Pembahasan Soal Buku BSE Matematika SMP Kelas 7 Halaman 85 (Kurikulum Merdeka)

Hai sobat semuanya, pada postingan kali ini mimin akan sajikan pembahasan beberapa soal buku BSE matematika SMP kelas 7 halaman 85 (kurikulum merdeka). Selamat belajar. Semoga bermanfaat.

Soal nomor 1 (Aljabar Linear)

$ \begin{aligned} (1).& 4a + 3a\\&= (4+3)a\\&=7a\\ \\ (2). &8a – 6a\\&=(8-6)a\\&=2a\\ \\ (3).&-2x – 4x\\&=(-2-4)x\\&=-6x\\ \\(4).& 9a – 10a\\&=(9-10)a\\&=-a\\ \\ (5).& -2x + 7x\\&=(-2+7)x\\&=5x\\ \\ (6). &4a + 6 + a + 3\\&=4a  + a+6 + 3\\&=(4+1)a+6 + 3\\&=5a+9\\ \\ (7).&-5x + 10 + 3x – 9\\&= -5x + 3x+10 – 9\\&=(-5+3)x+10 – 9\\&= -2x+1\\ \\ (8).&7 – 8a – a + 6\\&=  – 8a – a + 6+7\\&=(-8-1)a+ 6+7\\&=-9a+13\\ \\ (9). &2,7x – 1,4x\\&=(2,7-1,4)x\\&=1,3x\\ \\(10).& \frac{2}{3}y+\frac{5}{6}y\\&=(\frac{2}{3}+\frac{5}{6})y\\&=\frac{9}{6}y\\&=\frac{3}{2}y \end{aligned}$

Soal nomor 2 (Penjumlahan dan Pengurangan Bentuk Aljabar)

$ \begin{aligned} (1).&(6x + 2) + (2x – 9)\\&=6x + 2 + 2x – 9\\&= 6x  + 2x +2– 9\\&=(6+2)x +2– 9\\&=8x-7\\ \\ (2).& (5 – 6x) + (9x – 7)\\&=5 – 6x+ 9x – 7\\&=  – 6x+ 9x – 7+5\\&=(-6+9)x– 7+5\\&=3x-2\\ \\ (3).&(\frac{4}{9}x-\frac{5}{3})+(\frac{5}{9}x+\frac{4}{3})\\&=\frac{4}{9}x-\frac{5}{3}+\frac{5}{9}x+\frac{4}{3}\\&=\frac{4}{9}x+\frac{5}{9}x-\frac{5}{3}+\frac{4}{3}  \\&= (\frac{4}{9}+\frac{5}{9})x-\frac{5}{3}+\frac{4}{3}\\&= x-\frac{1}{3}\\ \\ (4).&(7x + 4) – (5x – 1) \\&= (7x + 4) + (-5x + 1)\\&= 7x + 4 -5x + 1\\&=7x  -5x + 4+ 1\\&=(7-5)x+ 4+ 1\\&=2x+5\\ \\ (5).&(-2y + 8) – (3y + 6)\\&= (-2y + 8) + (-3y - 6)\\&= -2y + 8 + -3y - 6\\&= -2y  + -3y+ 8 - 6 \\&=(-2-3)y+ 8 - 6\\&=-5y+2\\ \\ (6).&(14 – a) - (-9 – a)\\&= (14 – a) + (9 + a)\\&=14 – a + 9 + a\\&=  – a  + a+14+9\\&=23\\ \\ (7).&(\frac{1}{4}y+6)-(-\frac{1}{2}y-3)\\&= (\frac{1}{4}y+6)+(\frac{1}{2}y+3)\\&= \frac{1}{4}y+6+\frac{1}{2}y+3\\&= \frac{1}{4}y+\frac{1}{2}y+6+3\\&=(\frac{1}{4}+\frac{1}{2})y +6+3\\&= \frac{3}{4}y+9  \end{aligned}$

Soal nomor 3 (Aljabar Linear dan Perkalian serta Pembagian dengan Bilangan)

$ \begin{aligned} (1).&9a × 3\\&= 9×a × 3 \\&=9 × 3×a\\&=27a\\ \\(2).&(-5) × 8x\\&=(-5) × 8×x\\&= -40x\\ \\ (3).&-0,6y × 4\\&=-0,6×y × 4\\&= -0,6 × 4×y\\&=-2,4y\\ \\ (4).&12 \times \frac{4}{3}a\\&= 12 \times \frac{4}{3}×a\\&=16a\\ \\ (5).&-3(a + 7)\\&=(-3) ×  a + (-3) \times 7) \\&= -3a-21 \\ \\(6).&(6x – 5) × 4\\&= 6x × 4 – 5 × 4 \\&= 24x-20\\ \\ (7).& \frac{1}{2}(8a-6)\\&= \frac{1}{2}  \times 8a+\frac{1}{2} \times (-6)\\&= 4a-3\\ \\ (8).& \frac{12x-5}{3} \times 8 \\&= (12x-5) \times \frac{8}{3}  \\ &=  12x \times \frac{8}{3} +  (-5) \times  \frac{8}{3} \\&= 32x-(\frac{40}{3})\\&= 24x-10\\ \\ (9).&15y : 5\\&= 15y \times \frac{1}{5}\\&= 15 \times \frac{1}{5} \times y\\&= 3y\\ \\ (10).&21a : (-3)\\&= 21a \times (-\frac{1}{3})\\&=  21 \times (-\frac{1}{3}) \times a \\&= -7a\\ \\(11).& (-8x) : 20\\&= -8x \times \frac{1}{20}\\&= -8 \times \frac{1}{20} \times x \\&= -\frac{8}{20}x\\&= -\frac{2}{5}x\\ \\ (12). &10a : \frac{5}{12}\\&= 10a \times \frac{12}{5}\\&= 10 \times \frac{12}{5} \times a\\&=24a\\ \\ (13). & (10x – 35) : 5\\&= (10x – 35) \times \frac{1}{5}\\&= 10x  \times \frac{1}{5} – 35 \times \frac{1}{5}\\&= 2x-7\\ \\ (14).& (-6a + 9) : (-3)\\&= (-6a + 9) \times (-\frac{1}{3})\\&= -6a \times (-\frac{1}{3}) + 9 \times (-\frac{1}{3})\\&= 2a-3\\ \\ (15).& (12x+4) : \frac{2}{3}\\&= (12x+4) \times \frac{3}{2}\\&= 12x \times \frac{3}{2} + 4 \times \frac{3}{2} \\&= 18x+6  \end{aligned}$

Soal nomor 4 (Berbagai Penyederhanaan)

$ \begin{aligned} (1).& 4x + 5(2x – 7)\\ &=4x + 10x – 35\\ &= 14x-35\\ \\ (2).&7(2a – 1) + 6(-3a + 2)\\&= 14a –7 -18a + 12\\&= -4a+5\\ \\ (3).&-(4a + 7) + 3(a + 5)\\&= -4a - 7 + 3a + 15\\&= -a+8\\ \\ (4).& 9x – 2(x – 8)\\&=9x – 2x + 16\\&= 7x+16\\ \\ (5).&8(y – 1) - (7y + 2)\\&=8y – 8 - 7y - 2\\&= y-10\\ \\ (6). &-5(x – 1) - 4(2x + 1)\\&= -5x +5 - 8x -4\\&= -13x+1\\ \\(7).&6(2a + 4) - 8(3 - a)\\&= 12a + 24 - 24 +8a\\&= 20a\\ \\ (8).& \frac{1}{4}(x-8) + \frac{1}{2}(x-4)\\&=  \frac{1}{4}x-2 + \frac{1}{2}x-2 \\&= \frac{3}{4}x-4\\ \\ (9).&\frac{1}{9}(3x+7) - \frac{1}{3}(x+2)\\&= \frac{1}{3}x+\frac{7}{9}-\frac{1}{3}x-\frac{2}{3}\\&= \frac{1}{9}  \end{aligned}$

Demikianlah pembahasan beberapa soal buku BSE matematika SMP kelas 7 halaman 85 (kurikulum merdeka). Semoga bermanfaat. 

Referensi

Tim Gakko Tosho. 2021. Matematika untuk Sekolah Menengah Pertama Kelas VII. Jakarta: Kementerian Pendidikan, Kebudayaan, Riset, dan Teknologi.