for the equation i just gave it also says to solve it by letting y equal each side fo the equation and graphing. there is a graph off to the side. so what does all this mean?
you would end up with
y = w + 3.14 which is a straight line on the w-y grid, w along the horizontal.
Its slope is 1 and the y-intercept is 3.14
the second equation would be
y = w + 3.19 after you simplify the right side.
It also has a slope of 1 but a y-intercept of 3.19, slightly higher.
So you have two parallel lines slightly apart, obviously they cannot meet, as a result, no solution as noted in my previous reply to the other posting.
y=-3x+2,y+3x=-4
parallel lines
When you are asked to solve an equation by letting y equal each side of the equation and graphing, it means that you need to rearrange the equation so that one side is equal to y and then create a graph to visually represent the relationship between x and y.
To solve an equation by letting y equal each side, you generally start with an equation in the form y = ... or ... = y. Let's say your equation is:
y = x + 2
To create a graph, you need to choose different values for x, substitute them into the equation, and determine the corresponding y values. This will give you a set of ordered pairs (x, y) that you can plot on a graph.
For example, let's choose three values for x: -2, 0, and 2. Substituting these values into the equation, we get:
For x = -2: y = (-2) + 2 = 0
For x = 0: y = (0) + 2 = 2
For x = 2: y = (2) + 2 = 4
So, the ordered pairs are (-2, 0), (0, 2), and (2, 4). Plotting these points on a coordinate plane will give you a series of points.
Once you've plotted the points, you can connect them with a straight line. This line represents all the possible solutions to the equation. In this case, it will be a straight line with a slope of 1 and a y-intercept of 2.
By examining the graph, you can see that any point on the line will satisfy the equation y = x + 2. This means that by graphing the equation, you can visually determine the solutions to the equation.
Note: This method is particularly helpful for linear equations, but may not be suitable for more complex equations or equations involving higher degree polynomials.