Category: Systems of equations

Solve the system of equations (conics)

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Q:  Solve the system of equations (both of which are conic sections)
1:   x2 + y2 – 20x + 8y + 7 = 0
2:  9x2 + y2 + 4x + 8y + 7 = 0

A:  I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2:  9x2 + y2 + 4x + 8y + 7 = 0

1: -x2 – y2 + 20x – 8y – 7 = 0   [after it has been multiplied by -1]

Now, add them together to get equation 3:

3:  8x2 +24x = 0

Now, equation 3 only has x’s so we can solve for x (by factoring):

(8x)(x + 3) = 0

So, x = 0 or -3

But, we aren’t done… Find the solution to this system of conics means that we are finding the point(s) of intersection — and it turns out there are two points of intersection since there were two solutions for x.  So, our points (solutions) are:

(0, ?)  and (-3, ?)

We need to find the y-coordinate of each solution separately.  We do this by plugging in the x value to either of the original equations (both will lead us to the same correct solution):

First x-value:  (0, ?)

x2 + y2 – 20x + 8y + 7 = 0

(0)2 + y2 – 20(0) + 8y + 7 = 0

y2  + 8y + 7 = 0

Solve for y by factoring:

(y + 7)(y + 1)=0

y = -7, -1

OH!  So this really means that there are more than two solutions…. When x = 0, we found two answers for y.

(0, -7) and (0, -1)

Second x-value:  (-3, ?)

x2 + y2 – 20x + 8y + 7 = 0

(-3)2 + y2 – 20(-3) + 8y + 7 = 0

9 + y2 +60 + 8y + 7 = 0

y2 + 8y + 76 = 0

Since this does not factor, we need to solve for y by using the quadratic formula.  I did not show my work here, but we end up with a negative under the square root, so there are no real solutions for when x = -3.

Therefore, there are two real solutions to this equation (2 points of intersection) and they are:

(0, -7) and (0, -1)

Here is a visual of what we are solving for….. We had the equations for the two graphed conic sections and we found the two points of intersection shown below:

Graph the system of equations

StaceyLeave a comment

Q:  Graph the system of linear equations, determine the number of solutions, and find the solution:

y = 4x + 2 and y =- 2x – 3

A: Okay, for the sake of talking about the lines I will name them:

Line 1:  y = 4x + 2

Line 2:  y =- 2x – 3

A “solution” to a system of linear equations is the point where the lines intersect.  So, we need to graph both of these lines on the same grid and see if and where they intersect.

Step 1:  Graph Line 1 and Line 2 on the same grid:

Tricks for graphing line 1:  Since line 1 is in the form y = mx + b, we can use the method of finding the slope and the y-intercept to graph the line.

The slope of line 1 is 4 and the y-intercept of line 1 is 2.

On your x-y graph, go up 2 on the y-axis and put a point [this is the y-intercept, or the “b” value]

Now, starting from the y-intercept point 2:  go up 4 and right 1 (this is the slope 4/1).  Create another point.  Connect the dots to make a line.  You should have:

 

Now, we need the same strategy to graph line 2 on the same grid.

Line 2:  y =- 2x – 3

The y-intercept is -3 and the slope is -2.  Plot the y-intercept of -3 on the y-axis.  From that point:  go down 2, right 1 to create another point.  Connect the dots to make a line.  Your graph should now look like:

So, the question now is:  How many solutions are there and what is the solution.  Recall, the solution is (x, y) point of intersection.
Clearly, there is 1 solution.  The lines do cross at 1 point.

What is that point?  Well, you have to estimate that point from the graph.  This is one problem with solving a system by graphing — it can be inaccurate.  Depending on the “neatness” of your graph, your answer will vary.

To me, the answer appears to be around the point (-.8, -1.3)

*To get the accurate solution, we solve these types of problems algebraically.