Show that each pair is a solution of the given equation. Use the graph of the two points to determine another solution. \( y=\frac{1}{2} x+1 ;(6,4) \) and \( (-2,0) \) Use the points given above and the graphing tool to

To show that each pair is a solution of the equation \( y = \frac{1}{2}x + 1 \), substitute the x-values of each point into the equation. For the point \( (6, 4) \): \( y = \frac{1}{2}(6) + 1 = 3 + 1 = 4 \), which confirms that \( (6, 4) \) is indeed a solution. For the point \( (-2, 0) \): \( y = \frac{1}{2}(-2) + 1 = -1 + 1 = 0 \), verifying that \( (-2, 0) \) is also a solution. Now, drawing the line represented by the equation on a graph would show that the line slopes up because of the positive \( \frac{1}{2} \) coefficient. To find another solution, you could pick any x-value, say 2. Plugging it in gives \( y = \frac{1}{2}(2) + 1 = 1 + 1 = 2\), thus \( (2, 2) \) is another point on the line! Moreover, understanding the practicality of linear equations can be a game-changer! They help in a myriad of real-world scenarios, such as budgeting for expenses or understanding rates of change in science and economics. Just imagine creating a budget with the equation; suddenly, your finances are as organized as your closet after a good spring cleaning! If you’re interested in diving deeper into linear equations, consider exploring concepts like slope-intercept form versus point-slope form. Each provides a unique viewpoint on linear relationships and could help you describe a line in various contexts—perfect for those immersive math discussions with friends or in study groups!