WAEC - Mathematics (2017 - No. 29)
Make x the subject of the relation d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)
x = \(\frac{6 + 12}{d^2 + y}\)
x = \(\frac{12}{d^2 - y}\)
x = \(\frac{12}{y} - 2d^2\)
x = \(\frac{12}{2d^2 + y}\)
Explanation
d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)
\(d^2 = \frac{6}{x} - \frac{y}{2}\)
\(2xd^2 = 12 - xy\)
\(2xd^2 + xy = 12\)
x = \(\frac{12}{2d^2 + y}\)
Comments (0)
