Construct a system of two equations that has the solution (3, -5), prove to me that (3, -5) is the solution?

Question

I'm taking my online math class and I do not know how to do this: Construct a system of two equations that has the solution (3, -5), prove to me that (3, -5) is the solution by either solving the system or explaining how you constructed the system. With no infinite solution.

Answer 1

x = 3

y = -5

Answer 2

"Construct a system of two equations that has the solution (3, -5)"

 

Now, remember that two points define a line.  We need another point for each line.  We will keep this point as the intersection of the two lines (the "solution" to the system of equations).

 

To make it very easy, I like a slope of 3 and I like the Origin, so with point (0,0), one equation is:

        y = 3x -14            [the y-intercept b is -14]

   and I like a little larger x, but a little smaller y (slope -1), so point (4,-6) gives the other equation:

        y = -x -2             [the y-intercept b is -2]

 

Now, we are done playing teacher, let's play student and "solve" the system.

     y = 3x -14     (eq1)

     y = -x - 2     (eq2)

 

Let's use the elimination method:  multiply 3*eq2 and add the equations:

     y  =  3x - 14

    3y = -3x - 6

 -------------------

    4y =       -20

 

      y =       -5

 

Use either equation to solve for x:

     -5 = -(x) - 2

     -3 = -x                           [add 2 to each side]

      3 = x

 

A "solution" of (3,-5) means that the point is on both lines.

 

You know. it may be easier being a teacher than being a student.

Answer 3

I think what makes this problem hard is the fact that there are so many correct possibilities. 

What is the solution to a set of equations? It's where those two curves or lines intersect. 

So, make up equations whose graphs intersect at (3,-5). It's easiest if both equations are linear. (Although, it is possible for us to choose any type. There could be two parabolas who intersect at (3,-5) or even some oscillating curve that intersects another at (3,-5). The possibilities are infinite.)

Let's use linear equations to make our lives easier. 

consider point-slope form of a line: y-y1=m(x-x1) where (x1,y1) is a point on that line. 

For us, we are required to have (3,-5) be a point on our line, so our (x1,y1)=(3,-5).

So, we have y-(-5)=m(x-3). Let's simplify this as y+5=m(x-3). Now, fill in ANY slope (m) you want! Literally ANY slope. Do this twice to get two equations, and that's it. That's your system of equations. 

You could have y+5=10(x-3) and y+5=7(x-3). OR

You could have y+5=.5(x-3) and y+5=4.081(x-3). OR

You could have y+5=-23(x-3) and y+5=17(x-3). OR

You could have y+5=525,600(x-3) and y+5=0.0000314(x-3). OR

Any other slopes you can think of, as long as your (x1,y1) is (3,-5). 

To prove it, solve your system by substitution or elimination, and show that your two equations really do intersect at (3,-5). I challenge you to come up with your own slopes. Try not to use one of the ones I suggested.