# Math Help - solution sets

1. ## solution sets

find the solution sets

4x^2 + 9y^2 = 25
3y + 2x = 1

2. Originally Posted by eepyej
find the solution sets

4x^2 + 9y^2 = 25
3y + 2x = 1
Have you tried this yourself yet?

Hint: Get the second equation as y in terms of x or vice versa. Substitute into the first equation.

3. Hi eepyej

From ur second equation make either x or y the subject of formula as Prove It told,... 3y+2x=1
=> 2x=1-3y
=> x=(1-3y)/2

Then substitute this x in the first equation, u will get a quadratic equation in terms of y only, solve it

4. Originally Posted by eepyej
find the solution sets

4x^2 + 9y^2 = 25
3y + 2x = 1
Originally Posted by Kai
Hi eepyej

From ur second equation make either x or y the subject of formula as Prove It told,... 3y+2x=1
=> 2x=1-3y
=> x=(1-3y)/2

Then substitute this x in the first equation, u will get a quadratic equation in terms of y only, solve it

Or you can do it in a way to avoid fractions!

$4x^2 + 9y^2 = 25\implies (2x)^2+(3y)^2=25$
$3y + 2x = 1$

You can solve for either 2x or 3y, and then substitute

--Chris

5. Nice, didnt see it