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Solving an Equation with Variables on Both Sides

Find the valu...

Question

Find the value of $x$ :
$3 (x - 5) + 5 x = 7 (p - x) - 2 (x - q)$ , where $p$ and $q$ are constants.

x = \frac{15 + 7 p + 2 q}{17}

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x = \frac{17 - 2 p + 7 q}{15}

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x = \frac{17 - 7 p + 2 q}{11}

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x = \frac{15 - 7 p - 2 q}{13}

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Solution

The correct option is A $x = \frac{15 + 7 p + 2 q}{17}$
The given equation is,
$3 (x - 5) + 5 x = 7 (p - x) - 2 (x - q) \Rightarrow 3 x - 15 + 5 x = 7 p - 7 x - 2 x + 2 q \Rightarrow 8 x - 15 = - 9 x + 7 p + 2 q \Rightarrow 8 x + 9 x = 15 + 7 p + 2 q \Rightarrow 17 x = 15 + 7 p + 2 q \Rightarrow x = \frac{15 + 7 p + 2 q}{17}$

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Similar questions

P (x) = {5 x}^{5} + {3 x}^{3} - 7

g (x) = {2 x}^{2} + 1

p = {lim}_{x \to \infty} \frac{(2 x - 3) (3 x + 5) (4 x - 6)}{3 x^{3} + 2 x^{2} - x + 1}

q = {lim}_{x \to 0} \sqrt{x^{2} - 5 x + 9} + x

p - 2 q

p (x) = a x^{3} + 3 x - 13

q (x) = 2 x^{3} - 5 x + 1

x + 2

a

If the roots of the ploynomial $x^{2} = - x + q$ are double in value to the $2 x^{2} - 5 x - 3,$ find the values of $q .$

d

f (x) = 5 x^{2} - 2 x + \frac{26}{5}

f (x)

x = r

\sum_{n = 1}^{\infty} (1 + (n - 1) d) r^{n - 1}

\frac{p}{q}

p

q

(3 q - p)

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