මේ ගාණ හදන්නෙ කොහොමද?

[AnuradhaRa]

x + y = 210
xy = 69

x y සොයන්න

 

[McGayan]

x + y = 210
xy = 69

x y සොයන්න

ඔකේ අහන්න තියෙන්නේ මොකක්ද බන්?
හැබැයි ඔය ගානේ පොඩි ලස්සනක් තියනවා. මොකද x, y සමමිතිකයි.

 

[AnuradhaRa]

ඔකේ අහන්න තියෙන්නේ මොකක්ද බන්?
හැබැයි ඔය ගානේ පොඩි ලස්සනක් තියනවා. මොකද x, y සමමිතිකයි.

හදල පෙන්නන්න

 

[dilann]

To find the values of \( x \) and \( y \) that satisfy both equations \( x + y = 210 \) and \( xy = 69 \), we can solve the system of equations.

1. Given \( x + y = 210 \), solve for \( y \):

\[ y = 210 - x \]

2. Substitute \( y \) in the second equation \( xy = 69 \):

\[ x(210 - x) = 69 \]

\[ 210x - x^2 = 69 \]

3. Rearrange the equation to standard quadratic form:

\[ x^2 - 210x + 69 = 0 \]

4. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -210 \), and \( c = 69 \):

\[ x = \frac{210 \pm \sqrt{210^2 - 4 \cdot 1 \cdot 69}}{2 \cdot 1} \]

5. Calculate the discriminant:

\[ \Delta = 210^2 - 4 \cdot 69 = 44100 - 276 = 43824 \]

6. Calculate the roots:

\[ x = \frac{210 \pm \sqrt{43824}}{2} \]

\[ \sqrt{43824} \approx 209.15 \]

\[ x = \frac{210 \pm 209.15}{2} \]

So, we get two solutions for \( x \):

\[ x_1 = \frac{210 + 209.15}{2} = \frac{419.15}{2} \approx 209.575 \]

\[ x_2 = \frac{210 - 209.15}{2} = \frac{0.85}{2} \approx 0.425 \]

7. Corresponding values for \( y \):

For \( x_1 \approx 209.575 \):

\[ y_1 = 210 - x_1 \approx 210 - 209.575 = 0.425 \]

For \( x_2 \approx 0.425 \):

\[ y_2 = 210 - x_2 \approx 210 - 0.425 = 209.575 \]

So the pairs \((x, y)\) that satisfy the given equations are:

\[ (x, y) = (209.575, 0.425) \]

or

\[ (x, y) = (0.425, 209.575) \]

 

[AnuradhaRa]

To find the values of \( x \) and \( y \) that satisfy both equations \( x + y = 210 \) and \( xy = 69 \), we can solve the system of equations.

1. Given \( x + y = 210 \), solve for \( y \):

\[ y = 210 - x \]

2. Substitute \( y \) in the second equation \( xy = 69 \):

\[ x(210 - x) = 69 \]

\[ 210x - x^2 = 69 \]

3. Rearrange the equation to standard quadratic form:

\[ x^2 - 210x + 69 = 0 \]

4. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -210 \), and \( c = 69 \):

\[ x = \frac{210 \pm \sqrt{210^2 - 4 \cdot 1 \cdot 69}}{2 \cdot 1} \]

5. Calculate the discriminant:

\[ \Delta = 210^2 - 4 \cdot 69 = 44100 - 276 = 43824 \]

6. Calculate the roots:

\[ x = \frac{210 \pm \sqrt{43824}}{2} \]

\[ \sqrt{43824} \approx 209.15 \]

\[ x = \frac{210 \pm 209.15}{2} \]

So, we get two solutions for \( x \):

\[ x_1 = \frac{210 + 209.15}{2} = \frac{419.15}{2} \approx 209.575 \]

\[ x_2 = \frac{210 - 209.15}{2} = \frac{0.85}{2} \approx 0.425 \]

7. Corresponding values for \( y \):

For \( x_1 \approx 209.575 \):

\[ y_1 = 210 - x_1 \approx 210 - 209.575 = 0.425 \]

For \( x_2 \approx 0.425 \):

\[ y_2 = 210 - x_2 \approx 210 - 0.425 = 209.575 \]

So the pairs \((x, y)\) that satisfy the given equations are:

\[ (x, y) = (209.575, 0.425) \]

or

\[ (x, y) = (0.425, 209.575) \]

මීට වැඩිය හොදයි...

 

[McGayan]

To find the values of \( x \) and \( y \) that satisfy both equations \( x + y = 210 \) and \( xy = 69 \), we can solve the system of equations.

1. Given \( x + y = 210 \), solve for \( y \):

\[ y = 210 - x \]

2. Substitute \( y \) in the second equation \( xy = 69 \):

\[ x(210 - x) = 69 \]

\[ 210x - x^2 = 69 \]

3. Rearrange the equation to standard quadratic form:

\[ x^2 - 210x + 69 = 0 \]

4. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -210 \), and \( c = 69 \):

\[ x = \frac{210 \pm \sqrt{210^2 - 4 \cdot 1 \cdot 69}}{2 \cdot 1} \]

5. Calculate the discriminant:

\[ \Delta = 210^2 - 4 \cdot 69 = 44100 - 276 = 43824 \]

6. Calculate the roots:

\[ x = \frac{210 \pm \sqrt{43824}}{2} \]

\[ \sqrt{43824} \approx 209.15 \]

\[ x = \frac{210 \pm 209.15}{2} \]

So, we get two solutions for \( x \):

\[ x_1 = \frac{210 + 209.15}{2} = \frac{419.15}{2} \approx 209.575 \]

\[ x_2 = \frac{210 - 209.15}{2} = \frac{0.85}{2} \approx 0.425 \]

7. Corresponding values for \( y \):

For \( x_1 \approx 209.575 \):

\[ y_1 = 210 - x_1 \approx 210 - 209.575 = 0.425 \]

For \( x_2 \approx 0.425 \):

\[ y_2 = 210 - x_2 \approx 210 - 0.425 = 209.575 \]

So the pairs \((x, y)\) that satisfy the given equations are:

\[ (x, y) = (209.575, 0.425) \]

or

\[ (x, y) = (0.425, 209.575) \]

check if you have enters 89 instead of 69 somewhere,
0.425 * 209.575 = 89.069375

edit::
aah chat gpt ?

 

[hirashgeff]

x + (69/x)=210
x^2+69-210x=0
x^2-210x+69 =0

 

[Amew]

x + y = 210
xy = 69

x y සොයන්න

x(210-x) = 69
x^2 - 210x + 69 = 0

Put the values to -b දන හෝ රින වර්ග මූල b වර්ගය රින 4ac බෙදීම 2a

 

[Candid-B]

danne na.

 

[McGayan]

හදල පෙන්නන්න

y = 69/x
x + 69/x = 210
x^2 - 210x +69 = 0
x=0.329087 or 209.670913
y=209.670913 or 0.329087

 

[AnuradhaRa]

y = 69/x
x + 69/x = 210
x^2 - 210x +69 = 0
x=0.329087 or 209.670913
y=209.670913 or 0.329087

සමීකරණයෙන් විතරයි නේද බං හදන්න පුලුවන්
වර්ගපූර්ණයෙනුත් බෑ

 

[Rastapopoulos]

ඔබගේ සමානීකරණ දෙකේ අගයන් සොයා ගැනීමට x සහ y ගණනය කරමු.

x + y = 210 ---- (1)
xy = 69 ---- (2)
```

(1) සමානීකරණයෙන් y නිගමනය කරමු:
`
y = 210 - x
```

දෙවන සමානීකරණය තුළ y = 210 - x වෙනුවට දමමු:
``
x(210 - x) = 69
210x - x^2 = 69
x^2 - 210x + 69 = 0
```

මෙය x සඳහා සමීකරණයක් වන, x^2 - 210x + 69 = 0 ලෙස ලබා ගන්නා පරිදි, ගුණිත හඳුන්වා දිය යුතුය.

a = 1, b = -210, c = 69
```
x = [ -b ± sqrt(b^2 - 4ac) ] / 2a
```

මෙය ගුණිත සමීකරණයට යෙදවමු:
```
x = [ 210 ± sqrt(210^2 - 4*1*69) ] / 2*1
x = [ 210 ± sqrt(44100 - 276) ] / 2
x = [ 210 ± sqrt(43824) ] / 2
```

sqrt(43824) අගය සොයා ගැනීමේදී:
```
x = [ 210 ± 209.46 ] / 2
```

මෙම අගයන් සොයා ගන්නේ:
```
x1 = (210 + 209.46) / 2 ≈ 209.73
x2 = (210 - 209.46) / 2 ≈ 0.27
```

x1 සහ x2 ලෙස අගයන් ඇත. ඒවාය:
```
x ≈ 209.73, y ≈ 0.27
```
අවශ්‍ය අගයන් ඇත:
```
x ≈ 209.73
y ≈ 0.27
```

ගොඩක් අමාරු උනා හදන්න උඹ වෙනුවෙන්මයි හැදුවේ හරිද බලන්න ,