Solve the following system of linear equations by telling what the graph would look like and what this tells you about the solution to the system of linear graphs. Y=x+3. -4x+4y=28
To solve the system of linear equations, let's first graph each equation:
The first equation, y = x + 3, is in slope-intercept form, where the slope is 1 and the y-intercept is 3. We can start by graphing the y-intercept at point (0, 3), then use the slope to find another point. When x increases by 1, y increases by 1 as well. Thus, we can plot another point at (1, 4). Drawing a line through these two points, we have the graph of the first equation:
Graph of the equation y = x + 3:
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1 2 3 4 5 6
The second equation, -4x + 4y = 28, can be rewritten in slope-intercept form by dividing both sides by 4. This gives us y = x + 7. Following a similar process as before, we plot the y-intercept at (0, 7), then find another point using the slope of 1. Plotting a point at (1, 8), we can draw the graph of the second equation:
Graph of the equation y = x + 7:
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1 2 3 4 5 6
Looking at the two graphs, we observe that they are lines with different slopes. This means that the lines are not parallel and will intersect at a single point. Therefore, the system of linear equations has a unique solution. To find the coordinates of the point of intersection, we can solve the system algebraically using the substitution or elimination method:
Let's solve the system using the substitution method:
Substitute y in the second equation with its equivalent expression from the first equation:
-4x + 4(y) = 28
-4x + 4(x + 3) = 28
-4x + 4x + 12 = 28
12 = 28
This equation is not possible because 12 is not equal to 28. Therefore, there is no solution to the system of linear equations.
In summary, the graph of the system of linear equations consists of two lines that intersect at the same point. However, upon solving the system algebraically, we find that there is no solution.