Solve the simultaneous equations 3x-5=2y and y(x+y)=x(x+y)-5
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Updated 144 days ago|7/11/2023 12:50:09 AM
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User: Solve the simultaneous equations 3x-5=2y and y(x+y)=x(x+y)-5

Weegy: x = 3, y =2

Question
Updated 144 days ago|7/11/2023 12:50:09 AM
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To solve the simultaneous equations:

1) 3x - 5 = 2y
2) y(x + y) = x(x + y) - 5

Let's rearrange the second equation to simplify it:

2y(x + y) = x(x + y) - 5
2xy + 2y^2 = x^2 + xy - 5

Now, we have the following system of equations:

3x - 5 = 2y
2xy + 2y^2 = x^2 + xy - 5

To proceed, we can substitute the value of y from the first equation into the second equation:

2x(3x - 5) + 2(3x - 5)^2 = x^2 + x(3x - 5) - 5

Expanding and simplifying:

6x^2 - 10x + 2(9x^2 - 30x + 25) = x^2 + 3x^2 - 5x - 5

Combining like terms:

6x^2 - 10x + 18x^2 - 60x + 50 = 4x^2 - 5x - 5

Simplifying further:

24x^2 - 70x + 50 = 4x^2 - 5x - 5

Bringing all terms to one side:

20x^2 - 65x + 55 = 0

This quadratic equation can be solved using factoring, completing the square, or the quadratic formula. However, upon inspection, it seems that the equation does not have simple integer solutions. Therefore, a precise numerical solution would require using a numerical method such as factoring or the quadratic formula.
Added 144 days ago|7/11/2023 12:50:09 AM

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