解:①
![](http://thumb.1010pic.com/pic5/latex/1112.png)
x+
![](http://thumb.1010pic.com/pic5/latex/335.png)
=2.4,
![](http://thumb.1010pic.com/pic5/latex/1112.png)
x+
![](http://thumb.1010pic.com/pic5/latex/335.png)
-
![](http://thumb.1010pic.com/pic5/latex/335.png)
=2.4-
![](http://thumb.1010pic.com/pic5/latex/335.png)
,
![](http://thumb.1010pic.com/pic5/latex/1112.png)
x=2,
![](http://thumb.1010pic.com/pic5/latex/1112.png)
x÷
![](http://thumb.1010pic.com/pic5/latex/1112.png)
=2÷
![](http://thumb.1010pic.com/pic5/latex/1112.png)
,
x=
![](http://thumb.1010pic.com/pic5/latex/114038.png)
;
②
![](http://thumb.1010pic.com/pic5/latex/597.png)
:x=3:24,
3x=
![](http://thumb.1010pic.com/pic5/latex/597.png)
×24,
3x=9,
x=3.
分析:(1)利用等式的基本性质解答,即“方程两边同时加上,同时减去一个数,同时乘以(或除以相同的数(0)除外,结果仍然成立,方程两边同时减去
![](http://thumb.1010pic.com/pic5/latex/335.png)
得出
![](http://thumb.1010pic.com/pic5/latex/1112.png)
x=2,利用等式的基本性质方程两边同时除以
![](http://thumb.1010pic.com/pic5/latex/1112.png)
来解;
(2)先用比例的基本性质“两内项之积等于两外项之积”,把原算式转化为 3X=
![](http://thumb.1010pic.com/pic5/latex/597.png)
×24,再根据等式的基本性质,方程两边同时除以3来解;
点评:此题考查了利用等式的基本性质解方程,即“方程的两边同时加上或减去相同的数,同时乘上或除以相同的数(0除外),等式仍然成立”;以及比例的基本性质“两外项之积等于两内项之积”.