Question

calculați a).3(2x-3)=4(3x+5);b).5(3x+5)=7(3x-10);c).10(2x-5)=5(2x-9);d).5(3x-8)=4(4x+7)​

Answer 1

Răspuns:

$a)3(2x - 3) = 4(3x + 5) \\ 3 \times 2x - 3 \times 3 = 4 \times 3x + 4 \times 5 \\ 6x - 9 = 12x + 20 \\ 6x - 12x = 20 + 9 \\ - 6x = 29 \\ x = \frac{ - 29}{6}$

$b)5(3x + 5) = 7(3x - 10) \\ 5 \times 3x + 5 \times 5 = 7 \times 3x - 7 \times 10 \\ 15x + 25 = 21x - 70 \\ 15x - 21x = - 25 - 70 \\ - 6x = - 95 \\ x = \frac{ - 95}{ - 6} \\ x = \frac{95}{6}$

$c) 10(2x - 5) = 5(2x - 9) \\ 10 \times 2x - 10 \times 5 = 5 \times 2x - 5 \times 9 \\ 20x - 50 = 10x - 45 \\ 20x - 10x = 50 - 45 \\ 10x = 9 \\ x = \frac{9}{10}$

$d)5(3x - 8) = 4(4x + 7) \\ 5 \times 3x - 5 \times 8 = 4 \times 4x + 4 \times 7 \\ 15x - 40 = 16x + 28 \\ 15x - 16x = 40 + 28 \\ - x = 68 \: \: \: | \times ( - 1) \\ x = - 68$