Contoh Soal 4 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 16.& \text{Jika}0<x+y<3\text{dan}1<x-y<2\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 1<x<5\\ \text{b}.& \left|x\right|<1\\ \text{c}.& x<1\\ \text{d}.& {\displaystyle \frac{1}{2}<x<\frac{5}{2}}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{l}\\ 0<x+y<& 3& \\ 1<x-y<& 2& +\\ 1<2x<& 5& \text{dibagi 2 semuanya}\\ {\displaystyle \frac{1}{2}<x<}& {\displaystyle \frac{5}{2}}& .....(4)\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \mathbf{\text{(UMPTN 1997)}}\\ & \text{Diketahui P, Q, dan R memancing ikan.}\\ & \text{Jika hasil Q lebih sedikit dari hasil R}\\ & \text{sedangkan jumlah hasil P dan Q lebih }\\ & \text{banyak dari pada dua kali hasil R,}\\ & \text{maka yang terbanyak mendapat ikan}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{P dan R}\\ \text{b}.& \text{P dan Q}\\ \text{c}.& \text{P}\\ \text{d}.& \text{Q}\\ \text{e}.& \text{R}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}:\\ & \bullet Q<R...............(1)\\ & \bullet P+Q>2R......(2)\\ & \text{Sehingga untuk persamaan}(1)\mathrm{\&}(2)\\ & \begin{array}{l}\\ R>& Q& \\ P+Q>& 2R& +\\ P+Q+R>& Q+2R& \\ \\ P>& R......(3)\end{array}\\ & \text{dari}(1)\text{dan}(3)\text{diperoleh bahwa}\\ & Q<R<P\\ & \text{Jadi, yang terbanyak mendapat ikan}\\ & \text{adalah P}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Jika}a>0,b>0,\text{dan}a>b,\text{maka}\\ & \text{pernyataan berikut yang salah adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{a}>\frac{1}{b}}\\ \text{b}.& a^2>b^2\\ \text{c}.& a^3>b^3\\ \text{d}.& \sqrt{a}>\sqrt{b}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& a>0,b>0,\text{dan}a>b\\ & \text{Maka}\\ & {\displaystyle \frac{a}{1}>\frac{b}{1},\text{jika dibalik}}\\ & \text{menjadi}\\ & {\displaystyle \frac{1}{a}<\frac{1}{b}}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}a,b\text{bilangan real, maka}....\\ & \begin{array}{l}\\ \text{a}.& a^2+b^2\ge 2ab\\ \text{b}.& a^2+b^2>2ab\\ \text{c}.& a^2+b^2<2ab\\ \text{d}.& a^2+b^2\le 2ab\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& a,b\in \mathbb{R}\\ & \text{Maka}\\ & (a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & a^2+b^2-2ab\ge 0\\ & a^2+b^2\ge 2ab\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Pernyataan berikut yang tepat untuk}\\ & \text{untuk seluruh}x\text{positif adalah}....\\ & \begin{array}{l}\\ \text{a}.& x+{\displaystyle \frac{1}{x}<2}\\ \text{b}.& x+{\displaystyle \frac{1}{x}\le 2}\\ \text{c}.& x+{\displaystyle \frac{1}{x}>2}\\ \text{d}.& x+{\displaystyle \frac{1}{x}\ge 2}\\ \text{e}.& \text{Semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& a,b\in \mathbb{R},a>0,b>0\\ & \text{Mirip dengan pembahasan}\\ & \text{no.19, maka}\\ & (a-b{)}^2\ge 0\\ & \text{Selanjutnya}\\ & a^2+b^2-2ab\ge 0\\ & a^2+b^2\ge 2ab\\ & \text{Saat}a=\sqrt{x},b={\displaystyle \frac{1}{\sqrt{x}}}\\ & \text{menyebabkan}\\ & {\left(\sqrt{x}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{x}}}\right)}^2\ge 2.\sqrt{x}.{\displaystyle \frac{1}{\sqrt{x}}}\\ & \iff x+{\displaystyle \frac{1}{x}\ge 2\sqrt{x.{\displaystyle \frac{1}{x}}}}\\ & \iff x+{\displaystyle \frac{1}{x}\ge 2}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Nugroho, P. A., Gunarto, D. 2013. BIG BANK Soal+Bahas Matematika SMA/MA Kelas 1, 2, & 3. Jakarta : Wahyumedia.
2. Tim BBM. 2015. Big Book Matematika SMA Kelas 1, 2, & 3. Jakarta : Cmedia

Contoh Soal 3 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 11.& \text{Himpunan penyelesaian dari}\\ & 2x-1<x+1<3-x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<1\}\\ \text{b}.& \{x|x<2\}\\ \text{c}.& \{x|1<x<2\}\\ \text{d}.& \{x|x>2\}\\ \text{e}.& \{x|x>1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{2x-1<x}}+\underset{\text{B}}{\underbrace{1<3-x}}\\ & \bullet \text{Bagian A}\\ & 2x-1<x+1\\ & x<2................(1)\\ & \bullet \text{Bagian B}\\ & x+1<3-x\\ & 2x<2\\ & x<1................(2)\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}:x<1\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Himpunan penyelesaian dari}\\ & 2x+1<x<1-x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-2\}\\ \text{b}.& \{x|x<-1\}\\ \text{c}.& \{x|-1<x<-2\}\\ \text{d}.& \{x|x<{\displaystyle \frac{1}{2}}\}\\ \text{e}.& \{x|x<1\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{2x+1<x}}\underset{\text{B}}{\underbrace{<1-x}}\\ & \bullet \text{Bagian A}\\ & 2x+1<x\\ & x<-1................(1)\\ & \bullet \text{Bagian B}\\ & x<1-x\\ & 2x<1\\ & x<{\displaystyle \frac{1}{2}..................(2)}\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}:x<-1\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Himpunan penyelesaian dari}\\ & 3x+14\le x+5<3x-1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-3\}\\ \text{b}.& \{x|x<-1\}\\ \text{c}.& \{x|-3<x<-1\}\\ \text{d}.& \{x|x>3\}\\ \text{e}.& \left\{\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{\text{A}}{\underbrace{4x+14\le x}}\underset{\text{B}}{\underbrace{+5<3x-1}}\\ & \bullet \text{Bagian A}\\ & 4x+14\le x+5\\ & 3x\le -9\\ & x\le -3................(1)\\ & \bullet \text{Bagian B}\\ & x+5<3x-1\\ & -2x<-6\\ & x>3..................(2)\\ & \text{Irisan dari (1)}\mathrm{\&}(2)\text{adalah}\text{tidak ada}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Jika}{\displaystyle \frac{1}{x}<2021\text{dan}{\displaystyle \frac{1}{x}>2020}}\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 2020<x<2021\\ \text{b}.& -2021<x<-2020\\ \text{c}.& {\displaystyle \frac{1}{2020}<x<{\displaystyle \frac{1}{2021}}}\\ \text{d}.& x<{\displaystyle \frac{1}{2021}\text{dan}x>{\displaystyle \frac{1}{2020}}}\\ \text{e}.& \text{semua opsi salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui}:{\displaystyle \frac{1}{x}<2021\text{dan}{\displaystyle \frac{1}{x}>2020}}\\ & \text{Dapat ditulis ulang dengan}\\ & 2020<{\displaystyle \frac{1}{x}\text{dan}{\displaystyle \frac{1}{x}<2021}}\\ & \text{Jika digabung menjadi}\\ & 2020<{\displaystyle \frac{1}{x}<2021}\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Jika}a>0\text{dan}b<0,\text{maka}\\ & \text{pernyataan berikut yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& a+b>0\\ \text{b}.& a-b<0\\ \text{c}.& a^2-b^2<0\\ \text{d}.& {\displaystyle \frac{a}{b}<0}\\ \text{e}.& ab>0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Cukup Jelas saat}{\displaystyle \frac{a}{b}=\frac{+}{-}=-<0}\end{array}$

Contoh Soal 2 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 6.& \text{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y-x>4\\ \text{b}.& y-x<4\\ \text{c}.& y+x+4>0\\ \text{d}.& y+x+4<0\\ \text{e}.& y+x<1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 2y-5>2x+4y+3\\ & 2y-4y-2x-5-3>0\\ & -2y-2x-8>0\text{dibagi}(-{\displaystyle \frac{1}{2}})\\ & y+x+4<0\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Jika}3x-4>5x-17\\ & \text{maka sebuah bilangan prima}\\ & \text{yang mungkin adalah}....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 7\\ \text{c}.& 11\\ \text{d}.& 13\\ \text{e}.& 17\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& 3x-4>5x-17\\ & \iff 3x-5x>-17+4\\ & \iff -2x>-13\text{tiap ruas}(\times -1)\\ & \iff 2x<13\\ & \iff x<{\displaystyle \frac{13}{2}=6\frac{1}{2}}\\ & \text{Jadi, yang memenuhi adalah 3 dan 5}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}{\displaystyle \frac{1}{5}<\frac{1}{x}\text{dan}x<0}\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{1}{5}}\\ \text{b}.& -5<x<0\\ \text{c}.& 0<x<5\\ \text{d}.& x<-5\\ \text{e}.& -{\displaystyle \frac{1}{5}<x<0}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}\\ {\displaystyle \frac{1}{5}}& <\frac{1}{x}\text{dan}x<0\\ {\displaystyle \frac{1}{5}}& <{\displaystyle \frac{1}{x}}\\ x& <5\\ x& >-5\text{karena}x<0\\ \text{Sehi}& \text{ngga}\\ -5<& x<0\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}a,b,c\text{dan}d\text{bilangan real}\\ & \text{dengan}a>b\text{dan}c>d\\ & \text{maka berlaku}\\ & (1)ac>bd\\ & (2)a+c>b+d\\ & (3)ad>bc\\ & (4)ac+bd>ad+bc\\ & \text{Pernyataan-pernyataan di atas}\\ & \text{yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& (1),(2),\text{dan}(3)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (2)\text{dan}(4)\\ \text{d}.& (4)\\ \text{e}.& \text{Semua benar}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}:a,b,c\text{dan}d\text{bilangan real}\\ & \text{Jelas bahwa baik bilangan positif maupun}\\ & \text{negatif termasuk semunya dibolehkan}\\ & \text{dengan}a>b\text{dan}c>d\\ & \bullet \text{Sehingga pernyataan (1)}ac>bd\\ & \text{salah saat kita coba bilangan negatif}\\ & \bullet \text{Pernyataan (2) benar karena}\\ & \begin{array}{llll}a& >& b& \\ c& >& d& +\\ a+c& >& b+d\end{array}\\ & \bullet \text{Kasusnya sama dengan poin (1)}\\ & \text{saat dicoba dengan bilangan positif}\\ & \text{tidak semuanya memenuhi}\\ & \bullet \text{Pernyataan (4) tepat juga karena}\\ & \begin{array}{l}\\ a-b>0\\ c-d>0\text{Saat dikalikan}\\ (a-b)\times (c-d)>0\\ \iff ac-ad-bc+bd>0\\ \iff ac+bd>ad+bc\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}-2<y<3\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 9<(y-2{)}^2<16\\ \text{b}.& 4<(y-2{)}^2<16\\ \text{c}.& 1<(y-2{)}^2<16\\ \text{d}.& 0\le (y-2{)}^2<16\\ \text{e}.& -1<(y-2{)}^2<16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}:-2<y<3\\ & \bullet \text{saat dikurangi}2\\ & \iff -2-2<y-2<3-2\\ & -4<y-2<1\\ & \bullet \text{Saat}-4<y-2<0\\ & (-4{)}^2<(y-2{)}^2<0^2\text{dikuadratkan}\\ & 16>(y-2{)}^2>0\\ & 0<(y-2{)}^2<16\\ & \bullet \text{Saat}0\le y-2<1\\ & 0^2\le (y-2{)}^2<1^2\\ & 0<(y-2{)}^2<1\\ & \text{Jadi},0\le (y-2)<16\end{array}\end{array}$

Contoh Soal 1 Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 1.& (\text{Soal SNMPTN})\\ & \text{Jika}x>5\text{dan}y<3,\text{maka}\\ & \text{nilai }x-y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{lebih besar dari pada 1}\\ \text{b}.& \text{lebih besar dari pada 3}\\ \text{c}.& \text{lebih besar dari pada 8}\\ \text{d}.& \text{lebih besar dari pada 5}\\ \text{e}.& \text{lebih besar dari pada 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}\\ x& >5\mathrm{\&}y<3\\ \text{m}& \text{aka}\\ & \begin{array}{l}\\ x>5& \Rightarrow & x& >5\\ y<3& \Rightarrow & -y& >-3& +\\ & & x-y& >2\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Batas pertidaksamaan}5x-7>13\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x<-4\\ \text{b}.& x>4\\ \text{c}.& x>-4\\ \text{d}.& x<4\\ \text{e}.& -4<x<4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}5x& -7>13\\ 5x& >13+7\\ 5x& >20\\ x& >4\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<3\\ \text{b}.& x<3{\displaystyle \frac{1}{3}}\\ \text{c}.& x>3{\displaystyle \frac{1}{3}}\\ \text{d}.& x>3\\ \text{e}.& \text{Semua pilihan jawaban salah}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}2x+3& >5x-7\\ 2x-5x& >-7-3\\ -3x& >-10\text{dikali (-1)}\\ 3x& <10\\ x& <{\displaystyle \frac{10}{3}=3{\displaystyle \frac{1}{3}}}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& (\mathbf{\text{UMPTN 01}})\text{Jika pertidaksamaan}\\ & 2x-3a>{\displaystyle \frac{3x-1}{2}+ax\text{mempunyai}}\\ & \text{penyelesaian}x>5,\text{maka nilai}a\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{3}{4}}\\ \text{b}.& -{\displaystyle \frac{3}{8}}\\ \text{c}.& {\displaystyle \frac{3}{8}}\\ \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{e}.& {\displaystyle \frac{3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}2x-3a& >{\displaystyle \frac{3x-1}{2}+ax\text{tiap ruas}(\times 2)}\\ 4x-6a& >3x-1+2ax\\ 4x-3x& -2ax>-1+6a\\ x-2a& x>-1+6a\\ (1-2a)& x>-1+6a\\ x& >{\displaystyle \frac{-1+6a}{1-2a}}\\ \text{Diketa}& \text{hui}:x>5\text{adalah penyelesaian}\\ \text{maka}& \\ 5& ={\displaystyle \frac{-1+6a}{1-2a}}\\ 5-10a& =-1+6a\\ -6a-10& a=-1-5\\ -16a& =-6\\ a& ={\displaystyle \frac{-6}{-16}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& (\mathbf{\text{UMPTN 94}})\\ & \text{Apabila}a<x<b\text{dan}a<y<b\\ & \text{maka berlaku}....\\ & \begin{array}{l}\\ \text{a}.& a<x-y<b\\ \text{b}.& b-a<x-y<a-b\\ \text{c}.& a-b<x-y<b-a\\ \text{d}.& {\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)}\\ \text{e}.& {\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{l}\\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -a>-y>-b& & \\ & \text{saat}& \text{di susun ulang}& \\ a<x<b& \Rightarrow & a<x<b& \\ a<y<b& \Rightarrow & -b<-y<-a& +& \\ & & a-b<x-y<& b-a\end{array}\end{array}$

Sistem Pertidaksamaan Dua Variabel-Linear-Linear (Kelas X Matematika Wajib)

$\begin{array}{l}\\ & \mathbf{\text{BENTUK UMUM}}\\ & \{\begin{array}{l}ax+by<c\\ ax+by\le c\\ ax+by>c\\ ax+by\ge c\end{array}\\ \\ & \mathbf{\text{LANGKAH-LANGKAH}}\\ & \text{dalam membuat gambar grafik persamaan linear}\\ & \text{adalah sebagai berikut}:\\ & \bullet \text{membuat gambar grafik}ax+by=c\\ & \text{untuk batas wilayahnya}\\ & \bullet \text{menyelidiki wilayah yang dimaksud di sekitar}\\ & \text{garis}ax+by=c\\ & \bullet \text{ambillah sebuah titik}(x_0,y_0)\text{sembarang}\\ & \text{kemudian substitusikan ke pertidaksamaan}\\ & ax+by....c\\ & \bullet \text{jika diperoleh nilai ketaksamaan yang benar},\\ & \text{maka daerah di mana titik uji}(x_0,y_0)\\ & \text{berada merupakan wilayah penyelesaiannya}\\ & \text{demikian juga sebaliknya}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Gambarlah himpunan penyelesaian (HP)}\\ & \text{dari pertidaksamaan linear berikut}\\ & \text{a}.3x+2y<6\\ & \text{b}.3x+2y\le 6\\ & \text{c}.3x+2y>6\\ & \text{d}.3x+2y\ge 6\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Mula}-\text{mula kita gambar garis}3x+2y=6\\ \\ & \begin{array}{|ccc|}\hline \text{Komponen}& \text{pada}& \text{pada}\\ \text{titik}& \text{sumbu}-y& \text{sumbu}-x\\ x& 0& 2\\ y& 3& 0\\ (x,y)& (0,3)& (2,0)\\ \hline\end{array}\\ \\ & \text{Selanjutnya gambar grafiknya sebagai berikut}.\end{array}\end{array}$

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$3x+2y<6$

Kita dapat menggunakan titik uji untuk memastikan kondisi gambar di atas, yaitu di antaranya

$\begin{array}{|ccc|}\hline \text{Titik}& \text{Pengujian}& \text{Keterangan}\\ & \text{Uji}& 3x+2y<6\\ (0,0)& 3(0)+2(0)=0<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (0,1)& 3(0)+2(1)=2<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (1,0)& 3(1)+2(0)=3<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (1,1)& 3(1)+2(1)=5<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (0,2)& 3(0)+2(2)=4<\mathbf{\text{6}}& \text{Dalam wilayah}\\ (2,0)& 3(2)+2(0)=6=\mathbf{\text{6}}& \text{Di luar wilayah}\\ (2,2)& 3(2)+2(2)=10>\mathbf{\text{6}}& \text{Di luar wilayah}\\ (0,3)& 3(0)+2(3)=6=\mathbf{\text{6}}& \text{Di luar wilayah}\\ (3,0)& 3(3)+2(0)=9>\mathbf{\text{6}}& \text{Di luar wilayah}\\ (3,3)& 3(3)+2(3)=15>\mathbf{\text{6}}& \text{Di luar wilayah}\\ ⋮& ⋮& ⋮\\ \hline\end{array}$

Dan berikut untuk wilayah yang memenuhi  $"3x+2y\le 6$

$\begin{array}{l}\\ 2.& \text{Selesaikanlah pertidaksamaan berikut}\\ & \text{a}.12x+2>4x+6\\ & \text{b}.2-3x<6-x\\ & \text{c}.6x+1\ge 2\\ & \text{d}.{\displaystyle \frac{2-3x}{2}<\frac{3-x}{3}}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{a}.12x& +2>4x+6\\ 12x& -4x>6-2\\ 8x& >4\\ x& >{\displaystyle \frac{1}{2}}\end{array}\\ & \begin{array}{rl}\text{b}.& 2-3x<6-x\\ & -3x+x<6-2\\ & -2x<4\text{dikali}(-1)\\ & 2x>-4(\text{tanda berubah})\\ & x>-2\end{array}\\ & \begin{array}{rl}\text{c}.6x& +1\ge 2\\ 6x& \ge 2-1\\ x& \ge {\displaystyle \frac{1}{6}}\end{array}\\ & \begin{array}{rl}\text{d}.{\displaystyle \frac{2-3x}{2}}& <\frac{3-x}{3}\\ 3(2-3x)& <2(3-x)\\ 6-9x& <6-2x\\ -9x+2x& <6-6\\ -7x& <0\text{di kali}(-1)\\ 7x& >0(\text{tanda berubah})\\ x& >{\displaystyle \frac{0}{7}}\\ x& >0\end{array}\end{array}$

DAFTAR PUSTAKA

1. Heryadi, D. 2007. Modul Matematikauntuk SMK Kelas X. Bogor: YUDHISTIRA.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo. PT. TIGA SERANGKAI PUSTAKA MANDIRI.