Question

(3x-2)(x+4)=3(x+5)(x-1)​

Answer 1

Ответ:

вот пожалуйста обращайся я рада помочь)))))

Answer 2

To solve the equation (3x-2)(x+4) = 3(x+5)(x-1), we need to find the value(s) of x that satisfy the equation. We'll begin by expanding both sides of the equation and then simplifying to isolate x.

Expanding the left-hand side: (3x - 2)(x + 4) = 3x(x) + 3x(4) - 2(x) - 2(4) = 3x^2 + 12x - 2x - 8 = 3x^2 + 10x - 8

Expanding the right-hand side: 3(x + 5)(x - 1) = 3(x(x) + x(-1) + 5(x) + 5(-1)) = 3(x^2 - x + 5x - 5) = 3(x^2 + 4x - 5)

Now, the equation becomes: 3x^2 + 10x - 8 = 3x^2 + 12x - 15

Notice that the quadratic terms (3x^2) cancel out on both sides of the equation. Now, let's simplify the equation further:

3x^2 + 10x - 8 - (3x^2 + 12x - 15) = 0

Performing the subtraction: (3x^2 - 3x^2) + (10x - 12x) + (-8 + 15) = 0

Simplifying further: 0x - 2 + 7 = 0

Combine the constants: 5 = 0

The equation 5 = 0 is not valid; it means there is no value of x that satisfies the original equation. Therefore, there is no solution for this equation.

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