Question 451984: i am having troubles with solving the systems of equations in substitution. how do you solve the equation x-2y=0 and 4x-3y=15 in ordered pair form, (x,y)?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! x-2y=0 .............1
-3y=15-4x---------2
/-3 =
y=-5+1.33x
Plug the value of y in (1)
x-2(-5+ 1.33x)= 0
x+10-2.67x=0
x-2.67x =0-10
-1.67x=-10
/-1.67
x=6
Plug the value of x in (1)
x-2y=0
1*6-2y= 0
6-2y=0
-2y=-6
/-2
y=3
(2,3)
Answer by ikleyn(53617)  (Show Source):
You can put this solution on YOUR website! .
i am having troubles with solving the systems of equations in substitution. how do you solve the equation
x-2y=0 and 4x-3y=15 in ordered pair form, (x,y)?
~~~~~~~~~~~~~~~~~~~~~
When I look at the solution by @mananth, I see that it is done in anti-pedagogic way.
It is absolutely inappropriate to teach in this way.
I will solve it in a simple manner, as simple as possible and as clear and straightforward as it should be done.
The starting equations are
x - 2y = 0, (1)
4x - 3y = 15. (2)
From the first equation, express x = 2y.
Now substitute '2y' instead of 'x' into the second equation
4(2y) - 3y = 15,
8y - 3y = 15,
5y = 15,
y = 15/5 = 3.
Now x = 2y = 2*3 = 6.
ANSWER. x = 6, y = 3.
Solved correctly, in a simple, clear and straightforward manner.
|
|
|
