Question

Помогите решить систему х/(у + 3) + (10 - х) /у = 2 + 40/60; (10 - х) /(у + 3) + х/у = 2 + 20/60

Answer 1

Решение смотри на фото

Answer 2

Solving the System of Equations

To solve the system of equations:

\[ \frac{x}{y + 3} + \frac{10 - x}{y} = \frac{2 + 40}{60} \]

and

\[ \frac{10 - x}{y + 3} + \frac{x}{y} = \frac{2 + 20}{60} \]

we can follow these steps:

Step 1: Simplify the equations by getting rid of the fractions by multiplying through by the common denominator, which is 60.

Step 2: Expand and simplify the equations.

Step 3: Rearrange the equations to isolate the variables.

Step 4: Solve the resulting linear equations.

Let's go through each step in detail.

Step 1: Simplify the equations

Multiplying through by 60, we get:

Equation 1: \(60x + 20(10 - x) = 2(y + 3)(y)\)

Equation 2: \(60(10 - x) + 60x = 2(y + 3)(y + 3)\)

Step 2: Expand and simplify the equations

Expanding and simplifying the equations, we have:

Equation 1: \(60x + 200 - 20x = 2y^2 + 12y\)

Equation 2: \(600 - 60x + 60x = 2y^2 + 12y + 18\)

Simplifying further:

Equation 1: \(40x + 200 = 2y^2 + 12y\)

Equation 2: \(600 = 2y^2 + 12y + 18\)

Step 3: Rearrange the equations

Rearranging Equation 1, we have:

\(2y^2 + 12y - 40x - 200 = 0\)

Rearranging Equation 2, we have:

\(2y^2 + 12y - 582 = 0\)

Step 4: Solve the resulting linear equations

To solve the quadratic equations, we can use the quadratic formula:

For Equation 1: \(2y^2 + 12y - 40x - 200 = 0\)

The quadratic formula is given by:

\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

where \(a = 2\), \(b = 12\), and \(c = -40x - 200\).

For Equation 2: \(2y^2 + 12y - 582 = 0\)

The quadratic formula is given by:

\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

where \(a = 2\), \(b = 12\), and \(c = -582\).

By solving these quadratic equations, we can find the values of \(y\) in terms of \(x\).

Please note that without the specific value of \(x\), we cannot provide the exact solutions for \(y\).

I hope this helps! Let me know if you have any further questions.