Exercise Zone : Sistem Persamaan Linear Dua Variabel

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StandarSNBTHOTS

No. 1

Jika 2x + 3y = 4a + 1 dan 3x − 5y = a + p, maka 10x − 23y + 5 =....

1. 4p + 4
2. 4p + 5
3. 4p + 6

4. 4p + 7
5. 4p + 8

Ganesha Operation

ALTERNATIF PENYELESAIAN

CARA 1

$$\begin{array}{rcll}2x+3y& =& 4a+1& \times 3\\ 3x-5y& =& a+p& \times 2\end{array}$$ $$\begin{array}{rcrrcll}& 6x& +& 9y& =& 12a+3& \\ & 6x& -& 10y& =& 2a+2p& -\\ & & & 19y& =& 10a-2p+3\\ & & & y& =& {\displaystyle \frac{10a-2p+3}{19}}\end{array}$$ $$\begin{array}{rl}2x+3y& =4a+1\\ 2x+3\left({\displaystyle \frac{10a-2p+3}{19}}\right)& =4a+1\\ 2x+{\displaystyle \frac{30a-6p+9}{19}}& =4a+1\\ 2x& =4a+1-{\displaystyle \frac{30a-6p+9}{19}}\\ 2x& ={\displaystyle \frac{(4a+1)19-(30a-6p+9)}{19}}\\ 2x& ={\displaystyle \frac{76a+19-30a+6p-9}{19}}\\ 2x& ={\displaystyle \frac{46a+6p+10}{19}}\\ x& ={\displaystyle \frac{23a+3p+5}{19}}\end{array}$$ $$\begin{array}{rl}10x-23y+5& =10\left({\displaystyle \frac{23a+3p+5}{19}}\right)-23\left({\displaystyle \frac{10a-2p+3}{19}}\right)+5\\ & ={\displaystyle \frac{10(23a+3p+5)-23(10a-2p+3)+5(19)}{19}}\\ & ={\displaystyle \frac{230a+30p+50-230a+46p-69+95}{19}}\\ & ={\displaystyle \frac{76p+76}{19}}\\ & =4p+4\end{array}$$

CARA 2

$$\begin{array}{rcll}2x+3y& =& 4a+1& \times 1\\ 3x-5y& =& a+p& \times 4\end{array}$$ $$\begin{array}{rl}2x+3y& =4a+1\\ 12x-20y& =4a+4p-\\ -10x+23y& =-4p+1\\ 10x-23y& =4p-1\\ 10x-23y+5& =4p-1+5\\ & =4p+4\end{array}$$

Jadi, 10x − 23y + 5 = 4p + 4.
JAWAB: A

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No. 2

Misalkan (p, q) = (p₁, q₁) adalah penyelesaian sistem persamaan: $$\{\begin{array}{ll}5p+4q& =6\\ p-2q& =4\end{array}$$ Nilai 2p₁ + q₁ = ....

1. 3
2. 7
3. 8

4. 10
5. 13

SUMBER

ALTERNATIF PENYELESAIAN

CARA 1

$$\begin{array}{rlr}5p+4q& =6& \times 1\\ p-2q& =4& \times 2\end{array}$$ $$\begin{array}{rl}5p+4q& =6\\ 2p-4q& =8& +\\ 7p& =14\\ p& =2\end{array}$$ $$\begin{array}{rl}p-2q& =4\\ 2-2q& =4\\ -2q& =2\\ q& =-1\end{array}$$ $$\begin{array}{rl}2p_1+q_1& =2(2)+(-1)\\ & =4-1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$$

CARA 2

$$\begin{array}{rl}\left(\begin{array}{c}2p_1+q_1\end{array}\right)& =\left(\begin{array}{cc}2& 1\end{array}\right){\left(\begin{array}{cc}5& 4\\ 1& -2\end{array}\right)}^{-1}\left(\begin{array}{c}6\\ 4\end{array}\right)\\ & =\left(\begin{array}{cc}2& 1\end{array}\right){\displaystyle \frac{1}{5\cdot (-2)-4\cdot 1}}\left(\begin{array}{cc}-2& -4\\ -1& 5\end{array}\right)\left(\begin{array}{c}6\\ 4\end{array}\right)\\ & =\left(\begin{array}{cc}2& 1\end{array}\right){\displaystyle \frac{1}{-14}}\left(\begin{array}{c}-28\\ 14\end{array}\right)\\ & =\left(\begin{array}{cc}2& 1\end{array}\right)\left(\begin{array}{c}2\\ -1\end{array}\right)\\ & =\left(\begin{array}{c}3\end{array}\right)\end{array}$$

CARA 3

$$\{\begin{array}{ll}5p+4q=6& \times 5\\ p-2q=4& \times 3\end{array}$$ $$\begin{array}{rl}25p+20q& =30\\ 3p-6q& =12+\\ 28p+14q& =42\\ 2p+q& =3\\ 2p_1+q_1& =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$$

Jadi, 2p₁ + q₁ = 3.
JAWAB: A

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No. 3

Carilah himpunan penyelesaian dari:2\sqrt2x+\sqrt3y=7 \sqrt2x+2\sqrt3y=8

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}2\sqrt{2}x+\sqrt{3}y& =7\times 2\\ \sqrt{2}x+2\sqrt{3}y& =8\end{array}$$ $$\begin{array}{rl}4\sqrt{2}x+2\sqrt{3}y& =14\\ \sqrt{2}x+2\sqrt{3}y& =8-\\ 3\sqrt{2}x& =6\\ x& ={\displaystyle \frac{6}{3\sqrt{2}}}\cdot {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}}\\ & ={\displaystyle \frac{6\sqrt{2}}{6}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{2}}}}}\end{array}$$

$$\begin{array}{rl}2\sqrt{2}x+\sqrt{3}y& =7\\ 2\sqrt{2}(\sqrt{2})+\sqrt{3}y& =7\\ 4+\sqrt{3}y& =7\\ \sqrt{3}y& =3\\ y& ={\displaystyle \frac{3}{\sqrt{3}}}\\ y& =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{3}}}}}\end{array}$$

Jadi, Hp = \left\{\left(\sqrt2,\sqrt3\right)\right\}.

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No. 4

Jika x + 2y = 3a + 1 dan 3x − y = a + p, maka nilai 8x − 5y adalah

1. 3p + 2
2. 3p + 1
3. 3p − 1

4. 3p − 2
5. 3p − 3

SUMBER

ALTERNATIF PENYELESAIAN

3x − y = a + p       × 3 $$\begin{array}{rl}9x-3y& =3a+3p\\ x+2y& =3a+1& -\\ 8x-5y& =\boxed{{\displaystyle \boxed{{\displaystyle 3p-1}}}}\end{array}$$

Jadi, 8x − 5y = 3p − 2.
JAWAB: C

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No. 5

Himpunan penyelesaian dari sistem persamaan {\dfrac5x+\dfrac4y=13} dan {\dfrac3x-\dfrac2y=21} adalah {(x₀, y₀)}. Nilai x₀ − y₀ =

1. \dfrac4{15}
2. \dfrac13
3. \dfrac25

4. \dfrac7{15}
5. \dfrac8{15}

SUMBER

ALTERNATIF PENYELESAIAN

Misal
\dfrac1x=a

\dfrac1y=b $$\begin{array}{rl}5a+4b& =13\\ 3a-2b& =21& \times 2\end{array}$$ $$\begin{array}{rl}5a+4b& =13\\ 6a-4b& =42& +\\ 11a& =55\\ a& =5\\ x_0& ={\displaystyle \frac{1}{5}}\end{array}$$

$$\begin{array}{rl}3a-2b& =21\\ 3(5)-2b& =21\\ 15-2b& =21\\ -2b& =6\\ b& =-3\\ y_0& =-{\displaystyle \frac{1}{3}}\end{array}$$ $$\begin{array}{rl}{x}_0-y_0& ={\displaystyle \frac{1}{5}}-(-{\displaystyle \frac{1}{3}})\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{8}{15}}}}}}\end{array}$$

Jadi, x_0-y_0=\dfrac8{15}.
JAWAB: E

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No. 6

Diketahui dua persamaan {\dfrac2{x+y}+\dfrac6{x-y}=2} dan {\dfrac4{x+y}-\dfrac9{x-y}=-1}. Nilai \dfrac{x}y yang memenuhi kedua persamaan tersebut adalah

1. 11
2. 9
3. 1

4. −9
5. −11

SUMBER

ALTERNATIF PENYELESAIAN

CARA BIASA

$$\begin{array}{rlr}{\displaystyle \frac{2}{x+y}}+{\displaystyle \frac{6}{x-y}}& =2& \times 2\\ {\displaystyle \frac{4}{x+y}}-{\displaystyle \frac{9}{x-y}}& =-1\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{4}{x+y}}+{\displaystyle \frac{12}{x-y}}& =4\\ {\displaystyle \frac{4}{x+y}}-{\displaystyle \frac{9}{x-y}}& =-1& -\\ {\displaystyle \frac{21}{x-y}}& =5\\ x-y& ={\displaystyle \frac{21}{5}}\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{2}{x+y}}+{\displaystyle \frac{6}{x-y}}& =2\\ {\displaystyle \frac{2}{x+y}}+{\displaystyle \frac{6}{{\displaystyle \frac{21}{5}}}}& =2\\ {\displaystyle \frac{2}{x+y}}+{\displaystyle \frac{30}{21}}& =2\\ {\displaystyle \frac{2}{x+y}}+{\displaystyle \frac{10}{7}}& =2\\ {\displaystyle \frac{2}{x+y}}& =2-{\displaystyle \frac{10}{7}}\\ {\displaystyle \frac{2}{x+y}}& ={\displaystyle \frac{4}{7}}\\ {\displaystyle \frac{1}{x+y}}& ={\displaystyle \frac{2}{7}}\\ x+y& ={\displaystyle \frac{7}{2}}\end{array}$$ $$\begin{array}{rl}x+y& ={\displaystyle \frac{7}{2}}\\ x-y& ={\displaystyle \frac{21}{5}}& +\\ 2x& ={\displaystyle \frac{77}{10}}\\ x& ={\displaystyle \frac{77}{20}}\end{array}$$ $$\begin{array}{rl}x+y& ={\displaystyle \frac{7}{2}}\\ {\displaystyle \frac{77}{20}}+y& ={\displaystyle \frac{7}{2}}\\ y& ={\displaystyle \frac{7}{2}}-{\displaystyle \frac{77}{20}}\\ & =-{\displaystyle \frac{7}{20}}\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{x}{y}}& ={\displaystyle \frac{{\displaystyle \frac{77}{20}}}{-{\displaystyle \frac{7}{20}}}}\\ & =-{\displaystyle \frac{77}{7}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -11}}}}\end{array}$$

CARA CEPAT

$$\begin{array}{rlr}{\displaystyle \frac{2}{x+y}}+{\displaystyle \frac{6}{x-y}}& =2& :2\\ {\displaystyle \frac{1}{x+y}}+{\displaystyle \frac{3}{x-y}}& =1\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{1}{x+y}}+{\displaystyle \frac{3}{x-y}}& =1\\ {\displaystyle \frac{4}{x+y}}-{\displaystyle \frac{9}{x-y}}& =-1& +\\ {\displaystyle \frac{5}{x+y}}-{\displaystyle \frac{6}{x-y}}& =0\\ {\displaystyle \frac{5}{x+y}}& ={\displaystyle \frac{6}{x-y}}\\ 5x-5y& =6x+6y\\ -x& =11y\\ {\displaystyle \frac{x}{y}}& ={\displaystyle \frac{11}{-1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -11}}}}\end{array}$$

Jadi, \dfrac{x}y=-11.
JAWAB: E

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No. 7

Misalkan x dan y memenuhi persamaan $$\{\begin{array}{ll}4x+7y& =61\\ 2x+9y& =69\end{array}$$ Maka pernyataan yang benar adalah

1. x > y
2. x < y
3. x = y

4. x² ≤ y²
5. x dan y tidak dapat ditentukan

SKMP Juli 2020

ALTERNATIF PENYELESAIAN

$$\{\begin{array}{ll}4x+7y& =61\\ 2x+9y& =69\times 2\end{array}$$ $$\begin{array}{rl}4x+7y& =61\\ 4x+18y& =138& -\\ -11y& =-77\\ y& =7\end{array}$$

$$\begin{array}{rl}4x+7y& =61\\ 4x+7(7)& =61\\ 4x+49& =61\\ 4x& =12\\ x& =3\end{array}$$ x < y

Jadi, x < y.
JAWAB: B

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No. 8

Rini membeli 3 benda A dan 4 benda B dengan membayar Rp2.700,00. Sementara Anis membeli 6 benda A dan 2 benda B dengan membayar Rp3.600,00. Harga yang harus dibayarkan jika Latif membeli sebuah benda A dan sebuah benda B adalah ....

1. Rp540.00
2. Rp720.00
3. Rp800.00

4. Rp960.00
5. Rp1,100.00

Latihan Pra UKK Semester Genap Kelas X

ALTERNATIF PENYELESAIAN

Misal:
benda A = x, dan
benda B = y $$\{\begin{array}{ll}3x+4y& =2700\\ 6x+2y& =3600& :2\end{array}$$ $$\begin{array}{rl}3x+4y& =2700\\ 3x+y& =1800& -\\ 3y& =900\\ y& =\frac{900}{3}\\ & =300\end{array}$$

$$\begin{array}{rl}3x+y& =1800\\ 3x+300& =1800\\ 3x& =1800-300\\ 3x& =1500\\ x& =\frac{1500}{3}\\ & =500\end{array}$$ $$\begin{array}{rl}x+y& =500+300\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 800}}}}\end{array}$$

Jadi, harga yang harus dibayarkan jika Latif membeli sebuah benda A dan sebuah benda B adalah Rp800,00.

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No. 9

Nada membeli kue untuk lebaran. Harga satu kaleng kue nastar sama dengan 2 kali harga satu kaleng kue keju. Harga 3 kaleng kue nastar dan 2 kaleng kue keju Rp480.000,00. Uang yang harus dibayarkan Nada untuk membeli 2 kaleng kue nastar dan 3 kaleng kue keju adalah ....

1. Rp480.000,00
2. Rp420.000,00
3. Rp360.000,00

4. Rp240.000,00
5. Rp180.000,00

defantri

ALTERNATIF PENYELESAIAN

Kita misalkan harga kue nastar dengan n dan harga kue keju dengan k.

Harga satu kaleng kue nastar sama dengan 2 kali harga satu kaleng kue keju
n = 2k

Harga 3 kaleng kue nastar dan 2 kaleng kue keju Rp480.000,00, sehingga kita peroleh persamaan
3n + 2k = 480.000.

Dari kedua persamaan di atas dapat kita peroleh:
$\begin{array}{rl}3n+n& =480000\\ 4n& =480000\\ n& =120000\end{array}$

$\begin{array}{rl}n& =2k\\ 120000& =2k\\ k& =60000\end{array}$

$\begin{array}{rl}2n+3k& =2(120000)+3(60000)\\ & =240000+180000\\ & =420000\end{array}$

Jadi, uang yang harus dibayarkan Nada untuk membeli 2 kaleng kue nastar dan 3 kaleng kue keju adalah Rp420.000,00.
JAWAB: B

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No. 10

Adi membeli 8 benda A dan 6 benda B dengan membayar Rp62.000,- sementara Bambang membeli 7 benda A dan 4 benda B dengan membayar Rp43.000,- Harga yang harus dibayarkan jika Caca membeli sebuah benda A dan sebuah benda B adalah ....

Perhatian: angka pada soal ini akan berubah jika halaman ini direfresh/direload

ALTERNATIF PENYELESAIAN

Misal:
harga 1 benda A = x
harga 1 benda B = y

$\begin{array}{rlr}8x+6y& =62.000& \times 2\\ 7x+4y& =43.000& \times 3\end{array}$

$\begin{array}{rl}16x+12y& =124.000\\ 21x+12y& =129.000& -\\ -5x& =-5.000\\ x& ={\displaystyle \frac{-5.000}{-5}}\\ & =1.000\end{array}$

$\begin{array}{rl}8x+6y& =62.000\\ 8(1.000)+6y& =62.000\\ 8.000+6y& =62.000\\ 6y& =62.000-8.000\\ 6y& =54.000\\ y& ={\displaystyle \frac{54.000}{6}}\\ & =9.000\end{array}$

harga sebuah benda A dan sebuah benda B
$\begin{array}{rl}x+y& =1.000+9.000\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 10.000}}}}\end{array}$

Jadi, harga yang harus dibayarkan jika Caca membeli sebuah benda A dan sebuah benda B adalah Rp10.000,-.

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