Question

A.4(6x+1) - 3(4x+3) = 43

B. sistema
x+4y = 7
3x+7y = 6

C. x^2 -10x + 21 = 0

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Answer 1

Respuesta:

$A)4(6x+1) - 3(4x+3) = 43 \\ 24x + 4 - 12x - 9 = 43 \\ 12x - 5 = 43 \\ 12x = 43 + 9 \\ 12x = 52 \\ x = \frac{52}{12} \\ x = \frac{13}{3} \\ B)x+4y = 7 \\ 3x+7y = 6 \\ x = 7 - 4y \\ 3(7 - 4y) + 7y = 6 \\ 21 - 12y + 7y = 6 \\ - 12y + 7y = 6 - 21 \\ - 5y = - 15 \\ y = \frac{ - 15}{ - 5} \\ y = 3 \\ x = 7 - 4(3) \\ x = 7 - 12 \\ x = - 5 \\ C) x^2 -10x + 21 = 0 \\ (x - 7)(x - 3) = 0$