Math, asked by anindyaadhikari13, 4 months ago

 \sf  \large\left \{ {{ {x}^{ \sqrt{y} }  +  {y}^{ \sqrt{x} } =  \dfrac{49}{48}  } \atop {  \large\sqrt{x} +  \sqrt{y}  = \dfrac{7}{2} }} \right.
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Higher mathematics.

Plz plz plz solve it. ​

Answers

Answered by Anonymous
3

Answer:

\sf \large\left \{ {{ {x}^{ \sqrt{y} } + {y}^{ \sqrt{x} } = \dfrac{49}{48} } \atop { \large\sqrt{x} + \sqrt{y} = \dfrac{7}{2} }} \right.

 x =  \frac{49}{48}  - \frac{7}{2}  + y

y =  -  (2x - \frac{7}{2} )

hope \: it \: helps \: you \: dear

Answered by akumar41864
1

Answer:

question :-

Find the value of the expressions for the given value of x.

(a) -3x + 2 for x = -1

(b) 3x^2 + 2x - 5 for x = - 4

(c) 9x - 5 for x = 3

(d) - 2x^2 + 3x - 2 for x = 2

Solution 1 :-

put the value of x = -1 in the equation

\mathsf{(a) \: - 3 x\: + \: 2}(a)−3x+2

\mathsf{: \longrightarrow \: - 3 \: \times \: - 1 + \: 2}:⟶−3×−1+2

\mathsf{: \longrightarrow \: 3 \: + \: 2}:⟶3+2

\mathsf{: \longrightarrow \: 5}:⟶5

Solution 2 :-

Put the value of x = - 4 in the equation .

\mathsf{(b) \: \:3x^2 + 2x - 5 }(b)3x

2

+2x−5

\mathsf{: \longrightarrow \:3 \: ( - 4)^2 \: + \: 2 \: \times \: - 4 \: - 5 }:⟶3(−4)

2

+2×−4−5

\mathsf{: \longrightarrow \:3 \: \times \: 16 \: - \: 8 \: - \: 5 }:⟶3×16−8−5

\mathsf{: \longrightarrow \:48 \: - \: 8 \: - \: 5 }:⟶48−8−5

\mathsf{: \longrightarrow \:40 \: - \: 5 }:⟶40−5

\mathsf{: \longrightarrow \: 35 }:⟶35

Solution 3 :-

Put the value of x = 3

\mathsf{(c) \: \:9x \: - \: 5 }(c)9x−5

\mathsf{: \longrightarrow \:9 \: \times \: 3\: - \: 5 }:⟶9×3−5

\mathsf{: \longrightarrow \:23 \: - \: 5 }:⟶23−5

\mathsf{: \longrightarrow \: 18}:⟶18

Solution 4 :-

Put the value of x = 2 in the equation .

\mathsf{(d)\:- 2x^2 \: + \: 3x \: - \: 2 }(d)−2x

2

+3x−2

\mathsf{: \longrightarrow \:- 2(2)^2 \: + \: 3 \: \times \: 2 \: - \: 2 }:⟶−2(2)

2

+3×2−2

\mathsf{: \longrightarrow \:- 2 \: \times \: 4\: + \: 3 \: \times \: 2 \: - \: 2 }:⟶−2×4+3×2−2

\mathsf{: \longrightarrow \: - 8\: + \:6 \: - \: 2 }:⟶−8+6−2

\mathsf{: \longrightarrow \:-10 \: + \: 6}:⟶−10+6

\mathsf{: \longrightarrow \:- \: 4}:⟶−4

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