SOLUTION: using elimination method solve: 2y+5x=7 2x-4y=10 (the page isnt coming up right. i hope you get this anyways!) Thanks(:

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: using elimination method solve: 2y+5x=7 2x-4y=10 (the page isnt coming up right. i hope you get this anyways!) Thanks(:      Log On


   



Question 378670: using elimination method solve:
2y+5x=7
2x-4y=10
(the page isnt coming up right. i hope you get this anyways!)
Thanks(:

Found 2 solutions by richard1234, jim_thompson5910:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Given
2y + 5x = 7
-4y + 2x = 10
Multiply the first equation by 2:
4y + 10x = 14
-4y + 2x = 10
Add the two equations
12x = 24
x = 2
From this, we obtain y = -3/2.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Note: the first equation is also 5x%2B2y=7 (I just rearranged the terms)


Start with the given system of equations:
system%285x%2B2y=7%2C2x-4y=10%29


2%285x%2B2y%29=2%287%29 Multiply the both sides of the first equation by 2.


10x%2B4y=14 Distribute and multiply.


So we have the new system of equations:
system%2810x%2B4y=14%2C2x-4y=10%29


Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:


%2810x%2B4y%29%2B%282x-4y%29=%2814%29%2B%2810%29


%2810x%2B2x%29%2B%284y%2B-4y%29=14%2B10 Group like terms.


12x%2B0y=24 Combine like terms.


12x=24 Simplify.


x=%2824%29%2F%2812%29 Divide both sides by 12 to isolate x.


x=2 Reduce.


------------------------------------------------------------------


10x%2B4y=14 Now go back to the first equation.


10%282%29%2B4y=14 Plug in x=2.


20%2B4y=14 Multiply.


4y=14-20 Subtract 20 from both sides.


4y=-6 Combine like terms on the right side.


y=%28-6%29%2F%284%29 Divide both sides by 4 to isolate y.


y=-3%2F2 Reduce.


So the solutions are x=2 and y=-3%2F2.


Which form the ordered pair .


This means that the system is consistent and independent.


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim