Question

both 3rd and 4th. Jo log math me master hai sirf vohi ye question solve kar sakte hai! ab dekhte hai kitne bhagte hai aur kitne eska samna karte hai aur agar yaha answer full ho jaye toh mere id pe previous questions me ye same question h vaha solve karo baki ke. aur spammers dur raho Varna tumhare account delete ho jayega.​

Answer 1

3rd part :

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=1}}}$

$\green{\tt{\therefore{Value\:of\:y=2}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies \frac{148}{x} + \frac{231}{y} = \frac{527}{xy} \\ \\ \tt: \implies \frac{231}{x} + \frac{148}{y} = \frac{610}{xy} \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies Value \: of \: x = ? \\ \\ \tt: \implies Value \: of \: y= ?$

$\tt: \implies \frac{148}{x} + \frac{231}{y} = \frac{527}{xy} \\ \\ \tt: \implies \frac{148y + 231x}{xy} = \frac{527}{xy} \\ \\ \tt: \implies 231x + 148y = \frac{527}{xy} \times xy \\ \\ \tt: \implies 231x + 148y = 527 - - - - - (1) \\ \\ \tt: \implies \frac{231}{x} + \frac{148}{y} = \frac{610}{xy} \\ \\ \tt: \implies \frac{231y + 148x}{xy} = \frac{610}{xy} \\ \\ \tt: \implies 148x + 231y = 610 \\ \\ \tt: \implies 148x = 610 - 231y \\ \\ \tt: \implies x = \frac{610 - 231y}{148} - - - - - (2) \\ \\ \text{putting \: value \: of \: x \: in \: (1}) \\ \\ \tt: \implies 231 \times \frac{610 - 231y}{148} + {148y} = 527 \\ \\ \tt: \implies \frac{140910 - 53361y + 21904y}{148} = 527 \\ \\ \tt: \implies 140910 - 31457y = 77996 \\ \\ \tt: \implies 140910 - 77996 = 31457y \\ \\ \tt: \implies y = \frac{62914}{31457} \\ \\ \green{\tt: \implies y = 2} \\ \\ \text{Putting \: value \: of \: y \: in \: (1) } \\ \tt: \implies 148\times 2+231x = 527 \\ \\ \tt: \implies 231x = 527-296 \\ \\ \green{\tt: \implies x = \frac{231}{231}=1 }$

4th part :

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=1}}}$

$\green{\tt{\therefore{Value\:of\:y=1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies \frac{7x - 2y}{xy} = 5 \\ \\ \tt: \implies \frac{8x + 7y}{xy} = 15 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Value \: of \: x = ? \\ \\ \tt: \implies Value \: of \: y= ?$

• According to given question :

$\tt: \implies \frac{7x - 2y}{xy} = 5 \\ \\ \tt: \implies \frac{7x - 2y}{5} = xy - - - - - (1)\\ \\ \tt: \implies \frac{8x + 7y}{xy} = 15 - - - - - (2 )\\ \\ \text{Putting \: value \: of \: xy \: in \: (2)} \\ \tt: \implies \frac{8x + 7y}{ \frac{7x - 2y}{5} } = 15 \\ \\ \tt: \implies \frac{8x + 7y}{7x - 2y} = 3 \\ \\ \tt: \implies 8x + 7y = 21x - 6y \\ \\ \tt: \implies 21x - 8x= 7y +6y \\ \\ \tt: \implies y = x - - - - - (3) \\ \\ \text{Putting \: value \: of \: y \: in \: (1)} \\ \\ \tt : \implies \frac{7x - 2 \times x}{5 } = x \times x \\ \\ \tt: \implies \frac{5x}{5} = {x}^{2} \\ \\ \tt: \implies x = {x}^{2} \\ \\ \green{ \tt: \implies x = 1} \\ \\ \text{Putting \: value \: of \: x \: in \: (3)}\\ \\ \tt: \implies y = x \\ \\ \green{\tt: \implies y = 1 }$

Answer 2

Solution 3 :

Given Equation :

1. $\sf{\dfrac{148}{x}\:+\:\dfrac{231}{y}\:=\:\dfrac{527}{xy}}$
2. $\sf{\dfrac{231}{x}\:+\:\dfrac{148}{y}\:=\:\dfrac{610}{xy}}$

To Find :

• Value of x and y.

Solution :

To solve such question when the denominator in the LHS of equation is xy, it is preferable to multiply the equation throughout by xy.

Multiplying equation (1) by xy,

$\sf{\cancel{x}y\:\times\:\dfrac{148}{\cancel{x}}\:+\:x\cancel{y}\:\times\:\dfrac{231}{\cancel{y}}\:=\:\cancel{xy}\:\times\:\dfrac{527}{\cancel{xy}}}$

Multiply equation (2) by xy,

$\sf{\cancel{x}y\:\times\:\dfrac{231}{\cancel{x}}\:+\:x\cancel{y}\:\times\:\dfrac{148}{\cancel{y}}\:=\:\cancel{xy}\:\times\:\dfrac{610}{\cancel{xy}}}$

So, the new equation now will be without the complex denominator xy.

$\longrightarrow$ $\sf{148y\:+\:231x=527\:\:\:(3)}$

$\longrightarrow$ $\sf{231y+148x=610\:\:\:(4)}$

In the equation (3) and (4) we can see that the coefficient (number) of two variables (x and y) are interchanged in the equation.

To solve this type of equation, we first add the two equations formed and then subtract the two equation.

Adding equation (3) and (4),

$\sf{148y+231x+231y+148x=527+610}$

$\sf{148y+231y+231x+148x=1137}$

$\longrightarrow$ $\sf{379y+379x=1137}$

Now, you can take 379 in the RHS as common, the results will be,

$\longrightarrow$ $\sf{379(y+x)=1137}$

$\longrightarrow$ $\sf{(y+x)=\cancel\dfrac{1137}{379}}$

$\longrightarrow$ $\sf{y+x=3\:\:\:(5)}$

Now, we are done with adding the equation.

Subtract the two equation.

$\sf{231y+148x-(148y+231x)=610-527}$

$\sf{231y+148x-148y-231x=83}$

$\sf{231y-148y+148x-231x=83}$

$\longrightarrow$ $\sf{83y-83x=83}$

Again, 83 is common in the RHS,

$\longrightarrow$ $\sf{83(y-x) =83}$

$\longrightarrow$ $\sf{y-x=\cancel\dfrac{83}{83}}$

$\sf{y-x=1\:\:\:(6)}$

Now, we are left with two equation with just variables in the RHS.

Add equation (5) and (6),

$\longrightarrow$ $\sf{y+x+y-x=3+1}$

$\longrightarrow$ $\sf{2y=4}$

$\longrightarrow$ $\sf{y=\cancel\dfrac{4}{2}}$

$\longrightarrow$ $\sf{y=2}$

Substitute, y = 2 in equation (5),

$\longrightarrow$ $\sf{y+x=3}$

$\longrightarrow$ $\sf{2-3=-x}$

$\longrightarrow$ $\sf{\cancel{-}1=\cancel{-}x}$

$\longrightarrow$ $\sf{x=1}$

Values of x and y :

$\large{\boxed{\sf{\purple{(x,y)\:=\:(1,2)}}}}$

Solution 4 :

Given equation :

1. $\sf{\dfrac{7x-2y}{xy}=5}$
2. $\sf{\dfrac{8x+7y}{xy}=15}$

To Find :

• Value of x and y.

Solution :

In equation (1) and (2), we can see that xy is the common denominator for 7x - 2y and in equation (2) for 8x + 7y.

$\longrightarrow$ $\sf{\dfrac{7\cancel{x}}{\cancel{x}y}-\:\dfrac{2\cancel{y}}{x\cancel{y}}\:=5}$

$\longrightarrow$ $\sf{\dfrac{7}{y}\:-\:\dfrac{2}{x}=5\:\:\:(3)}$

Now in equation (2),

$\longrightarrow$ $\sf{\dfrac{8\cancel{x}}{\cancel{x}y}+\:\dfrac{7\cancel{y}}{x\cancel{y}}\:=5}$

$\longrightarrow$ $\sf{\dfrac{8}{y}+\dfrac{7}{x}=15\:\:\:(4)}$

Let, 1/x be m and 1/y be n.

•°• We get,

$\longrightarrow$ $\sf{7n\:-2m=5\:\:\:(5)}$

$\sf{8n+7m=15\:\:\:(6)}$

Now, multiply equation (5) by 7,

$\sf{49n\:-14m=35\:\:(7)}$

Now, multiply equation (6) by 2,

$\sf{16n+14m=30\:\:\:(8)}$

Add equation (7) and (8),

$\sf{49n-14m+16n+14m=35+30}$

$\longrightarrow$ $\sf{49n+16n-14m+14m=65}$

$\longrightarrow$ $\sf{65n=65}$

$\longrightarrow$ $\sf{n\:=\:\cancel\dfrac{65}{65}}$

$\longrightarrow$ $\sf{n=1}$

Substitute, n = 1 in equation (6),

$\longrightarrow$ $\sf{8n+7m=15}$

$\longrightarrow$ $\sf{8(1)+7m=15}$

$\longrightarrow$ $\sf{8+7m=15}$

$\longrightarrow$ $\sf{7m=15-8}$

$\longrightarrow$ $\sf{7m=7}$

$\longrightarrow$ $\sf{m=\cancel\dfrac{7}{7}}$

$\longrightarrow$ $\sf{m=1}$

Resubstitute now, m = 1/x = 1,

$\longrightarrow$ $\sf{\dfrac{1}{x}=1}$

$\longrightarrow$ $\sf{x=1}$

Now, resubstitute 1/y = n = 1,

$\longrightarrow$ $\sf{\dfrac{1}{y}=1}$

$\longrightarrow$ $\sf{y=1}$

$\large{\boxed{\sf{\red{(x,y)\:=\:(1,1)}}}}$