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Answered

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[tex]x {}^{2} \div a + y {}^{2} \div b = 1[/tex]
find x​

Sagot :

Answer:

[tex]x=\sqrt{\dfrac{a(b-y^2)}{b}}[/tex]

Step-by-step explanation:

The given equation is :

[tex]x^2\div a+y^2\div b=1[/tex]

We need to find the value of x.

It can also written as :

[tex]\dfrac{x^2}{a}+\dfrac{y^2}{b}=1[/tex]

Subtract [tex]\dfrac{y^2}{b}[/tex] from both sides,

[tex]\dfrac{x^2}{a}+\dfrac{y^2}{b}-\dfrac{y^2}{b}=1-\dfrac{y^2}{b}\\\\\dfrac{x^2}{a}=\dfrac{b-y^2}{b}\\\\\text{Cross multiplying both sides}\\\\x^2=\dfrac{a(b-y^2)}{b}\\\\x=\sqrt{\dfrac{a(b-y^2)}{b}}[/tex]

Hence, the value of x is equal to [tex]\sqrt{\dfrac{a(b-y^2)}{b}}[/tex].

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