a The exercise says that Jai is solving the system of equations shown below when something strange happened. We need to solve the system and explain what the solution should be.
{y=-2x+52y+4x=10
Notice that the variable y is already isolated in the first equation, so we can substitute it in the second equation. This will allow us to solve for x.
As we can see, the second equations becomes a trivial equality after the substitution. This means that the second equation does not provide any new information, and therefore we cannot find a unique pair of values as a solution. Instead the relation represents a family of solutions where the value of y depends on the value we give to x, and our system has infinite possible solutions.
b We are asked to graph both equations from the system used in Part A on the same set of axes and explain what is happening. Notice that the first equation is already in the slope-intersect form.
y=mx+b
Here m is the slope of the line and b is the y-intercept. By direct comparison, we can see that for the first equation m=-2 and b=5. We can plot the point (0,5) and use the slope to find a second point. The line joining both points would be our graph.
However, to graph the second equation in the same way we need to isolate y first. Let's give it a try.
As we can see, both equations are equivalent and therefore, the same. The graph containing both equations is shown below.
We can see that they overlap. This make sense since they are equivalent.
c We are asked to explain how the graph of Part B can help us to explain our conclusions from Part A. As we could see in Part B, the graphs of both equations overlap. Since every intersection point represents a possible answer, this means that there are infinite solutions for our system of equations, just as we concluded in Part A.
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