Question

plz solve these both questions guys.... ​

Answer 1

Step-by-step explanation:

see the attachment.,...

Answer 2

$\bold{\underline{\underline{\huge{\sf{ANSWER\:1:}}}}}$

Given:

• 4u-v= 14uv
• 3u+2v= 16uv

To find:

The value of u & v.

Explanation:

We have,

4u -v= 14uv........................(1)

3u +2v= 16uv.........................(2)

From equation (1), we get;

→ $\frac{4u-v}{uv} =14$

→ $\frac{4}{v} -\frac{1}{u} =14.......................(3)$

From equation (2),we get;

→ $\frac{3u+2v}{uv} =16$

→ $\frac{3}{v} +\frac{2}{u} =16...........................(4)$

Let us take $\frac{1}{v} =R$ and $\frac{1}{u} =M$ then equation becomes, we get;

→ 4R - M= 14.............................(5)

→ 3R+2M= 16.............................(6)

• Using Substitution Method:

From equation (5), we get;

⇒ 4R -M= 14

⇒ 4R = 14 +M

⇒ R = $\frac{14+M}{4} .............................(7)$

Putting the value of R in equation (6), we get;

⇒ $3(\frac{14+M}{4} )+2M=16$

⇒ $\frac{42+3M}{4} +2M=16$

⇒ 42 +3M +8M = 64

⇒ 11M = 64-42

⇒ 11M = 22

⇒ M= $\cancel{\frac{22}{11} }$

⇒ M= 2

Putting the value of M in equation (7),we get;

⇒ R= $\frac{14+2}{4}$

⇒ R= $\cancel{\frac{16}{4} }$

⇒ R= 4

Now,

→ $\frac{1}{v} =R$

→ $\frac{1}{v}= 4$

→ v= 4

&

→ $\frac{1}{u} =M$

→ $\frac{1}{u} =2$

→ u = 2.

Thus,

The value of v is 4 & value of u is 2.

$\bold{\underline{\underline{\huge{\sf{ANSWER\:2:}}}}}$

Given:

$\frac{5}{(x+y)} -\frac{2}{(x-y)} =-1$

$\frac{15}{(x+y)} -\frac{7}{(x-y)}=10$

To find:

The value of x and y.

Explanation:

We have,

$\frac{5}{(x+y)} -\frac{2}{(x-y)} =-1................................(1)$

$\frac{15}{(x+y)} -\frac{7}{(x-y)} =10..............................(2)$

Let $\frac{1}{x+y} =R$ and $\frac{1}{x-y} =M$,we get;

Therefore,

5R -2M =-1.............................(3)

15R -7M= 10...........................(4)

• Using Substitution Method:

From equation (3),we get;

→ 5R -2M= -1

→ 5R = -1 +2M

→ R = $\frac{-1+2M}{5} ...............................(5)$

Putting the value of R in equation (4), we get;

→ $15(\frac{-1+2M}{5} )-7M=10$

→ $\frac{-15+30M}{5} -7M =10$

→ -15+30M -35M =50

→ -15 -5M =50

→ -5M= 50 +15

→ -5M = 65

→ M= $\cancel{\frac{65}{-5} }$

→ M= -13

&

Putting the value of M in equation (5),we get;

→ R= $\frac{-1+2(-13)}{5}$

→ R= $\frac{-1+(-26)}{5}$

→ R= $\frac{-1-26}{5}$

→ R= $\frac{-27}{5}$

Now,

⇒ $\frac{1}{x+y} =R$

⇒ x+y = R

⇒ x+ y= $\frac{-27}{5}$

⇒ 5x +5y= -27

⇒ 5x= -27 -5y

⇒ x= $\frac{-27-5y}{5}$......................(6)

&

⇒ $\frac{1}{x-y} =M$

⇒ x-y= -13

Putting the equation (6), in the place of x,we get;

⇒ $\frac{-27-5y}{5} -y= -13$

⇒ -27 -5y -5y =-65

⇒ -27-10y= -65

⇒ -10y= -65+27

⇒ -10y= -38

⇒ y= $\cancel{\frac{-38}{-10} }$

⇒ y= $\frac{19}{5}$

Putting the value of y in equation (6),we get;

⇒ x= $\frac{-27-\cancel{5}(\frac{19}{\cancel{5}} )}{5}$

⇒ x= $\frac{-27-19}{5}$

⇒ x= $\frac{-46}{5}$

Thus,

The value of x is -46/5 & y is 19/5.