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 = 53x27-4xy = xNow 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 = uNow Rearranging both the equation53u-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=4Hence, 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=7Put v=7 in eq\left ( 5\right )\Rightarrow 7+u=5\Rightarrow u=-2Hence, \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 =yThen 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 =450Cost of 1 chair =150\therefore Cost of 1 table and 2 chair =450+150\times2=750
Let no of 20p coins =xNo of 25p coins =yAccording to the question x+y=50\Rightarrow x=50-y.....eq1And 20x+25y=1125\Rightarrow 4x+5y=225....eq2Put the value of x from eq1\Rightarrow 4\left ( 50-y \right )+5y=225......eq3\Rightarrow 200-4y+5y=225\Rightarrow y=25Put y=25 in eq1\Rightarrow x+25=50\Rightarrow x=25
No of 20p coins =25
No of 25p coins =25
Let the first number be x and second number y.
According to questionx+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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