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10. 解方程：$\frac {x-2}{6}-\frac {x+2}{3}= 1+\frac {x-1}{2}$。
答案
$x = - \frac { 9 } { 4 }$
11. 如图所示,直线$AB$,$CD相交于点O$,作$∠DOE= ∠1$,$OF平分∠AOE$,若$∠2= 28^{\circ }$,求$∠EOF$的度数。
解：$\because AB$，$CD$ 相交于点 $O$，$\therefore \angle 1 = \angle 2 =$
解：$\because AB$，$CD$ 相交于点 $O$，$\therefore \angle 1 = \angle 2 =$
$28^{\circ }$
，$\because \angle EOD = \angle 1$，$\therefore \angle EOB = 2 \angle 1 =$$56^{\circ }$
，$\because \angle AOE + \angle EOD = 180^{\circ }$，$\therefore \angle AOE = 180^{\circ } - \angle EOD = 180^{\circ } -$$56^{\circ }$
$=$$124^{\circ }$
。$\because OF$ 平分 $\angle AOE$，$\therefore \angle EOF = \frac { 1 } { 2 } \angle AOE = \frac { 1 } { 2 } ×$$124^{\circ }$
$=$$62^{\circ }$
。答案
解：$\because AB$，$CD$ 相交于点 $O$，$\therefore \angle 1 = \angle 2 = 28 ^ { \circ }$，$\because \angle EOD = \angle 1$，$\therefore \angle EOB = 2 \angle 1 = 56 ^ { \circ }$，$\because \angle AOE + \angle EOD = 180 ^ { \circ }$，$\therefore \angle AOE = 180 ^ { \circ } - \angle EOD = 180 ^ { \circ } - 56 ^ { \circ } = 124 ^ { \circ }$。$\because OF$ 平分 $\angle AOE$，$\therefore \angle EOF = \frac { 1 } { 2 } \angle AOE = \frac { 1 } { 2 } × 124 ^ { \circ } = 62 ^ { \circ }$。
12. 如图,直线$AB// CD// EF$,若$∠ABE= 32^{\circ }$,$∠DCE= 160^{\circ }$,求$∠BEC$的度数。
解：由 $CD // EF$，$\angle DCE = 160 ^ { \circ }$，$\therefore \angle CEF = 180 ^ { \circ } - \angle DCE = 180 ^ { \circ } - 160 ^ { \circ } =$
解：由 $CD // EF$，$\angle DCE = 160 ^ { \circ }$，$\therefore \angle CEF = 180 ^ { \circ } - \angle DCE = 180 ^ { \circ } - 160 ^ { \circ } =$
20°
，又因为 $AB // EF$，$\angle ABE = 32 ^ { \circ }$，$\therefore \angle BEF = \angle ABE =$32°
，故 $\angle BEC = \angle BEF - \angle CEF = 32 ^ { \circ } - 20 ^ { \circ } =$12°
。答案
解：由 $CD // EF$，$\angle DCE = 160 ^ { \circ }$，$\therefore \angle CEF = 180 ^ { \circ } - \angle DCE = 180 ^ { \circ } - 160 ^ { \circ } = 20 ^ { \circ }$，又因为 $AB // EF$，$\angle ABE = 32 ^ { \circ }$，$\therefore \angle BEF = \angle ABE = 32 ^ { \circ }$，故 $\angle BEC = \angle BEF - \angle CEF = 32 ^ { \circ } - 20 ^ { \circ } = 12 ^ { \circ }$。
