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

$\begin{array}{ll}\\ 16.&\textrm{Jika}\: \: 0<x+y<3\: \: \textrm{dan}\: \: 1<x-y<2\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1<x<5\\ \textrm{b}.&\left | x \right |<1\\ \textrm{c}.&x<1\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{2}<x<\frac{5}{2}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{array}{llll}\\ 0<x+y<&3&\\ 1<x-y<&2&+\\\hline \: \: 1<2x<&5&\color{black}\textrm{dibagi 2 semuanya}\\ \quad \displaystyle \frac{1}{2}<x<&\displaystyle \frac{5}{2}&\: .....\color{red}(4)\\ \end{array} \end{array}$

$\begin{array}{ll}\\ 17.&\textbf{(UMPTN 1997)}\\ &\textrm{Diketahui P, Q, dan R memancing ikan.}\\ & \textrm{Jika hasil Q lebih sedikit dari hasil R}\\ & \textrm{sedangkan jumlah hasil P dan Q lebih }\\ & \textrm{banyak dari pada dua kali hasil R,}\\ &\textrm{maka yang terbanyak mendapat ikan}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{P dan R}\\ \textrm{b}.&\textrm{P dan Q}\\ \color{red}\textrm{c}.&\textrm{P}\\ \textrm{d}.&\textrm{Q}\\ \textrm{e}.&\textrm{R} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\color{black}\textrm{Diketahui}:\\ &\bullet \: Q< R\: ...............\color{red}(1)\\ &\bullet \: P+Q> 2R\: ......\color{red}(2)\\ &\textrm{Sehingga untuk persamaan}\: \: \color{black}(1)\: \&\: (2)\\ &\begin{array}{llll}\\ \qquad\qquad R>&Q&\\ \qquad P+Q>&2R&+\\\hline P+Q+R>&Q+2R&\\\\ \qquad\quad\quad P>&R\: ......\color{red}(3)\\ \end{array}\\ &\textrm{dari} \: \color{red}(1)\: \color{blue}\textrm{dan}\: \color{red}(3)\: \color{blue}\textrm{diperoleh bahwa}\\ &Q<R< P\\ &\textrm{Jadi, yang terbanyak mendapat ikan}\\ &\color{red}\textrm{adalah P} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika}\: \: a>0,\: b>0,\: \: \textrm{dan}\: \: a>b,\: \: \textrm{maka}\\ &\textrm{pernyataan berikut yang salah adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{a}>\frac{1}{b}\\ \textrm{b}.&a^{2}>b^{2}\\ \textrm{c}.&a^{3}>b^{3}\\ \textrm{d}.&\sqrt{a}>\sqrt{b}\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&a>0,\: b>0,\: \: \textrm{dan}\: \: a>b\\ &\color{red}\textrm{Maka}\\ &\displaystyle \frac{a}{1}>\frac{b}{1},\: \: \textrm{jika dibalik}\\ &\color{red}\textrm{menjadi}\\ &\displaystyle \frac{1}{a}<\frac{1}{b} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan real, maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&a^{2}+b^{2}\geq 2ab\\ \textrm{b}.&a^{2}+b^{2}> 2ab\\ \textrm{c}.&a^{2}+b^{2}< 2ab\\ \textrm{d}.&a^{2}+b^{2}\leq 2ab\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&a,b\in \mathbb{R}\\ &\color{red}\textrm{Maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Pernyataan berikut yang tepat untuk}\\ &\textrm{untuk seluruh}\: \: x\: \: \textrm{positif adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x+\displaystyle \frac{1}{x}<2\\ \textrm{b}.&x+\displaystyle \frac{1}{x}\leq 2\\ \textrm{c}.&x+\displaystyle \frac{1}{x}>2\\ \color{red}\textrm{d}.&x+\displaystyle \frac{1}{x}\geq 2\\ \textrm{e}.&\textrm{Semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&a,b\in \mathbb{R},\: \: a>0,\: b>0\\ &\color{red}\textrm{Mirip dengan pembahasan}\\ &\color{red}\textrm{no.19, maka}\\ &(a-b)^{2}\geq 0\\ &\color{red}\textrm{Selanjutnya}\\ &a^{2}+b^{2}-2ab\geq 0\\ &a^{2}+b^{2}\geq 2ab\\ &\color{black}\textrm{Saat}\: \: a=\sqrt{x},\: \: b=\displaystyle \frac{1}{\sqrt{x}}\\ &\textrm{menyebabkan}\\ &\left ( \sqrt{x} \right )^{2}+\left ( \displaystyle \frac{1}{\sqrt{x}} \right )^{2}\geq 2.\sqrt{x}.\displaystyle \frac{1}{\sqrt{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2\sqrt{x.\displaystyle \frac{1}{x}}\\ &\Leftrightarrow \: x+\displaystyle \frac{1}{x}\geq 2 \end{aligned} \end{array}$

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)

$\begin{array}{ll}\\ 11.&\textrm{Himpunan penyelesaian dari}\\ &2x-1<x+1<3-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ x|x<1 \right \}\\ \textrm{b}.&\left \{ x|x<2 \right \}\\ \textrm{c}.&\left \{ x|1<x<2 \right \}\\ \textrm{d}.&\left \{ x|x>2 \right \}\\ \textrm{e}.&\left \{ x|x>1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x-1<x}}\, +\, \underset{\textrm{B}}{\underbrace{1<3-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x-1<x+1\\ &\qquad x<2\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+1<3-x\\ &\qquad 2x<2\\ &\qquad x<1\: ................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Himpunan penyelesaian dari}\\ &2x+1<x<1-x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-2 \right \}\\ \color{red}\textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-1<x<-2 \right \}\\ \textrm{d}.&\left \{ x|x<\displaystyle \frac{1}{2}\right \}\\ \textrm{e}.&\left \{ x|x<1 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\underset{\textrm{A}}{\underbrace{2x+1<x}} \underset{\textrm{B}}{\underbrace{\: <1-x}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 2x+1<x\\ &\qquad x<-1\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x<1-x\\ &\qquad 2x<1\\ &\qquad x<\displaystyle \frac{1}{2}\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}:\: \color{red}x<-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Himpunan penyelesaian dari}\\ &3x+14\leq x+5<3x-1\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x<-3 \right \}\\ \textrm{b}.&\left \{ x|x<-1 \right \}\\ \textrm{c}.&\left \{ x|-3<x<-1 \right \}\\ \textrm{d}.&\left \{ x|x>3\right \}\\ \color{red}\textrm{e}.&\left \{\: \: \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\underset{\textrm{A}}{\underbrace{4x+14\leq x}} \underset{\textrm{B}}{\underbrace{\, +\, 5 <3x-1}}\\ &\bullet \quad\color{black}\textrm{Bagian A}\\ &\qquad 4x+14\leq x+5\\ &\qquad 3x\leq -9\\ &\qquad x\leq -3\: ................\color{red}(1)\\ &\bullet \quad\color{black}\textrm{Bagian B}\\ &\qquad x+5<3x-1\\ &\qquad -2x<-6\\ &\qquad x>3\: ..................\color{red}(2)\\ &\textrm{Irisan dari (1)}\: \&\: (2)\: \: \textrm{adalah}\: \color{red}\textrm{tidak ada} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Jika}\: \: \displaystyle \frac{1}{x}<2021\: \: \textrm{dan}\: \: \displaystyle \frac{1}{x}>2020\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2020<x<2021\\ \textrm{b}.&-2021<x<-2020\\ \textrm{c}.&\displaystyle \frac{1}{2020}<x<\displaystyle \frac{1}{2021}\\ \textrm{d}.&x<\displaystyle \frac{1}{2021}\: \: \textrm{dan}\: \: x>\displaystyle \frac{1}{2020}\\ \textrm{e}.&\textrm{semua opsi salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}:\: \color{black}\displaystyle \frac{1}{x}<2021\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}>2020\\ &\textrm{Dapat ditulis ulang dengan}\\ &\color{black}2020<\displaystyle \frac{1}{x}\: \: \color{red}\textrm{dan}\: \: \color{black}\displaystyle \frac{1}{x}<2021\\ &\textrm{Jika digabung menjadi}\\ &\color{black}2020<\displaystyle \frac{1}{x}<\color{black}2021 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Jika}\: \: a>0\: \: \textrm{dan}\: \: b<0\: ,\: \textrm{maka}\\ &\textrm{pernyataan berikut yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a+b>0\\ \textrm{b}.&a-b<0\\ \textrm{c}.&a^{2}-b^{2}<0\\ \color{red}\textrm{d}.&\displaystyle \frac{a}{b}<0\\ \textrm{e}.&ab>0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\textrm{Cukup Jelas saat}\: \: \color{red}\displaystyle \frac{a}{b}=\frac{+}{-}=-<0 \end{array}$

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

$\begin{array}{ll}\\ 6.&\textrm{Bentuk sederhana dari}\\ & 2y-5>2x+4y+3\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y-x>4\\ \textrm{b}.&y-x<4\\ \textrm{c}.&y+x+4>0\\ \color{red}\textrm{d}.&y+x+4<0\\ \textrm{e}.&y+x<1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&2y-5>2x+4y+3\\ &2y-4y-2x-5-3>0\\ &-2y-2x-8>0\: \: \color{black}\textrm{dibagi}\: \left ( -\displaystyle \frac{1}{2} \right )\\ &y+x+4<0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 7.&\textrm{Jika}\: \: 3x-4>5x-17\\ &\textrm{maka sebuah bilangan prima}\\ &\textrm{yang mungkin adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&3\\ \textrm{b}.&7\\ \textrm{c}.&11\\ \textrm{d}.&13\\ \textrm{e}.&17 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&3x-4>5x-17\\ &\Leftrightarrow 3x-5x>-17+4\\ &\Leftrightarrow -2x>-13\quad \color{black}\textrm{tiap ruas}\: (\times -1)\\ &\Leftrightarrow 2x<13\\ &\Leftrightarrow x<\displaystyle \frac{13}{2}=6\frac{1}{2}\\ &\color{black}\textrm{Jadi, yang memenuhi adalah 3 dan 5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: \displaystyle \frac{1}{5}<\frac{1}{x}\: \: \textrm{dan}\: \: x<0\\ &\textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{1}{5}\\ \color{red}\textrm{b}.&-5<x<0\\ \textrm{c}.&0<x<5\\ \textrm{d}.&x<-5\\ \textrm{e}.&-\displaystyle \frac{1}{5}<x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\\ \displaystyle \frac{1}{5}&<\frac{1}{x}\: \: \: \textrm{dan}\: \: x<0\\ \displaystyle \frac{1}{5}&<\displaystyle \frac{1}{x}\\ x&<5 \\ x&>-5\qquad \color{black}\textrm{karena}\: \: x<0\\ \textrm{Sehi}&\textrm{ngga}\\ \color{red}-5<&\color{red}x<0 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: -2<y<3\: \: \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&9<(y-2)^{2}<16\\ \textrm{b}.&4<(y-2)^{2}<16\\ \textrm{c}.&1<(y-2)^{2}<16\\ \color{red}\textrm{d}.&0\leq (y-2)^{2}<16\\ \textrm{e}.&-1<(y-2)^{2}<16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}:\: -2<y<3\\ &\color{red}\bullet \quad \textrm{saat dikurangi}\: \: 2\\ &\qquad \Leftrightarrow \: -2-2<y-2<3-2\\ &\qquad -4<y-2<1\\ &\color{red}\bullet \quad \textrm{Saat}\: \: -4<y-2<0\\ &\qquad (-4)^{2}<(y-2)^{2}<0^{2}\quad \textrm{dikuadratkan}\\ &\qquad 16>(y-2)^{2}>0\\ &\qquad 0<(y-2)^{2}<16\\ &\color{red}\bullet \quad \textrm{Saat}\: \: 0\leq y-2<1\\ &\qquad 0^{2}\leq (y-2)^{2}<1^{2}\\ &\qquad 0<(y-2)^{2}<1\\ &\textrm{Jadi}\: ,\: \: \color{red}0\leq (y-2)<16 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 1.&(\textrm{Soal SNMPTN})\\ &\textrm{Jika}\: \: x>5\: \: \textrm{dan}\: \: y<3,\: \: \textrm{maka}\\ &\textrm{nilai }\: \: x-y\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{lebih besar dari pada 1}\\ \textrm{b}.&\textrm{lebih besar dari pada 3}\\ \textrm{c}.&\textrm{lebih besar dari pada 8}\\ \textrm{d}.&\textrm{lebih besar dari pada 5}\\ \color{red}\textrm{e}.&\textrm{lebih besar dari pada 2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}\\ x&>5\: \: \color{red}\&\: \: \color{blue}y<3\\ \textrm{m}&\textrm{aka}\\ &\begin{array}{llllll}\\ x>5&\Rightarrow &x&>5\\ y<3&\Rightarrow &\color{black}-y&\color{black}>-3&\color{red}+\\\hline &&\color{red}x-y&>\color{red}2 \end{array} \end{aligned} \end{array}$

$\begin{array}{l}\\ 2.&\textrm{Batas pertidaksamaan}\: \: 5x-7>13\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x<-4\\ \color{red}\textrm{b}.&x>4\\ \textrm{c}.&x>-4\\ \textrm{d}.&x<4\\ \textrm{e}.&-4<x<4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}5x&-7>13\\ 5x&>13+7\\ 5x&>20\\ x&\color{red}>4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Penyelesaian dari pertidaksamaan}\\ & 2x+3>5x-7\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x<3\\ \color{red}\textrm{b}.&x<3\displaystyle \frac{1}{3}\\ \textrm{c}.&x>3\displaystyle \frac{1}{3}\\ \textrm{d}.&x>3\\ \textrm{e}.&\textrm{Semua pilihan jawaban salah} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}2x+3&>5x-7\\ 2x-5x&>-7-3\\ -3x&>-10\quad \color{black}\textrm{dikali (-1)}\\ 3x&<10\\ x&<\color{red}\displaystyle \frac{10}{3}=3\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 01})\textrm{Jika pertidaksamaan}\\ & 2x-3a>\displaystyle \frac{3x-1}{2}+ax\: \: \textrm{mempunyai}\\ &\textrm{penyelesaian}\: \: x>5,\: \: \textrm{maka nilai}\: \: a\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{3}{4}\\ \textrm{b}.&-\displaystyle \frac{3}{8}\\ \color{red}\textrm{c}.&\displaystyle \frac{3}{8}\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}2x-3a&>\displaystyle \frac{3x-1}{2}+ax\quad \color{black}\textrm{tiap ruas}\: (\times 2)\\ 4x-6a&>3x-1+2ax\\ 4x-3x&-2ax>-1+6a\\ x-2a&x>-1+6a\\ (1-2a)&x>-1+6a\\ x&>\displaystyle \frac{-1+6a}{1-2a}\\ \textrm{Diketa}&\textrm{hui}:\: \: x>5\: \: \color{red}\textrm{adalah penyelesaian}\\ \color{red}\textrm{maka}\: \: &\\ 5&=\displaystyle \frac{-1+6a}{1-2a}\\ 5-10a&=-1+6a\\ -6a-10&a=-1-5\\ -16a&=-6\\ a&=\displaystyle \frac{-6}{-16}\\ &=\color{red}\displaystyle \frac{3}{8} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&(\textbf{UMPTN 94})\\ &\textrm{Apabila}\: \: a<x<b\: \: \textrm{dan}\: \: a<y<b\\ & \textrm{maka berlaku}\: \: \: ....\\ &\begin{array}{llll}\\ \textrm{a}.&a<x-y<b\\ \textrm{b}.&b-a<x-y<a-b\\ \color{red}\textrm{c}.&a-b<x-y<b-a\\ \textrm{d}.&\displaystyle \frac{1}{2}(b-a)<x-y<\frac{1}{2}(a-b)\\ \textrm{e}.&\displaystyle \frac{1}{2}(a-b)<x-y<\frac{1}{2}(b-a) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{array}{llllll}\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &-a>-y>-b&&\\\hline &\color{purple}\textrm{saat}&\color{black}\textrm{di susun ulang}&\\ a<x<b&\Rightarrow &a<x<b&\\ a<y<b&\Rightarrow &\color{black}-b<-y<-a&\color{red}+&\\\hline &&\color{red}a-b\color{blue}<\color{red}x-y\color{blue}<&\color{red}b-a \end{array} \end{array}$

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

$\begin{array}{ll}\\ &\textbf{BENTUK UMUM}\\ &\color{blue}\begin{cases} ax+by<c \\ ax+by\leq c \\ ax+by>c \\ ax+by\geq c \end{cases}\\\\ &\textbf{LANGKAH-LANGKAH}\\ &\color{purple}\textrm{dalam membuat gambar grafik persamaan linear}\\ &\: \textrm{adalah sebagai berikut}:\\ &\bullet\quad \textrm{membuat gambar grafik}\: \: \color{red}ax+by=c\\ &\quad \: \: \textrm{untuk batas wilayahnya}\\ &\bullet \quad \textrm{menyelidiki wilayah yang dimaksud di sekitar}\\ &\quad \: \: \textrm{garis} \: \: \color{red}ax+by=c\\ &\bullet \quad \textrm{ambillah sebuah titik}\: \color{red}\left ( x_{0},y_{0} \right )\: \color{black}\textrm{sembarang}\\ &\: \: \quad \textrm{kemudian substitusikan ke pertidaksamaan}\\ &\quad \: \: \color{red}ax+by\: ....\: c\\ &\bullet \quad \textrm{jika diperoleh nilai ketaksamaan yang benar},\\ &\: \: \quad \textrm{maka daerah di mana titik uji}\: \color{red}\left ( x_{0},y_{0} \right )\\ &\: \: \quad \textrm{berada merupakan wilayah penyelesaiannya}\\ &\: \: \quad \textrm{demikian juga sebaliknya} \end{array}$

$\LARGE\color{blue}\fbox{CONTOH SOAL}$

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

Dan berikut untuk wilayah dan juga batas-batas untuk pertidalsamaan

$\color{red}3x+2y<6$

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

$\begin{array}{|c|c|c|}\hline \color{black}\color{purple}\textrm{Titik}&\color{black}\textrm{Pengujian}&\color{black}\textrm{Keterangan}\\ &\color{blue}\textrm{Uji}&\color{red}3x+2y<6\\\hline (0,0)&3(0)+2(0)=0<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,1)&3(0)+2(1)=2<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,0)&3(1)+2(0)=3<\textbf{6}&\textrm{Dalam wilayah}\\\hline (1,1)&3(1)+2(1)=5<\textbf{6}&\textrm{Dalam wilayah}\\\hline (0,2)&3(0)+2(2)=4<\textbf{6}&\textrm{Dalam wilayah}\\\hline (2,0)&3(2)+2(0)=6=\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (2,2)&3(2)+2(2)=10>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (0,3)&3(0)+2(3)=6=\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (3,0)&3(3)+2(0)=9>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline (3,3)&3(3)+2(3)=15>\textbf{6}&\color{red}\textrm{Di luar wilayah}\\\hline \vdots &\vdots&\vdots \\\hline \end{array}$

Dan berikut untuk wilayah yang memenuhi  $"\color{red}3x+2y\leq 6$

$\begin{array}{ll}\\ 2.&\textrm{Selesaikanlah pertidaksamaan berikut}\\ &\textrm{a}.\quad 12x+2>4x+6\\ &\textrm{b}.\quad 2-3x<6-x\\ &\textrm{c}.\quad 6x+1\geq 2\\ &\textrm{d}.\quad \displaystyle \frac{2-3x}{2}<\frac{3-x}{3}\\\\ &\textrm{Jawab}\\ &\color{purple}\begin{aligned}\color{black}\textrm{a}.\: \: 12x&+2>4x+6\\ 12x&-4x>6-2\\ 8x&>4\\ x&>\displaystyle \frac{1}{2} \end{aligned}\\ &\color{blue}\begin{aligned}\color{black}\textrm{b}.\: \: &2-3x<6-x\\ &-3x+x<6-2\\ &-2x<4\: \: \color{red}\textrm{dikali}\: \: (-1)\\ &2x>-4\: \: (\color{black}\textrm{tanda berubah})\\ &x>-2 \end{aligned}\\ &\color{purple}\begin{aligned}\color{black}\textrm{c}.\: \: 6x&+1\geq 2\\ 6x&\geq 2-1\\ x&\geq \displaystyle \frac{1}{6} \end{aligned}\\ &\color{blue}\begin{aligned}\color{black}\textrm{d}.\: \: \: \: \color{blue}\displaystyle \frac{2-3x}{2}&<\frac{3-x}{3}\\ 3(2-3x)&<2(3-x)\\ 6-9x&<6-2x\\ -9x+2x&<6-6\\ -7x&<0\: \: \color{red}\textrm{di kali}\: \: (-1)\\ 7x&>0\: \: (\color{black}\textrm{tanda berubah})\\ x&>\displaystyle \frac{0}{7}\\ x&>0 \end{aligned} \end{array}$

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.