[tex]\displaystyle a) \\
\frac{4}{x-5}=7- \frac{3}{x-5}~~~\Longleftrightarrow~~~4\cdot \frac{1}{x-5}=7-3\cdot \frac{1}{x-5} \\ \\
Substitutie:~~\frac{1}{x-5} = a\\ \\
\Longrightarrow~~~ 4a=7-3a ~~~\Longrightarrow~~~4a+3a=7~~~\Longrightarrow~~~7a=7 ~~~\Longrightarrow~~~a=1 \\ \\
\text{Ne intoarcem la substitutie}\\ \\
\frac{1}{x-5} = a ~~~\Longleftrightarrow~~~\frac{1}{x-5} = 1 \Longleftrightarrow x-5=1 ~~~\Longrightarrow~~~ x=1+5=\boxed{6}[/tex]
[tex]\displaystyle b) \\
- \frac{3(x+2)}{x}= \frac{2(x+2)}{x}+5~~~~~~~Substitutie : ~~ \frac{x+2}{x}=b \\ \\
\Longrightarrow ~~~ 3b=2b+5 ~~~\Longleftrightarrow ~~~3b-2b=5 ~~~ \Longrightarrow ~~~b=5 \\ \\
\text{Revenim la substitutie: } \\ \\
\frac{x+2}{x}=b ~~~\Longleftrightarrow ~~~\frac{x+2}{x}=5 \\ \\ x+2=5x ~~~\Longleftrightarrow ~~~5x-x=2 ~~~\Longleftrightarrow ~~~4x=2 ~~~\Longrightarrow ~~~x= \frac{2}{4} = \boxed{\frac{1}{2} } [/tex]
[tex]\displaystyle c) \\
\frac{5t}{t+1} = - \frac{1}{2} + \frac{3t}{t+1}~~~~~~~~Substitutie: ~~ \frac{t}{t+1}=c \\ \\
5c = - \frac{1}{2} + 3x ~\Longleftrightarrow~ 5c-3c = - \frac{1}{2} ~\Longleftrightarrow~ 2c=- \frac{1}{2} ~\Longrightarrow~ c= \frac{- \frac{1}{2} }{2}=- \frac{1}{4} \\ \\
\text{Revenim la substitutie: } \\ \\
\frac{t}{t+1}=c ~\Longleftrightarrow~\frac{t}{t+1}=- \frac{1}{4} ~\Longleftrightarrow~4t = -(t+1)\\ \\
4t=-t-1 \\ 4t+t = -1 \\ 5t=-1 \\ t= \boxed{-\frac{1}{5}} [/tex]
[tex]\displaystyle d) \\
\frac{0,5a+1,5}{2a}=9- \frac{a+3}{a} ~~~~~~~\text{Amplificam prima fractie cu 2 } \\ \\
\frac{a+3}{4a}=9- \frac{a+3}{a} ~~~~~~~\text{Substitutie: } ~~ \frac{a+3}{a}=d \\ \\
\frac{d}{4}= 9-d \\ \\
d= 4(9-d) \\
d=36-4d \\
5d=36 \\ \\
d= \frac{36}{5}[/tex]
[tex]\displaystyle \\
\text{Ne intoarcem la substitutie: } \\ \\
\frac{a+3}{a}=d \\ \\
\frac{a+3}{a}=\frac{36}{5} \\ \\
36a=5(a+3) \\
36a=5a+15 \\
36a-5a=15 \\
31a = 15 \\ \\
a = \boxed{\frac{15}{31} } [/tex]
