does
shuffing feel
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to
stand an our
body
Answers
Answer:
Let us learn about cross multiplication formula first:
We consider two linear equations
\quad\quad \mathsf{a_{1}x+b_{1}y+c_{1}=0}a
1
x+b
1
y+c
1
=0
\quad\quad \mathsf{a_{2}x+b_{2}y+c_{2}=0}a
2
x+b
2
y+c
2
=0
Using the formula for cross multiplication, we get
\quad\mathsf{\dfrac{x}{b_{1}c_{2}-b_{2}c_{1}}=\dfrac{y}{c_{1}a_{2}-c_{2}a_{1}}=\dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}}
b
1
c
2
−b
2
c
1
x
=
c
1
a
2
−c
2
a
1
y
=
a
1
b
2
−a
2
b
1
1
Now we proceed to solve the given problem:
The given equations are
\quad\quad \mathsf{\dfrac{15}{x}+\dfrac{2}{y}=17}
x
15
+
y
2
=17
\quad\quad \mathsf{\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}}
x
1
+
y
1
=
5
36
Then the given equations become
\quad\quad \mathsf{15p+2q=17}15p+2q=17 15p+2q=17
\quad\quad \mathsf{p+q=\frac{36}{5}}p+q=
5
36
\quad\quad \mathsf{15p+2q-17=0}15p+2q−17=015p+2q−17=015p+2q−17=0
\quad\quad \mathsf{5p+5q-36=0}5p+5q−36=05p+5q−36=05p+5q−36=0
Now comparing these equations with the general equations, we have
\quad\quad \mathsf{a_{1}=15,\:b_{1}=2,\:c_{1}=-17}a
1
=15,b
1
=2,c
1
=−17
\quad\quad \mathsf{a_{2}=5,\:b_{2}=5,\:c_{2}=-36}a
2
=5,b
2
=5,c
2
=−36
Using the formula for cross multiplication, we get
\quad\mathsf{\dfrac{p}{b_{1}c_{2}-b_{2}c_{1}}=\dfrac{q}{c_{1}a_{2}-c_{2}a_{1}}=\dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}}
b
1
c
2
−b
2
c
1
p
=
c
1
a
2
−c
2
a
1
q
=
a
1
b
2
−a
2
b
1
1
\to \mathsf{\dfrac{p}{2\times (-36)-5\times (-17)}=\dfrac{q}{(-17)\times 5-(-36)\times 15}=\dfrac{1}{15\times 5-5\times 2}}→
2×(−36)−5×(−17)
p
=
(−17)×5−(−36)×15
q
=
15×5−5×2
1
\to \mathsf{\dfrac{p}{-72+85}=\dfrac{q}{-85+540}=\dfrac{1}{75-10}}→
−72+85
p
=
−85+540
q
=
75−10
1
\to \mathsf{\dfrac{p}{13}=\dfrac{q}{455}=\dfrac{1}{65}}→
13
p
=
455
q
=
65
1
\quad\quad \mathsf{\dfrac{p}{13}=\dfrac{1}{65}}
13
p
=
65
1
\implies \mathsf{p=\dfrac{13}{65}}⟹p=
65
13
⟹
\implies \mathsf{p=\dfrac{1}{5}}⟹p=
5
1
⟹p
\implies \mathsf{\dfrac{1}{x}=\dfrac{1}{5}\quad [\because p=\dfrac{1}{x}]}⟹
x
1
=
5
1
[∵p=
x
1
]
\implies \mathsf{\dfrac{1}{y}=7\quad [\because q=\dfrac{1}{y}]}⟹
y
1
=7[∵q=
y
1
]
Therefore the required solution is
\quad\quad \underline{\boxed{\bold{x=5,\:y=\dfrac{1}{7}}}}
x=5,y=
7
1