Q4. 1. Harga 2 burger dan 3 cola cola rp.

Berikut ini adalah pertanyaan dari xzvaa pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Q4.1. Harga 2 burger dan 3 cola cola rp. 80.000. Sedangkan harga 2 burger dan 1 cola cola rp. 60.000. berapakah harga burger dan cola cola?

2. Jika kamu membeli 5 burger dan 2 cola cola, berapa rupiah yg harus kamu bayar?

3. Harga 3 roti dan 2 es krim 650 yen dan harga 1 roti dan 2 es krim 350 yen. Tentukan harga roti dan harga es krim!

4. Jika kamu membeli 4 roti dan 4 es krim, berapa rupiah yang kamu bayar?

Good Luck_

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\mathbb{ \color{aqua}{ \underbrace{JAWABAN}}}$

$\boxed{ \: \begin{aligned} &\bf{harga \: burger = Rp. \: 25.000} \\ &\bf{harga \: coca \: cola = Rp. \: 10.000} \end{aligned} \: } \\$
$\boxed{ \bf{ \: Rp. \: 145.000 \: }} \\$
$\boxed{ \: \begin{aligned} &\bf{harga \: roti = 150 \: yen} \\ &\bf{harga \: es \: krim = 100 \: yen} \end{aligned} \: } \\$
$\boxed{ \bf{ \: 1.000 \: yen \: }}$

------------------

$\mathbb{ \color{orange}{ \underbrace{PENYELESAIAN}}}$

SOAL 1 :

misal :

burger = x
coca cola = y

$\\$

$\boxed{ \: \begin{aligned} &\begin{aligned} \tt{2 \: \sf{burger} + 3 \: \sf{coca \: cola}} &= \tt{80.000} \\ \tt{ \pink{2x + 3y}} & \pink{= {\tt{80.000}}}...... \tt{(1)} \end{aligned} \\ & \begin{aligned} \tt{2 \: \sf{burger} + 1 \: \sf{coca \: cola}} &= {\tt{60.000}} \\ \tt{ \pink{2x + y}} & \pink{= {\tt{60.000}}}...... \tt{(2)} \end{aligned} \end{aligned} \: } \\ \\$

eliminasi x

$\boxed{ \: \begin{aligned} \tt{ \bcancel{2x} + 3y} &= \tt{80.000} \\ \tt{ \bcancel{2x} + \: \: y} &= \tt{60.000} \\ \underline{~~~~~~~~~~~~~~~~}& \underline{~~~~~~~~~~~~~~~~~~~~~} - \\ \tt{2y} &= \tt{20.000} \\ \tt{y} &= \tt{ \frac{20.000}{2} } \\ \tt{y} &= \tt{ \red{ \boxed{ \bf{ \pink{10.000}}}}} \end{aligned} \: } \\ \\$

substitusikan y = 10.000 ke salah satu persamaan

$\boxed{ \: \begin{aligned} \tt{2x + y} &= \tt{60.000} \\ \tt{2x + 10.000} &= \tt{60.000} \\ \tt{2x} &= \tt{60.000 - 10.000} \\ \tt{2x} &= \tt{50.000} \\ \tt{x} &= \tt{ \frac{50.000}{2} } \\ \tt{x} &= \red{ \boxed{ \bf{ \pink{25.000}}}} \end{aligned} \: } \\ \\$

himpunan penyelesaian

$\boxed{ \bf{HP = \{x = \pink{25.000}~,~y = \pink{10.000} \}}} \\ \\$

SOAL 2 :

substitusikan x = 25.000 dan y = 10.000

$\boxed{ \: \begin{aligned} &\tt{5x + 2y} \\ & \tt{ = 5(25.000) + 2(10.000)} \\ & \tt{ = 125.000 + 20.000} \\ & \tt{ = \red{ \boxed{ \pink{ \bf{145.000}}}}} \end{aligned} \: } \\ \\$

SOAL 3 :

misal :

roti = a
es krim = b

$\\$

$\boxed{ \: \begin{aligned} &\begin{aligned} \tt{3 \: \sf{roti} + 2 \: \sf{es \: krim}} &= \tt{650} \\ \tt{ \pink{3a + 2b}} & \pink{= {\tt{650}}}...... \tt{(1)} \end{aligned} \\ & \begin{aligned} \tt{1 \: \sf{roti} + 2 \: \sf{es \: krim}} &= {\tt{350}} \\ \tt{ \pink{a + 2b}} & \pink{= {\tt{350}}}...... \tt{(2)} \end{aligned} \end{aligned} \: } \\ \\$

eliminasi b

$\boxed{ \: \begin{aligned} \tt{3a + \bcancel{2b}} &= \tt{650} \\ \tt{a + \bcancel{2b}} &= \tt{350} \\ \underline{~~~~~~~~~~~~~~~~}& \underline{~~~~~~~~~~~~~~~~~~~~~} - \\ \tt{2a} &= \tt{300} \\ \tt{a} &= \tt{ \frac{300}{2} } \\ \tt{a} &= \tt{ \red{ \boxed{ \pink{ \bf{150}}}}} \end{aligned} \: } \\ \\$

substitusikan a = 150 ke salah satu persamaan

$\boxed{ \: \begin{aligned} \tt{a + 2b} &= \tt{350} \\ \tt{150 + 2b} &= \tt{350} \\ \tt{2b} &= \tt{350 - 150} \\ \tt{2b} &= \tt{200} \\ \tt{b} &= \tt{ \frac{200}{2} } \\ \tt{b} &= \red{ \boxed{ \bf{ \pink{100}}}} \end{aligned} \: } \\ \\$

himpunan penyelesaian

$\boxed{ \bf{HP = \{a = \pink{150}~,~b = \pink{100} \}}} \\ \\$

SOAL 4 :

substitusikan a = 150 dan b = 100

$\boxed{ \: \begin{aligned} &\tt{4a + 4b} \\ & \tt{ = 4(150) + 4(100)} \\ & \tt{ = 600 + 400} \\ & \tt{ = \red{ \boxed{ \pink{ \bf{1.000}}}}} \end{aligned} \: }$

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