Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas X

$\begin{array}{l}\\ 11.& \text{Perhatikanlah ilustrasi grafik di bawah ini}\\ & \begin{array}{}\end{array}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan yang memenuhi rumus tersebut }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y=|-x-2|\\ \text{b}.& y=-2x-4\\ \text{c}.& y=-|2x-4|\\ \text{d}.& y=|-2x-4|\\ \text{e}.& y=|-2x+4|\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Dengan cara substitusi langsung kita}\\ & \text{akan mendapatkan}\\ & \bullet \text{untuk}x=4\text{menyebabkan nilai}y=4\\ & \text{dan sampai langkah di sini hanya ada }\\ & \text{1 persamaan yang memenuhi yaitu}:\\ & y=|-2x+4|\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Gambarlah garfik untuk persamaan}\\ & \left|x\right|+\left|y\right|=4\\ \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Dari soal diketahui}\\ & \left|x\right|+\left|y\right|=4\\ & \text{maka untuk}\end{array}\\ & \begin{array}{|cccc|}\hline x>0,y>0& & & x>0,y<0\\ x+y=4& & & x+(-y)=4\\ & & & \\ & & & \\ x<0,y>0& & & x<0,y<0\\ (-x)+y=4& & & (-x)+(-y)=4\\ \hline\end{array}\end{array}$.

$.\begin{array}{l}\\ & \text{Perhatikanlah ilustrasinya grafik di bawah ini}\\ & \begin{array}{}\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Tentukanlah nilai}x\text{dan}y\\ & \text{yang memenuhi}x+\left|x\right|+y=5\\ & \text{dan}x+\left|y\right|-y=10\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & \begin{array}{rl}& \left|x\right|+x+y=5\\ & \text{dan}\\ & x+\left|y\right|-y=10\\ & \end{array}\end{array}\end{array}$.

$.\begin{array}{r}\begin{array}{rl}& \text{untuk}:x>0,y>0\mathbf{\text{(kuadran I)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (x)+x+y=5\\ & \iff 2x+y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(y)-y=10\\ & \iff x=10(\text{ada})\end{array}\\ & \text{untuk}:x<0,y>0\mathbf{\text{(kuadran II)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (-x)+x+y=5\\ & \iff y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(y)-y=10\\ & \iff x=10\\ & \mathbf{\text{(tidak memenuhi)}}\end{array}\end{array}\end{array}$.

$.\begin{array}{r}\begin{array}{rl}& \text{untuk}:x<0,y<0\mathbf{\text{(kuadran III)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (-x)+x+y=5\\ & \iff y=5\\ & \mathbf{\text{(tidak memenuhi)}}\\ x+\left|y\right|-y=10& \iff x+(-y)-y=10\\ & \iff x-2y=10(\text{ada})\end{array}\\ & \text{untuk}:x>0,y<0\mathbf{\text{(kuadran IV)}}\\ & \{\begin{array}{ll}\left|x\right|+x+y=5& \iff (x)+x+y=5\\ & \iff 2x+y=5(\text{ada})\\ x+\left|y\right|-y=10& \iff x+(-y)-y=10\\ & \iff x-2y=10(\text{ada})\end{array}\end{array}\end{array}$.

$.\begin{array}{l}\\ & \text{Sebagai ilustrasinya, berikut grafiknya}\\ & \begin{array}{}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Gambarlah grafik fungsi mutlak berikut}\\ & \text{a}.y=|x-2|\\ & \text{b}.y=-|x-2|\\ & \text{c}.y=2+|x-2|\\ & \text{d}.y=2-|x-2|\\ & \text{e}.y=|2+|x-2||\\ & \text{f}.y=|2-|x-2||\end{array}$.

$.\begin{array}{l}\\ & \mathbf{\text{Jawab}}:\\ & \text{Berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{a}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{b}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut pula ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{c}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan berikut pula ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{d}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan sebagai ilustrasinya}\\ & \text{lihat soal no}\mathbf{\text{c}}\end{array}$.

$.\begin{array}{l}\\ & \text{Dan terakhir berikut ilustrasinya}\\ & \text{grafik dari soal}\mathbf{\text{f}}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Penyelesaian yang memenuhi untuk}\\ & |3x-(4x-7)|=6\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{-13,-1\}& & & \text{d}.& \{-13,1\}\\ \text{b}.& \{-1,13\}& \text{c}.& \{1,13\}& \text{e}.& \left\{13\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}|3x-(4x-7)|& =6\\ |3x-4x+7|& =6\\ |-x+7|& =6\\ (-x+7)& =\pm 6\\ -x+7& =\{\begin{array}{l}6\iff -x=6-7\iff x=1\\ \\ -6\iff -x=-6-7\iff x=13\end{array}\end{array}\end{array}$.