Given \(\frac { 3m-5 }{ m-3 } (4m-6)=2m-3\)
\(\Rightarrow\) \(\frac { 3m-5 }{ m-3 } +2m-3=2m-3\quad \Rightarrow \frac { 3m-5 }{ m-3 } =0\)
\(\Rightarrow\) \(3m-5 = Ox(m-3)\) \(\Rightarrow\) 3m - 5 =0
3m = 5  m= \(\frac { 5 }{ 3 } \)

\(x=\frac { -26 }{ 17 } \)

29 and 66

\(x=\frac { -8 }{ 5 } \) 

We have, \(\frac { 5 }{ 4 } -7y=\frac { -23 }{ 4 } \Rightarrow -7y=\frac { -23 }{ 4 } -\frac { 5 }{ 4 } \)        [ transpoting \(\frac { +5 }{ 4 } \) to RHS]
\(-7y=\frac { -28 }{ 4 } \) \(\Rightarrow -7y=-7\)           [\(\because \) - 7y = (-7)x y]
y = -7 \(\div \) (-7)                        [transpoting x (-7) to RHS]
y = -7 x \(\left( -\frac { 1 }{ 7 } \right) \) \(\Rightarrow \) y =1
So, Y=1 is the solution of the given liner Equation

Suppose, Anushka and Aarushi have their equal amount of sum, which is Rs. x
After giving \(\frac { 1 }{ 3 } \) of the money to Aarushi, 
Anushka has the amount  =Rs. \(\left( x-\frac { x }{ 3 } \right) \)
and then amount of Aarushi = Rs. \(\left( x+\frac { x }{ 3 } \right) \)
Now, as per the given condition, we have
\(\left( x+\frac { x }{ 3 } \right) -\frac { 1 }{ 2 } x\left( x+\frac { x }{ 3 } \right) =1600\)
\(\Rightarrow\) \(\left( x+\frac { x }{ 3 } \right) \left( 1+\frac { 1 }{ 2 } \right) =1600\)
\(\Rightarrow\)\(\left( x+\frac { x }{ 3 } \right) \times \frac { 1 }{ 2 } =1600\)
\(\Rightarrow\) \(\frac { 3x+x }{ 3 } =1600x2=3200\)
\(\Rightarrow\) \(\frac { 4x }{ 3 } =3200\Rightarrow x=3200x\frac { 3 }{ 4 } =2400\)
So, money gifted by Anushka = \(\frac { 1 }{ 3 } \)of 2400
\(\frac { 1 }{ 3 } \times 2400\) = Rs = 800

we have \(\frac { \left( 2+y \right) \left( 7-y \right) }{ \left( 5-y \right) \left( 4+y \right) } =1\)
By cross-multiplication, we get
(2 + y)(7 - y) =(5 - y)(4 + y)
\(\Rightarrow\)   14 - 2y + 7y - y² = 20 + 5y - 4y - y²
\(\Rightarrow\) 14 + 5y = 20 + Y   \(\Rightarrow\) 5y - y = 20-14
\(\Rightarrow\) 4y = 6   \(\Rightarrow\) y = \(\frac { 6 }{ 4 } =\frac { 3 }{ 2 } \)
Thus, solution of the given equation is \(\frac { 3 }{ 2 } \)

We have, 3(5y - 7) - 2(9y - 11) = 4(8y - 13) - 17 
15y - 21-18y + 22 = 32y - 52 -17
\(\Rightarrow\)  -3y + 1 = 32y-69
1+ 69 = 32y + 3y \(\Rightarrow\) 70 = 35y \(\Rightarrow\) Y = 2
On putting y = 2 in both sides of the given equation, we get
3(5 x 2 - 7) - 2(9 x 2 -11) = 4(8 x 2 -13)-17
\(\Rightarrow\) 3(10 -7) - 2(18 -11) = 4(16 -13) -17
\(\Rightarrow\) 3 x 3 - 2 x 7= 4 x 3 - 17
\(\Rightarrow\) 9-14 = 12 - 17 - 5 = -5
LHS = RHS
So, y = 2 is solution of the given linear equation

We have\(\frac { 17-3x }{ 5 } -\frac { 4x+2 }{ 3 } =5-6x+\frac { 7x+14 }{ 3 } \)
Multiplying both sides by 15 i.e. the LCM of 5 and 3, we get
3 (17 - 3x) - 5 (4x + 2) = 15 (5 - 6x) + 5 (7x + 14)
\(\Rightarrow\) 51- 9x - 20x - 10 = 75 - 90x + 35x + 70  \(\Rightarrow\) 41- 29x = 145 - 55x [transposing -55x to RHS and 41 to LHS
\(\Rightarrow\) -29x + 55x = 145 - 4
\(\Rightarrow\) 26x = 104 \(\Rightarrow\) \(\frac { 26x }{ 26 } =\frac { 104 }{ 26 } \) \(\Rightarrow\) x = 4     [dividing both sides by 26]
Hence, x = 4 is the solution of the given equation

Let the present ages of Rahul and Haroon be 5x yr and 7x yr, respectively.
4 yr later, age of Rahul = (5x + 4)yr
and age of Haroon=(7x + 4)yr
According to the question,
(5x + 4) + (7x + 4) = 56
\(\Rightarrow\) 5x + 4 + 7x + 4 = 56 \(\Rightarrow\) 12x = 56 - 8
\(\Rightarrow\) 12x = 56 -8  [transposing 8 to RHS]
\(\Rightarrow\) 12x = 48 \(\Rightarrow\) x=\(\frac { 48 }{ 12 } \)=4 [dividing both sides by 12]
\(\therefore\)  Present age of Rahul = 5x = 5 x 4 = 20 yr
and present age of Haroon = 7x = 7 x 4 = 28 yr
Hence, their present ages are 20 yr and 28 yr.

Let Sahil's monthly salary be Rs. x
Then, Suraj's monthly salary = Rs. 3x
After giving Rs.  6000 to Sahil, Sahil has money
= x + 6000 and Suraj has money = 3x - 6000
Then, according to the question,
2(x + 6000) = (3x -6000)
\(\Rightarrow\) 2x + 12000 = 3x - 6000
\(\Rightarrow\) 3x - 2x =12000+ 6000 =18000
x =18000
So, Sahil's monthly salary = Rs.18000
and Suraj's monthly salary = Rs. 54000.
(b) The value depicted by Suraj is their helpful nature. He helps his friend in the need

Let Abdul buys 2x m of trouser material 
Then, the shirt material bought by him = 3x m
Sale price of 1 m of trouser material
Rs (90 + 12% of 90)
\(\left( 90+\frac { 12\times 90 }{ 100 } \right) =Rs.100.80\)
Sale price of 2x m of trouser material = Rs.(2xx 100.80) = Rs. 201.60x
Sale price of 1m of shirt material
Rs. 50 +20% of Rs 50 = \(\left( 50+\frac { 20\times 50 }{ 100 } \right) =Rs.60\)
Sale price of 3x m of shirt material = Rs 3x x 60
= Rs. 180x
Total sale = Rs. (201.60+180) x = Rs. 381.60x
\(\therefore\)  381.60 = 38160
\(x=\frac { 38160 }{ 38.160 } =100\)
So,Abdul bought 2 x 100 = 200 m of trouser material.