Explanation
Given:
10x+y+2x−y=4
15x+y+5x−y=−2
let 1x+y=a and 1x−y=b
10a+2b=4.......eq115a−5b=−2........eq2
Multiply eq1 with 3 and eq2 with 2 we get30a+6b=12......eq330a−10b=−4......eq4
subtracting eq3 and eq4 we get16b=16b=1
Substituting b=1 in eq1 we get10a+2×1=410a=2a=15
As 1x+y=a=15
x+y=5............eq5
and 1x−y=b=1
x−y=1...........eq6
subtracting eq5 and eq6 we get2x=6x=3
putting x=3 in eq 5 we get3+y=5y=2
hence x=3 and y=2
6x+3y=6xy .........Given
Multiply by 1xy we get
6xxy+3yxy=6xyxy
6y+3x=6............eq1
2x+4y=5xy ...........Given
Multiply with 1xywe get
2xxy+4yxy=5xyxy
2y+4x=5........eq2
Let 1x=a and 1y=b
a+2b=2.......eq34a+2b=5.....eq4
Subtracting eq3 and eq4 we get−3a=−3a=1
Substituting a=1 in eq3 we get 1+2b=22b=1b=12
Thus if a=1x=1⇒x=1
If b=1y=12⇒y=2
Hence x=1 and y=2
The given pair of equations are:3x+y=1...(1)(2k−1)x+(k−1)y=2k+1..(2)Now rearranging eq1 and eq2 will get3x+y−1=0...(3)(2k−1)x+(k−1)y−(2k+1)=0..(4)Now compare witha1=3,b1=1,c1=−1a2=2k−1,b2=k−1,c2=−(2k+1)Now we geta1a2=32k−1,b1b2=1k−1,c1c2=−1−(2k+1)Now will takea1a2=b1b2⇒32k−1=1k−1⇒3k−3=2k−1⇒3k−2k=−1+3⇒k=2Hence k=2 is the value.
{\textbf{Step -1: Framing equations}}
\text{Without loss of generality, suppose,} x {\text{and}} y {\text{be two numbers and }} x > y.
{\text{Given that, sum of two numbers is 35}}{\text{.}}
\Rightarrow x + y = 35 \ldots \left( 1 \right)
{\text{Difference between these two numbers is 13}}{\text{.}}
\Rightarrow x - y = 13 \ldots \left( 2 \right)
{\textbf{Step -2: Adding equation }}\left( \mathbf 1 \right) {\textbf{and equation }}\left(\mathbf 2 \right)\textbf .
\Rightarrow x + y + \left( {x - y} \right) = 35 + 13
\Rightarrow 2x = 48
\Rightarrow x = 24
{\text{Substituting the value of}} x {\text{in equation }}\left( 1 \right) {\text{we get,}}
\Rightarrow 24 + y = 35
\Rightarrow y = 35 - 24
\Rightarrow y = 11
{\text{Hence, these two numbers are 24 and 11}}{\text{.}}
{\textbf{ Hence, option (C) is correct answer.}}
\dfrac {1}{(3x+y)}+\dfrac {1}{(3x-y)}=\dfrac {3}{4}, \dfrac {1}{2(3x+y)}+\dfrac {1}{2(3x-y)}=\dfrac {-1}{8}
let \dfrac{1}{3x+y}=a and \dfrac{1}{3x-y}=b
Thena+b=\dfrac{3}{4}\Rightarrow 4a+4b=3........eq1
\dfrac{a}{2}-\dfrac{b}{2}=\dfrac{-1}{8}
\dfrac{a-b}{2}=\dfrac{-1}{8}
a-b=\dfrac{-2}{8}\Rightarrow a-b=\dfrac{-1}{4}
4a-4b=-1..............eq2
Adding eq1 and eq2 we get
8a=2
a=\dfrac{2}{8}\Rightarrow \dfrac{1}{4}
Substituting the value of a in eq1 we get
4\times \dfrac{1}{4}+4b=3
4b=2
b=\dfrac{2}{4}\Rightarrow \dfrac{1}{2}
Then \dfrac{1}{3x+y}=\dfrac{1}{4}
3x+y=4................eq3
\dfrac{1}{3x-y}=\dfrac{1}{2}
3x-y=2...............eq4
Adding eq3 and eq4 we get6x=6x=1
Substituting x=1 in eq3 we get3\times 1 +y=4y=1
Thus x=1 and y=1
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