hàm số: y = 4(1/4) - 3 = 1 - 3 = -2 Vậy điểm cần tìm là (1/4, -2), nằm trên đồ thị của hàm số đã cho.
Completing Ordered Pairs for a Linear Equation
The given equation is: y = (4/5)x + 1 We need to complete the given ordered pairs for x values -5, 0, and 5. For x = -5: y = (4/5)(-5) + 1 y = -4 + 1 y = -3 So the ordered pair is (-5, -3). For x = 0: y = (4/5)(0) + 1 y = 0 + 1 y = 1 So the ordered pair is (0, 1). For x = 5: y = (4/5)(5) + 1 y = 4 + 1 y = 5 So the ordered pair is (5, 5). To graph the solution set, you can take any two of the completed ordered pairs and plot them on the coordinate plane. Then draw a straight line through those points, which represents the equation y = (4/5)x + 1.
Completing Ordered Pairs and Graphing Linear Equations
The image shows a linear equation \( y = -\frac{2}{5}x+1 \) and we are asked to complete the given ordered pairs (0, _), (5, _), and (10, _). Also, we're supposed to use two of the ordered pairs to graph the solution set for the equation. To find the missing y-values for each x-value, we'll substitute the x-values into the equation. 1. When x = 0: \( y = -\frac{2}{5}(0) + 1 \) \( y = 0 + 1 \) \( y = 1 \) So the ordered pair is (0, 1). 2. When x = 5: \( y = -\frac{2}{5}(5) + 1 \) Because \( \frac{2}{5} \times 5 = 2 \), we have: \( y = -2 + 1 \) \( y = -1 \) So the ordered pair is (5, -1). 3. When x = 10: \( y = -\frac{2}{5}(10) + 1 \) Because \( \frac{2}{5} \times 10 = 4 \), we have: \( y = -4 + 1 \) \( y = -3 \) So the ordered pair is (10, -3). To graph these points, you would plot (0, 1), (5, -1), (10, -3) on a coordinate plane and draw a straight line through these points, which would represent the solution set for the equation \( y = -\frac{2}{5}x+1 \).
Graphing Linear Equations to Find Intersection Points
To graph each pair of lines from the provided sets A, B, and C, follow these steps for each set: **Set A:** Equations: 1. \(y = 2x + 1\) 2. \(y = -x + 7\) For each equation, find two points by choosing values for \(x\) and calculating the corresponding \(y\) value. Then plot these points on a graph and draw a line through them. Equation 1: Let \(x = 0\), then \(y = 2(0) + 1 = 1\). So, one point is (0,1). Let \(x = 1\), then \(y = 2(1) + 1 = 3\). Another point is (1,3). These two points will help you graph the first line. Equation 2: Let \(x = 0\), then \(y = -(0) + 7 = 7\). One point is (0,7). Let \(x = 1\), then \(y = -(1) + 7 = 6\). Another point is (1,6). These two points will help you graph the second line. **Set B:** Equations: 1. \(2y + 8 = x\) 2. \(4y - 2x = -16\) First, rearrange these equations into slope-intercept form (\(y = mx + b\)). Equation 1: \(x = 2y + 8\) can be written as \(y = \frac{1}{2}x - 4\) Equation 2: \(4y - 2x = -16\) can be simplified to \(2y - x = -8\), which can be written as \(y = \frac{1}{2}x + 4\) Now, find two points for each of these equations using the method shown in set A, and graph them. **Set C:** Equations: 1. \(6x - 2y = 20\) 2. \(y + 3x - 5 = 0\) First, rearrange these equations into slope-intercept form. Equation 1: \(6x - 2y = 20\) can be simplified to \(3x - y = 10\), which can be written as \(y = 3x - 10\) Equation 2: \(y + 3x - 5 = 0\) can be rearranged to \(y = -3x + 5\) Again, find two points for each equation and graph each line. Please note that I cannot directly graph them here, but this explanation should help you plot them on your own graph paper. Once graphed, you can identify the intersection point of each pair of lines, which is the solution to each set of simultaneous equations.
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