(a) Express \(\frac{5 + \sqrt{2}}{3 - \sqrt{2}} - \frac{5 - \sqrt{2}}{3 + \sqrt{2}}\) in the form \(a + b\sqrt{2}\). (b) Solve the following equations simul...

(a) Simplify \(\dfrac{5+\sqrt2}{3-\sqrt2}-\dfrac{5-\sqrt2}{3+\sqrt2}\).

Rationalise each fraction. For the first, multiply by \(\dfrac{3+\sqrt2}{3+\sqrt2}\):

\[\frac{(5+\sqrt2)(3+\sqrt2)}{(3-\sqrt2)(3+\sqrt2)}=\frac{15+5\sqrt2+3\sqrt2+2}{9-2}=\frac{17+8\sqrt2}{7}.\]

For the second, multiply by \(\dfrac{3-\sqrt2}{3-\sqrt2}\):

\[\frac{(5-\sqrt2)(3-\sqrt2)}{9-2}=\frac{15-5\sqrt2-3\sqrt2+2}{7}=\frac{17-8\sqrt2}{7}.\]

Subtracting:

\[\frac{17+8\sqrt2}{7}-\frac{17-8\sqrt2}{7}=\frac{16\sqrt2}{7}.\]

So in the form \(a+b\sqrt2\): \(a=0,\ b=\dfrac{16}{7},\) i.e. \(\dfrac{16}{7}\sqrt2.\)

(b) Solve by the determinant (Cramer's) method:

\(3x-y-z=-2,\quad x+5y+2z=5,\quad 2x+3y+z=0.\)

Main determinant:

\[D=\begin{vmatrix}3&-1&-1\\1&5&2\\2&3&1\end{vmatrix}=3(5-6)+1(1-4)-1(3-10)=-3-3+7=1.\]

\[D_x=\begin{vmatrix}-2&-1&-1\\5&5&2\\0&3&1\end{vmatrix}=-2(5-6)+1(5-0)-1(15-0)=2+5-15=-8.\]

\[D_y=\begin{vmatrix}3&-2&-1\\1&5&2\\2&0&1\end{vmatrix}=3(5-0)+2(1-4)-1(0-10)=15-6+10=19.\]

\[D_z=\begin{vmatrix}3&-1&-2\\1&5&5\\2&3&0\end{vmatrix}=3(0-15)+1(0-10)-2(3-10)=-45-10+14=-41.\]

Therefore

\[x=\frac{D_x}{D}=-8,\quad y=\frac{D_y}{D}=19,\quad z=\frac{D_z}{D}=-41.\]

Check in equation (1): \(3(-8)-19-(-41)=-24-19+41=-2.\) Correct.