SOLUTION: Solve each equation by either substitution or addition method. x2 + 4y2 = 20 x + 2y = 6

Click here to see ALL problems on Expressions-with-variables

Question 1043288: Solve each equation by either substitution or addition method.
x2 + 4y2 = 20
x + 2y = 6

Found 3 solutions by Alan3354, stanbon, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Not clear.
----
Use ^ (Shift 6) for exponents.
eg, x^2, y^3, etc

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
Solve each equation by either substitution or addition method.
x^2 + 4y^2 = 20
x + 2y = 6
-----------
x = 6-2y
-----
Substitute for "x" and solve for "y"::
(6-2y)^2 + 4y^2 = 20
--------
36 - 24y + 4y^2 + 4y^2 = 20
--------
8y^2 - 24y +16 = 0
--------
y^2 - 3y + 2 = 0
--------
(y-1)(y-2) = 0
-------
Solve for "x"
If y = 1, x = 6-2y = 4
If y = 2, x = 6-2y = 2
----------------------------
Cheers,
Stan H.
----------------------

Answer by ikleyn(52884)   (Show Source):
You can put this solution on YOUR website!
.
Solve a system of two equations by the substitution method.
x2 + 4y2 = 20
x + 2y = 6
~~~~~~~~~~~~~~~~~~~

Answer. There are two solution: (x,y) = (2,2) and (x,y) = (4,1).

Solution

= , (1) x + 2y = 6. (2) From equation (2) express x = 6-2y and substitute it into the equation (1). You will get single equation for the unknown "y": = . Simplify and solve it. Then restore "x".

For details on how this methodology works see the lesson
    - Solving systems of algebraic equations of degree 2 and degree 1
in this site.

The ellipse (red + green) and the straight line (blue)

----------------------------------
Do you see my editing of your condition?
It says me something . . . Think about it.