Chapter Closure
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Notice that in the first equation y is already isolated.
Recall the slope-intercept form of a line.
The variables are eliminated and we are left with a false statement. This means our equation has no solution.
The lines do not intersect, meaning there is no solution.
We are given the system of equations shown below. We are asked to describe what happens when trying to solve them.
(II):y= 3x +2
(II):Distribute 2
(II):Multiply
(II):Add terms
As we can see, the variable x was eliminated and we were left with a false statement. This means that no matter which values we use, the equality will not hold. This means the system of equations has no solution.
We are asked to graph the system used in Part A and to tell how the graph of the system explains what happened with the equations. 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 = 3 and b=2. We can plot the point (0,2) 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 that a try.
LHS-6x=RHS-6x
Subtract terms
.LHS /-2.=.RHS /-2.
Write as a sum of fractions
Calculate quotient
Now we can graph this equation as well by following the same procedure we did before. We are going to show how both graphs would look when graphed together.
As we can see, the lines are parallel. Recall that the points of intersection represent the solution for a system of equations. In this case, there is no solution as the lines never touch. This is the same result we found at Part A.