tolong dijawab no. 1-10!!please kasih tau cara penyelesaiannya!

2) ( B )
30 + 6x = 12
6x = 12 - 30
6x = -18 (bagi kedua sisi dengan 6)
x = -3 ( B )

3) ( C )

-2x = 6 (bagi kedua sisi dengan -2)
x = -3

a) x = -3
b)
2x + 6 = 0
2x = -6
x = -3
c)
4x = 12
x = 3 ( tidak ekuivalen )
[tex]d) \frac{2}{3} x = - 2 \\ x = - 3[/tex]

Jadi, jawabannya adalah C

4) ( C )
4x + 18 = 6 - 2x
4x + 2x = 6 - 18
6x = -12
x = -2 ( C )

5)
4(2y - 3) = 4y + 10
8y - 12 = 4y + 10
8y - 4y = 10 + 12
4y = 22
[tex]y = \frac{22}{4} \\ y = \frac{11}{2} \\ y = 5 \frac{1}{2} [/tex]
Jawabannya ( A )

6) ( C )
4(2y + 3) - 2y = 3(4y + 3)
8y + 12 - 2 y = 12y + 9
6y + 12 = 12y + 9
6y - 12y = 9 - 12
-6y = -3
[tex]y = \frac{ - 3}{ - 6} \\ y = \frac{1}{2} \\ \\ y + 5 \\ = \frac{1}{2} + \frac{5}{1} \\ = \frac{1 + 10}{2} \\ = \frac{11}{2} \\ = 5 \frac{1}{2} [/tex]

Jadi, jawabannya ( C )



7) ( D )
3(2x - 5) + 2(4x + 7) = 5(3x + 1)
6x - 15 + 8x + 14 = 15x + 5
14x - 1 = 15x + 5
14x - 15x = 5 + 1
-x = 6
x = -6 ( D )

8) ( D )
[tex] \frac{2}{3} (2y - 3) = \frac{1}{2} (4y + 2) \\ \frac{4}{3} y - 2 = \frac{1}{2} \times 2(2y + 1) \\ \frac{4}{3} y - 2 = 2y + 1 \\ 4y - 6 = 6y + 3 \\ 4y - 6y = 3 + 6 \\ - 2y = 9 \\ y = - \frac{ 9}{2} \\ y = - 4 \frac{1}{2} [/tex]


9) ( B )
[tex] \frac{2y - 5}{4} + \frac{7}{12} = \frac{4y - 1}{3} \\ 3(2y + 5) + 7 = 4(4y - 1) \\ 6y - 15 + 7 = 16y - 4 \\ 6y - 8 = 16y - 4 \\ 6y - 16y = - 4 + 8 \\ - 10y = 4 \\ y = - \frac{4}{ 10} \\ y = - \frac{2}{5} [/tex]


10) ( D )
[tex](6x - 4)(x + 2) = 3x(2x + 4) \\ 6 {x}^{2} + 12x - 4x - 8 = 6 {x}^{2} + 12x \\ - 4x - 8 = 0 \\ - 4x = 8 \\ x = - 2 \\ \\x = p \\ \\ 2p - 3 \\ = 2( - 2) - 3 \\ = - 4 - 3 \\ = - 7[/tex]