solving system of equation

[joshuamichaelsanders]

My son has been asked to complete the following problem. "Use substitution to solve each system of equations. If the system does not have exactly one solution, state whether it has no solution or infinitely many solutions." x=-4y 3x+2y=20 Using substitution he's got the following pair (8,2). When I have him graph the lines to check his work the first line doesn't land on 8 and 2. The second line does. 3x+2y=20 2y=-3x+20 y=-3/2x+10 This line lands on the point 8 and 2. The first line does not x=-4y divide by -4 1/-4x=y The y intercept is 0 so I start him there and the I go over to the right one and down 4. if the numbers were reversed and it was over right 4 and down one it would land on 8,2 but no matter how I work the equation I can't get it there. -Joshua

 

[joshuamichaelsanders]

solved it. When I was looking at y=-1/4x I was starting at 0 and going over one and down four. I forgot it was rise over run so it should have been down 1 and over four.
-Joshua

 

[Deleted member 4993]

I hope you found out the solution is (8,-2)

Another (algebraic) way to check it is to use the given equations:

x=-4y

(8) = -4*(-2)

8 = 8............................checks

3x+2y=20

3*(8) + 2*(-2) = 20

24 - 4 = 20

20 = 20........................Checks

 

[joshuamichaelsanders]

I hope you found out the solution is (8,-2)

Another (algebraic) way to check it is to use the given equations:

x=-4y

(8) = -4*(-2)

8 = 8............................checks

3x+2y=20

3*(8) + 2*(-2) = 20

24 - 4 = 20

20 = 20........................Checks

Oh...yea my bad, the solution was (8,-2), I just didn't right it correctly on the board. One of my many failings....I never recheck my work. Thanks. We just couldn't graph the line correctly. It never landed on (8,-2). It wasn't until I remembered rise over run that the lightbulb lit up.

-Joshua