(a) Solve the equation : \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\) (b) In the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°....

Answer Details

(a) Solve the equation.

\[\frac{2}{3}(3x-5)-\frac{3}{5}(2x-3)=3\]

Multiply every term by \(15\) (the LCM of \(3\) and \(5\)):

\[15\cdot\frac{2}{3}(3x-5)-15\cdot\frac{3}{5}(2x-3)=15\cdot3\]\[10(3x-5)-9(2x-3)=45\]\[30x-50-18x+27=45\]\[12x-23=45\]\[12x=68\]\[x=\frac{68}{12}=\frac{17}{3}=\mathbf{5\tfrac{2}{3}}\]

(b) Find \(m+n\).

From the diagram, \(P,Q,R\) lie on a straight line and \(P,U,T,S\) lie on a straight line, with \(\angle STQ=m\), \(\angle TUQ=80^{\circ}\), \(\angle UPQ=r\), \(\angle PQU=n\) and \(\angle RQT=88^{\circ}\).

At \(U\): \(P,U,T\) are collinear, so \(\angle QUP\) and \(\angle QUT\) are angles on a straight line:

\[\angle QUP=180^{\circ}-80^{\circ}=100^{\circ}\]

At \(Q\): \(P,Q,R\) are collinear, so the three angles on that line satisfy

\[\angle PQU+\angle UQT+\angle TQR=180^{\circ}\]\[n+\angle UQT+88^{\circ}=180^{\circ}\Rightarrow\angle UQT=92^{\circ}-n\]

Triangle \(UQT\):

\[\angle QUT+\angle UQT+\angle UTQ=180^{\circ}\]\[80^{\circ}+(92^{\circ}-n)+\angle UTQ=180^{\circ}\Rightarrow\angle UTQ=8^{\circ}+n\]

At \(T\): \(U,T,S\) are collinear, so \(\angle UTQ\) and \(\angle STQ(=m)\) are angles on a straight line:

\[m+\angle UTQ=180^{\circ}\]\[m+(8^{\circ}+n)=180^{\circ}\]\[m+n=\mathbf{172^{\circ}}\]