Question 1028985
.
help me to solve this two equations algebraically please 4x^2+3y^2=4 and 2x-3y^2=-3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
{{{4x^2}}} + {{{3y^2}}} = {{{4}}},       (1)
{{{2x}}}   - {{{3y^2}}} = {{{-3}}}.    (2)

Add the equations (1) and (2) (both sides). You will get

{{{4x^2 + 2x}}} = {{{1}}},   or

{{{4x^2 + 2x - 1}}} = {{{0}}}.

Apply the quadratic formula. You will get two roots for x

{{{x[1,2]}}} = {{{(-2 +- sqrt(2^2 + 4*4))/8}}} = {{{(-2 +- sqrt(20))/8}}} = {{{(-1 +- sqrt(5))/4}}}.

Having this, substitute the found values for x into (2) and then find the corresponding values of y.
</pre>