Question 206033
Make x the subject of the following:
Do you mean solve for x?
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y = 2 - x² / 2x² + 3
{{{y*(2x^2 + 3) = 2 - x^2}}}
{{{2x^2y + 3y = 2 - x^2}}}
{{{2x^2y + x^2 = 2 - 3y}}}
{{{x^2(2y+1) = 2-3y}}}
{{{x^2 = (2-3y)/(2y+1)}}}
{{{x = sqrt((2-3y)/(2y+1))}}}
Sqrt can be ±
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y = √2x- 3y (the sign before 2x is of square root)
How much is under the radical? The 2x, or both terms?
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1/a + 1/√x = y 
{{{1/a + 1/sqrt(x) = y}}}
{{{1 + a/sqrt(x) = ay}}}
{{{a/sqrt(x) = ay-1}}}
{{{sqrt(x)/a = 1/(ay-1)}}}
{{{sqrt(x) = a/(ay-1)}}}
{{{x = a^2/(ay-1)^2}}}