![\left ( (3)/(2)* (4)/(5) \right )+\left ( (9)/(5)- (10)/(3) \right )-\left ( (1)/(2)* (3)/(4) \right )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/u8ffl2j5lfxa83t2wv97g23qr8xlwqtkuv.png)
In the first parenthesis, multiply the numerators so we get -12 and after multiply the denominator we will get 10
So the fraction in the first parenthesis will be
![\left ( (-12)/(10) \right )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/38qdq2sw95a1hb04l2kx9ihaa8pdnqa6nx.png)
To solve the 2nd parenethesis, find the LCD of the denominator.
So the LCD of 5 and 3 = 15,
Now make the each denominator same as that of LCD.
So the new fractions in the 2nd parenthesis =
=
![\left ( (-23)/(15) \right )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ga3thf61r2vm2v6c3gaf5n00ufzaucz2i4.png)
In the third parenthesis, multiply the numerators so we get 3 and after multiply the denominator we will get 8
So third parenthesis =
![\left ( (3)/(8)\ \right )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/akvt195x4qm2hvdszextezkm9yc9xmtcz8.png)
=
![\left ( (-12)/(10) \right ) + \left ( (-23)/(15) \right ) -\left ( (3)/(8)\ \right )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/2uss1zhhayv576d37gmok1ywjexx90rlpu.png)
LCD of 12,15 and 8 is 120
=
![\left ( (-144)/(120) \right ) + \left ( (-184)/(120) \right ) -\left ( (45)/(120)\ \right )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/hb6nsqgvvyvn9k222muozz35f8ab9ow3ud.png)
=
![\left ( (-373)/(120) \right )](https://img.qammunity.org/2019/formulas/mathematics/middle-school/e5cvihchkkh6z4gzmdkli281462g5qb9sc.png)