Question

(1/3+x)÷7=(3/y+x)÷9 ай нид хелппппаппр​

Answer 1

Ответ:

dy/dx=-2/21y^2

Объяснение:

Answer 2

Certainly, I can help you solve the equation. Let's break it down step by step.

The given equation is:

$\frac{1}{3+x} \div 7 = \frac{3}{y+x} \div 9$

To simplify this equation, we can start by multiplying both sides of the equation by their respective denominators to eliminate the fractions.

Multiply both sides by 7 to get rid of the fraction on the left side:

$7 \cdot \frac{1}{3+x} \div 7 = 7 \cdot \frac{3}{y+x} \div 9$

Simplify the left side:

$\frac{1}{\cancel{3+x}} \cdot \cancel{7} = \frac{21}{y+x} \div 9$

Now, multiply both sides by 9 to eliminate the fraction on the right side:

$9 \cdot \frac{1}{3+x} = \frac{21}{y+x} \cdot \frac{1}{\cancel{9}}$

Simplify both sides:

$\frac{9}{3+x} = \frac{21}{y+x}$

Cross-multiply to get rid of the fractions:

$9 \cdot (y+x) = 21 \cdot (3+x)$

Distribute on both sides:

$9y + 9x = 63 + 21x$

Now, let's isolate the variables. Move the $21x$ term to the left side by subtracting $21x$ from both sides:

$9y - 21x + 9x = 63$

Combine like terms:

$-12x + 9y = 63$

Divide by the common factor (-3):

$4x - 3y = -21$

So, the simplified form of the given equation is $4x - 3y = -21$. If you have specific values for $x$ and $y$, you can substitute them into this equation to check if it holds true. If you need further assistance or clarification, feel free to ask!