SBMPTN Zone : Sistem Persamaan

Berikut ini adalah kumpulan soal mengenai Sistem Persamaan tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tipe:

Standar SBMPTN HOTS

No. 1

Jika A dan B memenuhi \begin{cases}\dfrac{3A}{2A+3B}+\dfrac{6B}{2A-3B}=3\\[8pt]\dfrac{-6A}{2A+3B}+\dfrac{3B}{2A-3B}=-1\end{cases} maka \dfrac{AB}{4A^2-9B^2}= ....

1. -\dfrac23
2. -\dfrac13
3. -\dfrac19

4. \dfrac19
5. \dfrac13

SBMPTN 2017

Misal {x=\dfrac{A}{2A+3B}} dan {y=\dfrac{B}{2A-3B}}

\(\begin{array}{rl|l} 3x+6y&=3&\times2\\ -6x+3y&=-1&\times1 \end{array}\)

\(\eqalign{ 6x+12y&=6\\ -6x+3y&=-1\qquad+\\\hline 15y&=5\\ y&=\dfrac13\\ \dfrac{B}{2A-3B}&=\dfrac13 }\)

\(\eqalign{ 3x+6y&=3\\ 3x+6\left(\dfrac13\right)&=3\\ 3x+2&=3\\ x&=\dfrac13\\ \dfrac{A}{2A+3B}&=\dfrac13 }\)

\(\eqalign{ \dfrac{AB}{4A^2-9B^2}&=\left(\dfrac{A}{2A+3B}\right)\left(\dfrac{B}{2A-3B}\right)\\[4pt] &=\left(\dfrac13\right)\left(\dfrac13\right)\\[4pt] &=\dfrac19 }\)

JAWAB: D

No. 2

Jika x dan y memenuhi {\dfrac{2x+y}{3x-2y+3}=\dfrac1{15}} dan {\dfrac1{x+y}=\dfrac7{-2x+y}} maka nilai {x-y=} ....

\(\eqalign{ \dfrac{2x+y}{3x-2y+3}&=\dfrac1{15}\\ 15(2x+y)&=3x-2y+3\\ 30x+15y&=3x-2y+3\\ 27x+17y&=3 }\)

\(\eqalign{ \dfrac1{x+y}&=\dfrac7{-2x+y}\\ -2x+y&=7x+7y\\ -9x&=6y\\ 3x+2y&=0 }\)

\(\begin{array}{rl|l} 27x+17y&=3&\times1\\ 3x+2y&=0&\times9 \end{array}\)

\(\eqalign{ 27x+17y&=3\\ 27x+18y&=0\qquad-\\\hline -y&=3\\ y&=-3 }\)

\(\eqalign{ 3x+2y&=0\\ 3x+2(-3)&=0\\ x&=2 }\)

\(\eqalign{ x-y&=2-(-3)\\ &=5 }\)

Jadi, x-y=5.

No. 3

Jika {ax+y=4}, {x+by=7}, dan {ab=2}, maka nilai {2ax-3y=}

1. {28-32a}
2. {28-35a}
   
3. {28-38a}
   

4. {28-40a}
5. {28-46a}
   

\(\begin{aligned} x+by&=7&\qquad\color{red}{\times a}\\ ax+aby&=7a\\ ax+2y&=7a \end{aligned}\)

(\begin{aligned} ax+y&=4&\qquad\color{red}{\times 7}\\ ax+2y&=7a&\qquad\color{red}{\times 5} \end{aligned}\)

\(\begin{aligned} 7ax+7y&=28\\ 5ax+10y&=35a&\qquad\color{red}{-}\\\hline 2ax-3y&=\boxed{\boxed{28-35a}} \end{aligned}\)

Jadi, 2ax-3y=28-35a.
JAWAB: B

No. 4

Jika 3x-2y = 4 dan \dfrac{216^x}{72^y} = 162, maka nilai dari y adalah . . .

1. -1
2. 0
3. 1

4. 2
5. 3

\(\begin{aligned} \dfrac{216^x}{72^y}&= 162\\ \dfrac{\left(6^3\right)^x}{\left(2\cdot6^2\right)^y}&= 162\\ \dfrac{6^{3x}}{2^y\cdot6^{2y}}&= 162\\ \dfrac{6^{3x-2y}}{2^y}&= 2\cdot81\\ \dfrac{6^4}{2^y}&= 2\cdot3^4\\ 2^y&=\dfrac{6^4}{2\cdot3^4}\\ &=\dfrac{2^4\cdot3^4}{2\cdot3^4}\\ &=2^3\\ y&=3 \end{aligned}\)

Jadi, y=3.
JAWAB: E

No. 5

Jika a dan b memenuhi \begin{cases}\dfrac2{2a-b}+\dfrac7{2a+b}&=3\\\dfrac1{2a-b}-\dfrac7{2a+b}&=0\end{cases} maka {a+b=}

1. 5
2. 6
3. 7

4. 8
5. 9

SKMP Juli 2020

\(\eqalign{ \dfrac2{2a-b}+\dfrac7{2a+b}&=3\\\dfrac1{2a-b}-\dfrac7{2a+b}&=0\qquad&{\color{red}+}\\\hline \dfrac3{2a-b}&=3\\ 2a-b&=1 }\)

\(\eqalign{ \dfrac2{2a-b}+\dfrac7{2a+b}&=3\\ \dfrac21+\dfrac7{2a+b}&=3\\ 2+\dfrac7{2a+b}&=3\\ \dfrac7{2a+b}&=1\\ 2a+b&=7 }\)

\(\eqalign{ 2a-b&=1\\ 2a+b&=7\qquad&{\color{red}+}\\\hline 4a&=8\\ 2a&=4\\ a&=2 }\)

\(\eqalign{ 2a+b&=7\\ 4+b&=7\\ b&=3 }\)

\(\eqalign{ a+b&=2+3\\ &=\boxed{\boxed{5}} }\)

Jadi, {a+b=5}.