Explanation
The given equations are$$\Rightarrow 6x+5y=8xy....eq\left (1 \right )$$
$$\Rightarrow 8x+3y=7xy....eq\left (2 \right )$$
$$Divide \ both \ equations \ by \ \ xy $$
$$\Rightarrow \dfrac{6}{y}+\dfrac{5}{x}=8....eq\left ( 3 \right )$$
$$\Rightarrow \dfrac{8}{y}+\dfrac{3}{1x}=7....eq\left ( 4 \right )$$
$$Put\ \dfrac{1}{x}=u\ and\ \dfrac{1}{y}=v\ in\ eq\left (3 \right ) and \left ( 4 \right )$$
$$\Rightarrow 6v+5u=8....eq\left ( 5 \right )$$
$$\Rightarrow 8v+3u=7 ....eq\left ( 6\right )$$
$$Multiply \ eq \ \left (5 \right )\ by \ 3 \ and \ eq\left ( 6 \right ) \ by\ 5 \ and \ subtract \ both$$
$$\Rightarrow \left (18v+15u=24 \right )-\left ( 40v+15u=35 \right )$$
$$\Rightarrow -22v=-11\Rightarrow v=\dfrac{1}{2}$$
$$Put \ v=\dfrac{1}{2} \ in \ eq\left ( 5\right )$$
$$\Rightarrow 6\left ( \dfrac{1}{2} \right )+5u=8\Rightarrow u=1$$
$$Hence,\dfrac{1}{x}= u=1\Rightarrow x=1$$
$$ \dfrac{1}{y}= v=\dfrac{1}{2}\Rightarrow y=2$$
Given systems of equations are:$$9+25xy = 53x$$$$27-4xy = x$$Now using reducible method will getLet convert this equation in $$\dfrac{1}{x}\ and\ \dfrac{1}{y}$$ formWill take $$xy$$ as $$L.C.M$$ as $$xy$$ is commonFrom eq will get $$\Rightarrow \dfrac{9}{xy} +25 = \dfrac{53}{y}$$ ...3$$\dfrac{27}{xy} - 4 = \dfrac{1}{y}$$ ...4Now let $$\dfrac{1}{xy} = v and \dfrac{1}{y} = u $$Now will get $$9v+25 = 53u $$ and $$27v-4 = u$$Now Rearranging both the equation$$53u-9v = 25 ...(5)$$$$u-27v = -4 ....(6)$$Now multiplying eq5 by 3 $$\Rightarrow 159u-27v = 75 ...(7)$$Now subtracting eq 7 and eq6$$\Rightarrow 159u-27v = 75$$$$\Rightarrow -u +27v = 4$$$$\Rightarrow 158u = 79$$$$\Rightarrow u = \dfrac{1}{2}$$Now putting $$u = \dfrac{1}{2}$$ in eq 6$$\Rightarrow \dfrac{1}{2} -27\left ( 2 \right ) v= -4\left ( 2 \right )$$$$\Rightarrow 1 - 54v = -8$$$$\Rightarrow 54v = 9$$$$\Rightarrow v = \dfrac{9}{54}$$$$\Rightarrow v = \dfrac{1}{6}$$Now putting u and v values$$\dfrac{1}{y} = u$$$$ \Rightarrow \dfrac{1}{y} = \dfrac{1}{2}$$$$\Rightarrow y = 2$$$$\dfrac{1}{xy} = v$$$$\Rightarrow \dfrac{1}{xy} = \dfrac{1}{6}$$$$\Rightarrow \dfrac{1}{2x} = \dfrac{1}{6}$$$$ \Rightarrow x = \dfrac{6}{2}$$$$\Rightarrow x = 3$$$$\Rightarrow x = 3 , y = 2$$
The given equations are$$\Rightarrow \dfrac{11}{2x}-\dfrac{9}{2y}=-\dfrac{23}{2}....eq\left ( 1 \right )$$
$$\Rightarrow \dfrac{3}{4x}+\dfrac{7}{15y}=-\dfrac{23}{6}....eq\left ( 2 \right )$$
Put $$u=\dfrac{1}{x} $$ and $$ v=\dfrac{1}{y} $$ in $$ eq\left (1 \right )$$ and $$ \left ( 2 \right )$$
$$\Rightarrow \dfrac{11}{2}u-\dfrac{9}{2}v=-\dfrac{23}{2}\Rightarrow 11u-9v=-23....eq\left ( 3 \right )$$
$$\Rightarrow \dfrac{3}{4}u+\dfrac{7}{15}v=\dfrac{23}{6}\Rightarrow 45u+28v=230 ....eq\left ( 4 \right )$$
Multiply $$eq \left (3 \right )$$ by 28 and $$eq\left ( 4 \right )$$ by 9 and subtract both
$$\Rightarrow \left ( 308u-252v=-644 \right )-\left ( 405u+252v=2070 \right )$$
$$\Rightarrow 713u=1426\Rightarrow u=2$$
$$Put\ u=2\ in\ eq\left ( 3 \right )$$
$$\Rightarrow 11\times2-9v=-23\Rightarrow v=5$$
$$Hence, \dfrac{1}{x}= u=2\Rightarrow x=\dfrac{1}{2}$$ $$ \dfrac{1}{y}= v=5\Rightarrow y=\dfrac{1}{5}$$
The given equations are$$\Rightarrow \dfrac{2}{x-1}+\dfrac{y-2}{4}=2......eq(1)$$$$\Rightarrow \dfrac{3}{2\left (x-1 \right )}\dfrac{2\left (y-2 \right )}{4}=\dfrac{47}{20}......eq(2)$$$$Put \dfrac{1}{x-1}=u \ and \ y-2=v \ in \ eq(1)\ and \ eq(2)$$$$\Rightarrow 2u+\dfrac{v}{4}=2\Rightarrow 8u+v=8....eq(3)$$$$\Rightarrow \dfrac{3u}{2}+\dfrac{2v}{5}=\dfrac{47}{20}\Rightarrow 30u+8v=47.....eq(4)$$$$Multiply \ eq(3) \ by \ 8$$$$\Rightarrow 64u+8v=64.....eq(5)$$$$Subtract \ eq(4) \ and \ eq(5)$$$$\Rightarrow \left (64u+8v=64 \right )-\left (30u+8v+47 \right )$$$$\Rightarrow 34u=17\Rightarrow u=\dfrac{1}{2}$$$$put \ u=\dfrac{1}{2} \ in \ eq(3)$$$$\Rightarrow 8\times \dfrac{1}{2}+v=8\Rightarrow v=4$$$$Hence, u=\dfrac{1}{x-1} =\dfrac{1}{2}\Rightarrow x-1=2\Rightarrow x=3$$$$ v=y-2=4\Rightarrow y=6 $$
The given equations are$$\Rightarrow \dfrac{x-y}{xy}=9\Rightarrow x-y=9xy....eq\left (1 \right )$$$$\Rightarrow \dfrac{x+y}{xy}=5\Rightarrow x+y=5xy....eq\left (2 \right )$$Divide both eq by $$xy$$$$\Rightarrow \displaystyle \frac{1}{y}-\frac{1}{x}=9....eq\left ( 3 \right )$$$$\Rightarrow \displaystyle \frac{1}{y}+\frac{1}{x}=5....eq\left ( 4 \right )$$Put $$\dfrac{1}{x}=u$$ and $$\dfrac{1}{y}=v$$ in eq $$\left (3 \right )$$ and $$\left ( 4 \right )$$$$\Rightarrow v-u=9 ....eq\left ( 5 \right )$$$$\Rightarrow v+u=5 ....eq\left ( 6\right )$$Subtract eq $$\left (5 \right )$$ and eq$$\left ( 6 \right )$$$$\Rightarrow \left (v-u=9 \right )-\left ( v+u=5 \right )$$$$\Rightarrow 2v=-14\Rightarrow v=7$$Put $$v=7$$ in eq$$\left ( 5\right )$$$$\Rightarrow 7+u=5\Rightarrow u=-2$$Hence, $$\dfrac{1}{x}= u=-2\Rightarrow x=-\dfrac{1}{2}$$$$ \dfrac{1}{y}= v=7\Rightarrow y=\dfrac{1}{7}$$
Suppose the cost of $$1$$ table $$=x$$ and cost of $$1$$ chair $$=y$$Then according to the question$$\Rightarrow 4x+3y=2250$$...........(1)$$\Rightarrow 3x+4y=1950$$.............(2)
Multiply (1) by $$4$$ and (2) by $$3$$ and Subtract both$$\Rightarrow (16x+12y=9000)- (9x+12y=5850)$$
$$\Rightarrow 7x=3150\Rightarrow x=450$$
Put $$x=450$$ in $$eq2$$
$$\Rightarrow 450\times4+3y=2250\Rightarrow 3y=2250-1800\Rightarrow 3y=450$$
$$\Rightarrow y=150$$
Cost of $$1$$ table $$=450$$Cost of $$1$$ chair $$=150$$$$\therefore$$ Cost of $$1$$ table and $$2$$ chair $$ =450+150\times2=750$$
Let no of $$20$$p coins $$=x$$No of $$25$$p coins $$=y$$According to the question$$ x+y=50$$$$\Rightarrow x=50-y.....eq1$$And $$20x+25y=1125\Rightarrow 4x+5y=225....eq2$$Put the value of $$x$$ from $$eq1$$$$\Rightarrow 4\left ( 50-y \right )+5y=225......eq3$$$$\Rightarrow 200-4y+5y=225\Rightarrow y=25$$Put $$y=25$$ in $$eq1$$$$\Rightarrow x+25=50\Rightarrow x=25 $$
No of $$20$$p coins $$=25$$
No of $$25$$p coins $$=25$$
Let the first number be $$x$$ and second number $$y.$$
According to question$$x+y=8\quad\quad\quad\dots(i)$$
$$4(x-y)=8$$
$$\Rightarrow x-y=2\quad\quad\quad\dots(ii)$$
Add equations $$(i)$$ and $$(ii),$$$$\begin{aligned}{}\left( {x + y} \right) + \left( {x - y} \right)& = 8 + 2\\2x &= 10\\x &= 5\end{aligned}$$
Substitute $$x=5$$ in $$(i),$$$$\begin{aligned}{}\left( 5 \right) + y &= 8\\y&=8-5\\y &= 3\end{aligned}$$
So, the numbers are $$5$$ and $$3.$$
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