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)

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

Soal nomor 2 (Penjumlahan dan Pengurangan Bentuk Aljabar)

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

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

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

Soal nomor 4 (Berbagai Penyederhanaan)

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

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.