System of Linear Equations in Two Variables: Complete Problems and Step-by-Step Solutions

The equation $4x-12=3\color{red}{xy}$ is not classified as a linear equation in two variables because it contains a term that is the product of two different variables (marked in red).
(Answer C)

Given $2x + 4y = 8$.
This equation can be simplified and rewritten as follows.
$\begin{aligned} 2x + 4y & = 8 \\ \text{Divide both sides}&~\text{by}~2 \\ x + 2y & = 4 \\ 2y & = 4-x \\ y & = \dfrac{4-x}{2} \end{aligned}$
If $x = 0$, then $y = \dfrac{4-0}{2} = 2$.
If $x = 1$, then $y = \dfrac{4-1}{2} = \dfrac32$.
If $x = 2$, then $y = \dfrac{4-2}{2} = 1$.
If $x = 3$, then $y = \dfrac{4-3}{2} = \dfrac12$.
If $x = 4$, then $y = \dfrac{4-4}{2} = 0$.
If $x = 5$, then $y = \dfrac{4-5}{2} = -\dfrac12$.
Since $y \in$ integers, the solution set of the equation is $\{(0, 2), (2, 1), (4, 0)\}.$
(Answer D)

Given the system of linear equations:
$\begin{cases} 2x-3y & = -13 && (\cdots 1) \\ x+2y & = 4 && (\cdots 2) \end{cases}$
Using the elimination method, we obtain
$\begin{aligned} \! \begin{aligned} 2x-3y & = 13 \\ x + 2y & = 4 \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 2 \end{aligned} \right| & \! \begin{aligned}~2x-3y & = -13 \\ 2x+4y & = 8 \end{aligned} \\ & \rule{3 cm}{0.6pt} – \\ & \! \begin{aligned} -7y & = -21 \\ y & = 3 \end{aligned} \end{aligned}$
Substitute $y = 3$ into one of the equations, for example Equation $(2)$.
$\begin{aligned} x+2\color{red}{y} & = 4 \\ x+2(3) & = 4 \\ x+6 & = 4 \\ x & = -2 \end{aligned}$
Therefore, the solution of the system of equations is $x=-2$ and $y=3$.
(Answer B)

Given the system of linear equations in two variables
$\begin{cases} 2x-y & = 7 && (\cdots 1) \\ x+3y& = 14 && (\cdots 2) \end{cases}$
Eliminate $y$ from Equation $(1)$ and Equation $(2)$.
$\begin{aligned} \! \begin{aligned} 2x -y & = 7 \\ x + 3y & = 14 \end{aligned} \left| \! \begin{aligned} \times 3 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~6x -3y & = 21 \\ x+3y & = 14 \end{aligned} \\ & \rule{2.8 cm}{0.6pt} + \\ & \! \begin{aligned} 7x & = 35 \\ x & = 5 \end{aligned} \end{aligned}$
Substitute $x = 5$ into one of the equations, for example into Equation $(1)$.
$\begin{aligned} 2\color{red}{x} -y & = 7 \\ 2(5) -y & = 7 \\ 10 -y & = 7 \\ y & = 3 \end{aligned}$
The value of $y = 3$ is obtained, therefore $\boxed{x+2y=5+2(3)=11}.$
(Answer C)

Given the system of linear equations in two variables
$\begin{cases} 2x+3y & = 3 && (\cdots 1) \\ 3x-y & = 10 && (\cdots 2) \end{cases}$
Eliminate $y$ from Equation $(1)$ and Equation $(2)$.
$\begin{aligned} \! \begin{aligned} 2x + 3y & = 3 \\ 3x -y & = 10 \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 3 \end{aligned} \right| & \! \begin{aligned}~2x+3y & = 3 \\~9x-3y & = 30 \end{aligned} \\ & \rule{2.8 cm}{0.6pt} + \\ & \! \begin{aligned} 11x & = 33 \\ x & = 3 \end{aligned} \end{aligned}$
Substitute $x = 3$ into one of the equations, for example into Equation $(1)$.
$\begin{aligned} 2\color{red}{x} + 3y & = 3 \\ 2(3) + 3y & = 3 \\ 6 + 3y & = 3 \\ 3y & = -3 \\ y & = -1 \end{aligned}$
The value of $y = -1$ is obtained, therefore $\boxed{2x-y = 2(3)-(-1) = 7}.$
(Answer D)

Given the system of linear equations in two variables
$\begin{cases} 7x+3y & =-5 && (\cdots 1) \\ 5x+2y & =1 && (\cdots 2) \end{cases}$
Eliminate $y$ from Equation $(1)$ and Equation $(2)$.
$\begin{aligned} \! \begin{aligned} 7x+3y & = -5 \\ 5x+2y & = 1 \end{aligned} \left| \! \begin{aligned} \times 2 \\ \times 3 \end{aligned} \right| & \! \begin{aligned}~14x+6y & = -10 \\~15x+6y & = 3 \end{aligned} \\ & \rule{3.2 cm}{0.6pt} – \\ & \! \begin{aligned} -x & = -13 \\ x & = 13 \end{aligned} \end{aligned}$
Substitute $x = 13$ into one of the equations, for example into Equation $(1)$.
$\begin{aligned} 7\color{red}{x}+3y & = -5 \\ 7(13) + 3y & = -5 \\ 3y & = -96 \\ y & = -32 \end{aligned}$
Therefore, the solution set of the system of linear equations in two variables is $\boxed{\{(13, -32)\}}.$
(Answer A)

Given the system of linear equations
$\begin{cases} x- y & = 5 && (\cdots 1) \\ 3x -5y & = 5 && (\cdots 2) \end{cases}$
Eliminate $x$ from Equations $(1)$ and $(2)$.
$\begin{aligned} \! \begin{aligned} x-y & = 5 \\ 3x -5y & = 5 \end{aligned} \left| \! \begin{aligned} \times 3 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~3x-3y & = 15 \\~3x-5y & = 5 \end{aligned} \\ & \rule{2.7 cm}{0.6pt} – \\ & \! \begin{aligned} 2y & = 10 \\ y & = 5 \end{aligned} \end{aligned}$
Substitute $y = 5$ into one of the equations, for example Equation $(1)$.
$\begin{aligned} x-\color{red}{y} & = 5 \\ x-5 & = 5 \\ x & = 10 \end{aligned}$
Therefore, the solution set of the system of linear equations is $\boxed{\{(10, 5)\}}.$
(Answer B)

Given the system of linear equations: $\begin{cases} \dfrac{p}{2}+\dfrac{q}{4} & = \dfrac74 && (\cdots 1) \\ \dfrac{p}{4}+\dfrac{q}{3} & = \dfrac14 && (\cdots 2) \end{cases}$
Multiply both sides by $4$ in the first equation, while both sides are multiplied by $12$ in the second equation so that we obtain
$\begin{cases} 2p + q & = 7 && (\cdots 1) \\ 3p+4q & = 3 && (\cdots 2) \end{cases}$
Using the elimination method, we obtain
$\begin{aligned} \! \begin{aligned} 2p+q & = 7 \\ 3p+4q & = 3 \end{aligned} \left| \! \begin{aligned} \times 4 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~8p+4q & = 28 \\ 3p+4q & = 3 \end{aligned} \\ & \rule{2.8 cm}{0.6pt} – \\ & \! \begin{aligned} 5p & = 25 \\ p & = 5 \end{aligned} \end{aligned}$
Substitute $p=5$ into one of the equations, for example Equation $(1)$.
$\begin{aligned} 2\color{red}{p}+q & = 7 \\ 2(5)+q & = 7 \\ 10+q & = 7 \\ q & = -3 \end{aligned}$
Therefore, the solution of the system of equations is $p=5$ and $q=-3.$
(Answer B)

Given the system of linear equations: $\begin{cases} \dfrac{x+3}{4}-\dfrac{y-2}{3} & = \dfrac{37}{12} && (\cdots 1) \\ \dfrac{x-3}{2}-\dfrac{y+4}{3} & = -\dfrac16 && (\cdots 2) \end{cases}$
In Equation $(1)$, multiply both sides by $12$ to obtain
$\begin{aligned} 3(x+3)-4(y-2) & = 37 \\ 3x+9-4y+8 & = 37 \\ 3x-4y+17 & = 37 \\ 3x-4y & = 20 \end{aligned}$
In Equation $(2)$, multiply both sides by $6$ to obtain
$\begin{aligned} 3(x-3)-2(y+4) & = -1 \\ 3x-9-2y-8 & = -1 \\ 3x-2y-17 & = -1 \\ 3x-2y & = 16. \end{aligned}$
We obtain a simpler system of linear equations.
$\begin{cases} 3x-4y & = 20 && (\cdots 1) \\ 3x-2y & = 16 && (\cdots 2) \end{cases}$
Eliminate $x$ from the two equations above so that we obtain
$\begin{aligned} -4y-(-2y) & = 20-16 \\ -2y & = 4 \\ y & = -2. \end{aligned}$
Substitute $y=-2$ into one of the equations, for example Equation $(2)$.
$\begin{aligned} 3x-2\color{red}{y} & = 16 \\ 3x-2(-2) & = 16 \\ 3x+4 & = 16 \\ 3x & = 12 \\ x & = 4 \end{aligned}$
Therefore, the roots (solutions) of the system of equations are $x = 4$ and $y = -2.$
(Answer C)

Given the system of linear equations in two variables: $\begin{cases} 2p+3q & = 2 && (\cdots 1) \\ 4p-q & = 18 && (\cdots 2) \end{cases}$.
Using the elimination method, we obtain
$\begin{aligned} \! \begin{aligned} 2p+3q & = 2 \\ 4p-q & = 18 \end{aligned} \left| \! \begin{aligned} \times 2 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~4p+6q & = 4 \\ 4p-q & = 18 \end{aligned} \\ & \rule{2.8 cm}{0.6pt} – \\ & \! \begin{aligned} 7q & = -14 \\ q & = -2. \end{aligned} \end{aligned}$
Substitute $q = -2$ into one of the equations, for example Equation $(2)$.
$\begin{aligned} 4p-\color{red}{q} & = 18 \\4p-(-2) & = 18 \\ 4p & = 16 \\ p & = 4 \end{aligned}$
Thus, the roots (solutions of the system of equations) are $p=4$ and $q=-2$.
Therefore, the value of $\boxed{\begin{aligned} 5p-2q^2 & =5(4)-2(-2)^2 \\ & =20-8=12. \end{aligned}}$
(Answer B)

The system of equations above is not actually a system of linear equations in two variables, but it can be transformed into one by letting $x^2 = a$ and $y^2 = b$, so that we obtain
$\begin{cases} a-2b &= -2 && (\cdots 1) \\ 3a+b & = 57 && (\cdots 2) \end{cases}$
Using the elimination method, we obtain
$\begin{aligned} \! \begin{aligned} a-2b & = -2 \\ 3a+b & = 57 \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 2 \end{aligned} \right| & \! \begin{aligned}~a-2b & = -2 \\~6a+2b & = 114 \end{aligned} \\ & \rule{3 cm}{0.6pt} + \\ & \! \begin{aligned} 7a & = 112 \\ a & = 16 \end{aligned} \end{aligned}$
Substitute $a = 16$ into one of the equations, for example Equation $(2)$.
$\begin{aligned} 3\color{red}{a}+b & = 57 \\ 3(16) + b & = 57 \\ b & = 9 \end{aligned}$
Therefore, the value of $\boxed{\begin{aligned} 2x^2-3y^2 & = 2a-3b \\ & = 2(16)-3(9) \\ &= 32-27=5. \end{aligned}}$
(Answer C)

Let $x = \dfrac{1}{a+b}$ and $y = \dfrac{1}{a-b}$ so that we obtain the following system of linear equations in two variables
$\begin{cases} 7x+6y & = 3 && (\cdots 1) \\ 7x-3y & = 0 && (\cdots 2) \end{cases}$
We will find the value of $a^2-b^2=(a+b)(a-b) = \dfrac{1}{xy}$, which requires us to first determine the values of $x$ and $y$.
From the system of linear equations above, we can directly eliminate $x$ by subtracting the two equations.
$\begin{aligned} (7x+6y)-(7x-3y) & = 3-0 \\ 9y & = 3 \\ y & = \dfrac13 \end{aligned}$
Substitute $y = \dfrac13$ into one of the equations, for example Equation $(2)$.
$\begin{aligned} 7x-3\color{red}{y} & = 0 \\ 7x-3\left(\dfrac13\right) & = 0 \\ 7x-1 & = 0 \\ x & = \dfrac17 \end{aligned}$
Therefore, we obtain $\dfrac{1}{xy} = \dfrac{1}{\frac17 \cdot \frac13} = 21$. Thus, the value of $\boxed{a^2-b^2=21}.$
(Answer C)

We will determine the two line equations shown in the graph above.
The first line passes through the points $(2, 0)$ and $(0, 4)$. Since we know the coordinates of the intercepts on the coordinate axes, it will be easier to determine the equation of the line.
The equation of the first line is $2x + y = 4$.
The second line passes through the points $(-3, 0)$ and $(1, 2)$. To find the equation of the line, we can use the following quick method.

The equation of the second line is $x-2y=-3.$
Therefore, the point $(1, 2)$ is the solution of the system of equations $x-2y=-3$ and $2x+y=4.$
(Answer D)

Suppose the whole numbers are $a$ and $b$ with $a > b$ so that we obtain the following system of linear equations
$\begin{cases} a+b & = 27 && (\cdots 1) \\ a-b & = 3 && (\cdots 2) \end{cases}$
Add the two equations and we obtain $2a = 30$, which means $a = 15$, and $b = 12$.
The product of $a$ and $b$ is $ab = 15(12) = 180$.
Therefore, the product of the two numbers is $\boxed{180}.$
(Answer C)

The price of $5$ pencils and $2$ books is Rp26,000, which we write as $5a + 2b = 26.000.$
The price of $3$ pencils and $4$ books is Rp38,000, which we write as $3a + 4b = 38.000.$
Therefore, the corresponding system of linear equations is
$\begin{cases} 5a+2b=26.000 \\ 3a+4b=38.000 \end{cases}$
(Answer B)

Suppose $x$ = the price of $1$ notebook and $y$ = the price of $1$ pencil so that a mathematical model in the form of a system of linear equations can be formed as follows.
$\begin{cases} 2x + 3y & = 8.500 && (\cdots 1) \\ 3x + 2y & = 9.000 && (\cdots 2) \end{cases}$
Add Equations $(1)$ and $(2)$.
$\begin{aligned} \! \begin{aligned} 2x+3y & = 8.500 \\ 3x+2y & = 9.000 \end{aligned} \\ \rule{3.4 cm}{0.6pt} + \\ \! \begin{aligned} 5x + 5y& = 17.500 \\ x + y & = 3.500 \end{aligned} \end{aligned}$
Therefore, Anita must pay Rp3,500 to buy $1$ notebook and $1$ pencil.
(Answer D)

Let Amar’s age = $A$ and Bondan’s age = $B$.
We obtain the following system of linear equations in two variables.
$$\begin{cases} A & = \dfrac23B && (\cdots 1) \\ (A+6)+(B+6) & = 42 && (\cdots 2) \end{cases}$$Substitute Equation $(1)$ into Equation $(2)$.
$\begin{aligned} (\color{red}{A}+6)+(B+6) & = 42 \\ \dfrac23B+6+B+6 & = 42 \\ \dfrac53B & = 30 \\ B & = 30 \times \dfrac35 = 18 \end{aligned}$
Bondan’s current age is $18$ years, which means Amar’s current age is $\dfrac23(18) = 12$ years.
The difference between their ages is $\boxed{18-12=6~\text{years}}.$
(Answer D)

Let $x$ = the price of granulated sugar per kg and $y$ = the price of rice per kg so that a mathematical model in the form of a system of linear equations in two variables can be formed as follows.
$\begin{cases} 5x + 30y & = 410.000 && (\cdots 1) \\ 2x + 60y & = 740.000 && (\cdots 2) \end{cases}$
Eliminate $y$ from Equation $(1)$ and Equation $(2)$.
$$\begin{aligned} \! \begin{aligned} 5x+30y & = 410.000 \\ 2x+60y & = 740.000 \end{aligned} \left| \! \begin{aligned} \times 2 \\ \times 1 \end{aligned} \right| & \! \begin{aligned} 10x+60y & = 820.000 \\ 2x+60y & = 740.000 \end{aligned} \\ & \rule{4 cm}{0.6pt} – \\ & \! \begin{aligned} 8x & = 80.000 \\ x & = 10.000 \end{aligned} \end{aligned}$$Substitute $x = 10.000$ into one of the equations, for example Equation $(1)$.
$\begin{aligned} 5\color{red}{x} +30y & = 410.000 \\ 5(10.000) + 30y & = 410.000 \\ 50.000 + 30y & = 410.000 \\ 30y & = 360.000 \\ y & = 12.000 \end{aligned}$
Thus, the price of $1$ kg of granulated sugar is Rp10,000 and the price of $1$ kg of rice is Rp12,000.
Therefore, the price of $2$ kg of granulated sugar and $5$ kg of rice is
$2 \times 10.000 + 5 \times 12.000 =$ $\boxed{\text{Rp}80.000}.$
(Answer B)

Let $x$ = the price of granulated sugar per kg and $y$ = the price of rice per kg so that a mathematical model in the form of a system of linear equations in two variables can be formed as follows.
$\begin{cases} 2x + 3y & = 27.000 && (\cdots 1) \\ 3x + 3y & = 33.000 && (\cdots 2) \end{cases}$
Eliminate $y$ from Equation $(1)$ and Equation $(2)$.
$\begin{aligned} \! \begin{aligned} 2x+3y & = 27.000 \\ 3x+3y & = 33.000 \end{aligned} \\ \rule{3.3 cm}{0.6pt} – \\ \! \begin{aligned} -x & = -6.000 \\ x & = 6.000 \end{aligned} \end{aligned}$
Substitute $x = 6.000$ into one of the equations, for example Equation $(1)$.
$\begin{aligned} 2\color{red}{x} +3y & = 27.000 \\ 2(6.000) + 3y & = 27.000 \\ 12.000 + 3y & = 27.000 \\ 3y & = 15.000 \\ y & = 5.000 \end{aligned}$
Thus, the price of $1$ kg of granulated sugar is Rp6,000 and the price of $1$ kg of rice is Rp5,000.
(Answer A)

It is known that the perimeter of the rectangle is $58$ meters, so we write
$2(p + l) = 58 \Leftrightarrow p + l = 29.$
It is also known that the difference between the length and width is $9$ meters, so we write $p -l = 9.$
Thus, we obtain the following system of linear equations in two variables
$\begin{cases} p + l &= 29 && (\cdots 1) \\ p -l &= 9 && (\cdots 2) \end{cases}$
Eliminate $l$ from Equations $(1)$ and $(2).$
$\begin{aligned} \! \begin{aligned} p + l &= 29 \\ p -l&= 9 \end{aligned} \\ \rule{2.3 cm}{0.6pt} + \\ \! \begin{aligned} 2p &= 38 \\ p &= 19 \end{aligned} \end{aligned}$
For $p=19$, we obtain $19-l = 9$, which means $l = 10$.
Therefore, the area is $\boxed{L = pl = 19(10) = 190~\text{m}^2}.$
(Answer B)

Suppose $x =$ the price of one jar of nastar cake and $y =$ the price of one jar of cheese cake. Thus, we obtain the following system of linear equations in two variables
$\begin{cases} x &= 2y \\ 3x + 2y &= 480.000 \end{cases}$
Substitute $2y = x$ into Equation $2$ so that it becomes
$\begin{aligned} 3x + \color{red}{x} &= 480.000 \\ 4x &= 480.000 \\ x &= 120.000 \end{aligned}$
This means, $y = \dfrac{1}{2} \cdot 120.000 = 60.000$
The price of $2$ jars of nastar cake and $3$ jars of cheese cake is
$\begin{aligned} 2x + 3y &= 2(120.000) + 3(60.000) \\ &= 240.000 + 180.000 \\ &= 420.000 \end{aligned}$
Therefore, the amount of money Sukardi must pay is Rp420.000.
(Answer B)

Suppose $x, y$ respectively represent the prices of $1$ notebook and $1$ pen so that the following system of linear equations in two variables is formed
$\begin{cases} 10x + 4y &= 36.000 && (\cdots 1) \\ 5x + 8y &= 27.000 && (\cdots 2) \end{cases}$
Eliminate $x$ from Equations $(1)$ and $(2)$.
$$\begin{aligned} \! \begin{aligned} 10x + 4y &= 36.000 \\ 5x + 8y &= 27.000 \end{aligned} \left| \! \begin{aligned} \div 2 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~5x+2y &= 18.000 \\~5x+8y &= 27.000 \end{aligned} \\ & \rule{3.5 cm}{0.6pt} – \\ & \! \begin{aligned} 6y &= 9.000 \\ y &= 1.500 \end{aligned} \end{aligned}$$Substitute $y = 1.500$ into one of the equations, for example the first equation.
$\begin{aligned} 5x + 2\color{red}{y} &= 18.000 \\ 5x + 2(1.500) &= 18.000 \\ 5x + 3.000 &= 18.000 \\ 5x &= 15.000 \\ x &= 3.000 \end{aligned}$
Therefore, the prices of $1$ notebook and $1$ pen respectively are Rp3.000 and Rp1.500.
(Answer D)

Suppose the amount of money owned by the younger sibling is represented by $x$ and the amount of money owned by the older sibling is represented by $y$, so that the following system of linear equations in two variables is obtained
$\begin{cases} x -y & = 10.000 && (\cdots 1) \\ x + 2y & = 40.000 && (\cdots 2) \end{cases}$
Using the combination method, we obtain
$\begin{aligned} \! \begin{aligned} x + 2y & = 40.000 \\ x -y & = 10.000 \end{aligned} \\ \rule{3.2 cm}{0.6pt} – \\ \! \begin{aligned} 3y & = 30.000 \\ y & = 10.000 \end{aligned} \end{aligned}$
For $y=10.000$, we obtain $x = 10.000 + 10.000$, which means $x = 20.000.$
The total amount of their money can be written as
$\boxed{x+y=20.000+10.000=30.000}.$
Therefore, the total amount of their money is Rp30,000.
(Answer B)

Notice that
$\begin{aligned} \! \begin{aligned} 6x+2y & = 12 \\ 3x+y & = 6 \end{aligned} \left| \! \begin{aligned} \times \frac12 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~3x+y & = 6 \\ 3x+y & = 6 \end{aligned} \end{aligned}$
The system has two equations that are actually equivalent (the same). This means that the system contains two variables in a single equation, so there are $\infty$ (infinitely many) solutions.
(Answer D)

Since $x=2$ and $y=1$ are solutions of the system above, substitution gives
$\begin{cases} 2a-b = 6 \\ 4a+3b=2 \end{cases}$
We will determine the value of $b$ using the elimination method.
$\begin{aligned} \! \begin{aligned} 2a-b & = 6 \\ 4a+3b & = 2 \end{aligned} \left| \! \begin{aligned} \times 2 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~4a-2b & = 12 \\ 4a+3b & = 2 \end{aligned} \\ & \rule{2.5 cm}{0.8pt} – \\ & \! \begin{aligned} -5b & = 10 \\ b & = -2 \end{aligned} \end{aligned}$
Substitute $b=-2$ into one of the equations, for example the equation $2a-b=6$, so that we obtain
$2a-(-2)=6 \Leftrightarrow 2a=4 \Leftrightarrow a = 2$
Therefore, the value of $\boxed{a^2+b^2=(2)^2+(-2)^2=4+4=8}.$
(Answer D)

Suppose
$\begin{aligned} x & = \text{number of students} \\ y & = \text{number of computers}. \end{aligned}$
Based on the second statement of the problem, we can form a mathematical model in the form of a system of linear equations in two variables.
$\begin{cases} x & = 2y + 3 && (\cdots 1) \\ x & = 3(y -4) = 3y -12 && (\cdots 2)\end{cases}$
Substitute the value of $x$ from one equation into the other equation so that we obtain
$\begin{aligned} 2y + 3 & = 3y-12 \\ 3y-2y & = 12+3 \\ y & = 15 \end{aligned}$
Therefore, the number of computers at ABC School is $\boxed{15~\text{units}}.$
(Answer C)

Suppose $S$ and $K$ respectively represent the number of students and the number of rooms in the dormitory. Based on the given information, the following system of linear equations in two variables is obtained.
$$\begin{cases} S & = 2K + 12 && (\cdots 1) \\ S & = 3(K-2) = 3K-6 && (\cdots 2) \end{cases}$$Substitute the value of $S$ from one equation into the other equation so that we obtain
$\begin{aligned} 2K+12 & = 3K-6 \\ 3K-2K & = 6+12 \\ K & = 18 \end{aligned}$
Therefore, there are $\boxed{18}$ rooms in the school dormitory.
(Answer B)

Suppose $x$ and $y$ respectively represent the number of chairs and the number of classrooms. From the given information, we can form the following mathematical model in the form of a system of linear equations in two variables.
$\begin{cases} x & = 36y + 96 && (\cdots 1) \\ x & = 42y-48 && (\cdots 2) \end{cases}$
Subtract the two equations and obtain
$\begin{aligned} 6y-144 & = 0 \\ 6y & = 144 \\ y & = \dfrac{144}{6} = 24 \end{aligned}$
Therefore, the number of classrooms in the school is $\boxed{24}.$
(Answer B)

Given the system of linear equations in two variables
$\begin{cases} 2R_1+3R_2 & = 8 && (\cdots 1) \\ R_1-3R_2& = 1 && (\cdots 2) \end{cases}$
Eliminate $R_2$ from the two equations above.
$\begin{aligned} \! \begin{aligned} 2R_1+3R_2 & = 8 \\ R_1-3R_2 & = 1 \end{aligned} \\ \rule{3.1 cm}{0.6pt} + \\ \! \begin{aligned} 3R_1 & = 9 \\ R_1 & = 3 \end{aligned} \end{aligned}$
Substitute $R_1 = 3~\Omega$ into Equation $(2)$.
$\begin{aligned} \color{red}{R_1}-3R_2 & = 1 \\ 3-3R_2 & = 1 \\ -3R_2 & = -2 \\ R_2 & = \dfrac23 \end{aligned}$
Therefore, the values of $R_1$ and $R_2$ respectively are $3~\Omega$ and $\dfrac23 ~\Omega$.
(Answer B)

It is known that
$\begin{cases} mx+3y & = 21 && (\cdots 1) \\ 4x-3y & = 0 && (\cdots 2) \end{cases}$
From Equation $(2)$, we obtain
$-3y = -4x \Leftrightarrow y = \dfrac43x.$
In order for $y$ to be an integer, $x$ must be divisible by $3$.
Substitute $y = \dfrac43x$ into Equation $(1)$.
$\begin{aligned} mx+3\color{red}{y} & = 21 \\ mx + \cancel{3}\left(\dfrac{4}{\cancel{3}}x\right) & = 21 \\ mx + 4x & = 21 \\ (m+4)x & = 21 \end{aligned}$
The form $(m+4)x$ can be considered as the multiplication of two integers that results in $21$. The factors of $21$ are $1, 3, 7$, and $21$ (only $3$ and $21$ are possible values for $x$ because both are divisible by $3$).
Suppose $x = 3$ is chosen. As a result, $m = 3$ and $y = 4$ so that $\boxed{m+x+y = 3+3+4 = 10}.$ Suppose $x = 21$ is chosen. As a result, $m = -3$ and $y = 28$ so that
$\boxed{m+x+y = -3+21+28 = 46}.$
Therefore, the possible values of $m+x+y$ are $10$ or $46.$
(Answer C)

It is known that
$\begin{cases} (a+3)x + y & = 0 && (\cdots 1) \\ x + (a+3)y & = 0 && (\cdots 2) \end{cases}$
Multiply both sides of Equation $(2)$ by $(a+3)$ to obtain
$(a+3)x + (a+3)^2y = 0~~~~~(\cdots 3)$.
Subtract $(1)$ and $(3)$, then solve to find the value of $a$.
$\begin{aligned} y-(a+3)^2y & = 0 \\ y(1-(a+3)^2) & = 0 \\ 1-(a+3)^2 & = 0 && (\text{Divide by}~y) \\ 1-(a^2+6a+9) & = 0 \\ a^2+6a+8 & = 0 \\ (a+4)(a+2) & = 0 \end{aligned}$
The values obtained are $a=-4$ or $a=-2$.
Substitute $a=-4$ and $a=-2$ into the expression $a^2+6a+17$.
$$\begin{aligned} a = -4 & \Rightarrow (-4)^2 + 6(-4) + 17 = 9 \\ a = -2 & \Rightarrow (-2)^2 + 6(-2) + 17 = 9 \end{aligned}$$Therefore, the value of $\boxed{a^2+6a+17 = 9}.$
(Answer D)

Suppose $L$ and $N$ respectively represent wages during overtime days and wages during normal days.
Mr. Dede worked for $6$ days with $4$ of them being overtime days ($2$ remaining days were normal days), and he received a wage of Rp74,000. Mathematically, this is written as $\boxed{4L + 2N = 74.000}.$
Mr. Asep worked for $5$ days with $2$ of them being overtime days ($3$ remaining days were normal days), and he received a wage of Rp55,000. Mathematically, this is written as $\boxed{2L + 3N = 55.000}.$
Thus, the following system of linear equations is obtained
$\begin{cases} 4L + 2N & = 74.000 && (\cdots 1) \\ 2L+3N & = 55.000 && (\cdots 2) \end{cases}$
Equation $(1)$ can be simplified into $2L + N = 37.000$.
The value of $L$ will be found by eliminating $N$.
$$\begin{aligned} \! \begin{aligned} 2L + N & = 37.000 \\ 2L+3N & = 55.000 \end{aligned} \left| \! \begin{aligned} \times 3 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~6L + 3N & = 111.000 \\~2L + 3N & = 55.000 \end{aligned} \\ & \rule{4.2 cm}{0.6pt} – \\ & \! \begin{aligned} 4L & = 56.000 \\ L & = 14.000 \end{aligned} \end{aligned}$$Therefore, the wage for one overtime day is Rp14,000.
It is known that Mr. Dian worked for $4$ days and all of them were overtime days. The wage he received is
$\boxed{4L = 4(14.000) = \text{Rp}56.000}.$
(Answer C)

Suppose the first solution needed is $A$ liters and the second solution needed is $B$ liters.
The total amount of solution is $8$ liters. Mathematically, this is written as $\boxed{A+B = 8}.$
The first solution has an acid concentration of $25\%$ and the second solution contains $65\%$ acid. Their mixture produces $8$ liters of a new solution with an acid concentration of $40\%$. Mathematically, this is written as
$25\%A + 65\%B = 40\% \cdot 8.$
Simplify it into $\boxed{5A + 13B = 64}.$
Thus, the following system of linear equations is obtained
$\begin{cases} A+B & = 8 && (\cdots 1) \\ 5A +13B & = 64 && (\cdots 2) \end{cases}$
Equation $(1)$ is equivalent to $A=8-B$.
Substitute $A=8-B$ into Equation $(2)$.
$\begin{aligned} 5\color{red}{A} +13B &= 64 \\ \Rightarrow 5(8-B)+13B & = 64 \\ 40-5B+13B & = 64 \\ 8B & = 24 \\ B & = 3 \end{aligned}$
Substitute $B = 3$ into Equation $(1).$
$\begin{aligned} A+\color{red}{B} & =8 \\ A+3 & = 8 \\ A & = 5 \end{aligned}$
Therefore, the first solution needed is $5$ liters and the second solution needed is $3$ liters.
(Answer A)

Suppose $A$ and $B$ respectively represent Elvand’s rowing speed and the river current speed in km/hour.
Thus, the following system of linear equations can be formed
$\begin{cases} 2A+2B & = 9 && (\cdots 1) \\ 6A-6B & = 9 && (\cdots 2) \end{cases}$
Equation $(2)$ can be simplified into $2A-2B = 3$.
Eliminate $A$ from Equations $(1)$ and $(2)$.
$\begin{aligned} \! \begin{aligned} 2A+2B & = 9 \\ 2A-2B & = 3 \end{aligned} \\ \rule{3 cm}{0.6pt} + \\ \! \begin{aligned} 4A & = 12 \\ A & = 3 \end{aligned} \end{aligned}$
Therefore, Elvand’s rowing speed is $3$ km/hour.
(Answer D)

The system of linear equations $\begin{cases} a_1x + b_1y & = c_1 \\ a_2x+b_2y & = c_2 \end{cases}$ has infinitely many solutions if $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.$
Verification of the First Equation:
$\begin{aligned} \dfrac{a_1}{a_2} & = \dfrac{b_1}{b_2} \\ \dfrac{p+1}{3p-1} & = \dfrac{3p-2}{4p+2} \\ (p+1)(4p+2) & = (3p-1)(3p-2) \\ 4p^2+6p+2 & = 9p^2-9p+2 \\ 5p^2-15p & = 0 \\ 5p(p-3) & = 0 \\ p = 0 &~\text{or}~p=3 \end{aligned}$
Verification of the Second Equation:
$\begin{aligned} \dfrac{a_1}{a_2} & = \dfrac{c_1}{c_2} \\ \dfrac{p+1}{3p-1} & = \dfrac{\cancel{p}}{2\cancel{p}} \\ (p+1)(2) & = 3p-1 \\ 2p+2 & = 3p-1 \\ p & = 3 \end{aligned}$
Clearly, $p=3$ will cause the system above to have infinitely many solutions. Now, test $p = 0$.
$\begin{cases} (0+1)x+(3(0)-2)y & = 0 \\ (3(0)-1)x + (4(0)+2)y & = 2(0) \end{cases}$
Simplify into
$\begin{cases} x-2y & = 0 && (1) \\ -x+2y & = 0 && (2) \end{cases}$
It can be seen that Equations $(1)$ and $(2)$ are equivalent so there will be infinitely many solutions.
Therefore, the values of $p$ that satisfy are $p=0$ or $p=3$.
(Answer D)

Given the system of linear equations
$\begin{cases} 3x+2y & = 12 && (\cdots 1) \\ 2x-y & = 1 && (\cdots 2) \\ kx + 2y & = 16 && (\cdots 3) \end{cases}$
Solve Equations $1$ and $2$, meaning find the values of $(x, y)$ that satisfy both equations.
$\begin{aligned} \! \begin{aligned} 3x+2y & = 12 \\ 2x-y & = 1 \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 2 \end{aligned} \right| & \! \begin{aligned}~3x+2y & = 12 \\~4x-2y & = 2 \end{aligned} \\ & \rule{3 cm}{0.6pt} + \\ & \! \begin{aligned} 7x & = 14\\ x & = 2 \end{aligned} \end{aligned}$
For $x = 2$, substitute into Equation $2$ to obtain
$\begin{aligned} 2(\color{red}{2})-y & = 1 \\ 4-y & = 1 \\ y & = 3 \end{aligned}$
We obtain $(x, y) = (2, 3)$ as the solution for Equations $1$ and $2$, meaning that for the system of equations to have a solution, Equation $3$ must also have the same solution, namely $(2, 3)$.
$\begin{aligned} kx+2y & = 16 \\ \Rightarrow k(2) + 2(3) & = 16 \\ 2k + 6 & = 16 \\ 2k & = 10 \\ k & = 5 \end{aligned}$
Therefore, the value of $k$ is $\boxed{5}.$
(Answer E)

Given $$\begin{cases} kx + y & = 1 && (\cdots 1) \\ 4x + ky & = 2 && (\cdots 2) \end{cases}$$First, make the constants on the right-hand side equal. Multiply both sides of Equation $(1)$ by $2$ so that we obtain
$$\begin{cases} 2kx + 2y & = 2 && (\cdots 1) \\ 4x + ky & = 2 && (\cdots 2) \end{cases}$$In order to have infinitely many solutions, the coefficients of $x$ and $y$ must be equal so that
$$\begin{cases} 2k & = 4 \\ 2 & = k \end{cases}$$Clearly, $k = 2$ satisfies the condition.
Therefore, there is only $\boxed{1}$ possible value of $k$.
(Answer B)

Suppose the length of the buried part of pole 1 is $x.$ This means the length of the visible part above the ground is $5x.$ Also suppose the length of the buried part of pole 2 is $y$ so that the part above the ground is $3y.$
It is known that the total length of the two poles is $13$ meters and the total length of the buried parts is $2,\!5$ meters, so we obtain the following system of linear equations in two variables (SPLDV).
$$\begin{cases} 6x + 4y = 13 \\ x + y = 2,\!5 \end{cases}$$Using the elimination method, we obtain
$$\begin{aligned} \! \begin{aligned} 6x+4y & = 13 \\ x+y & = 2,\!5 \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 4 \end{aligned} \right| & \! \begin{aligned}~6x+4y & = 13 \\ 4x+4y & = 10 \end{aligned} \\ & \rule{3 cm}{0.6pt} – \\ & \! \begin{aligned} 2x & = 3 \\ x & = 1,\!5. \end{aligned} \end{aligned}$$Substituting $x = 1,\!5$ into Equation $2$ will give $y = 2,\!5-1,\!5 = 1.$
Check Statement 1:
The length of the first pole is $6x,$ namely $6(1,\!5) = 9$ meters. Therefore, Statement 1 is true.
Check Statement 2:
The length of the second pole is $4y,$ namely $4(1) = 4$ meters. Therefore, Statement 2 is true.
Check Statement 3:
The length of the buried part of the first pole is $x,$ namely $1,\!5$ meters, not $2$ meters. Therefore, Statement 3 is false.
Check Statement 4:
The length of the buried part of the second pole is $y,$ namely $1$ meter, not $0,\!5$ meter. Therefore, Statement 4 is false.

Therefore, put a check mark on the following two statements because they are true.

• The length of the first pole is $9$ meters.
• The length of the second pole is $4$ meters.
  
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Answer a)
Known
$$\begin{cases} \dfrac13(x-5)+\dfrac34(y+2) &=-2\dfrac12&& (\cdots 1) \\ \dfrac12(2x+3)-\dfrac23(2y+1) & = 8\dfrac16 && (\cdots 2) \end{cases}$$First, simplify Equation $(1)$ by multiplying both sides by $12$.
$$\begin{aligned} \dfrac13(x-5)+\dfrac34(y+2) &=-2\dfrac12 && (\times 12) \\ 4(x-5)+9(y+2) & = -30 \\ 4x-20+9y+18 & = -30 \\ 4x+9y-2 & = -30 \\ 4x+9y & = -28 && (\cdots 3) \end{aligned}$$Also simplify Equation $(2)$ by multiplying both sides by $6$.
$$\begin{aligned} \dfrac12(2x+3)-\dfrac23(2y+1) & = 8\dfrac16 && (\times 6) \\ 3(2x+3)-4(2y+1) & = 49 \\ 6x+9-8y-4 & = 49 \\ 6x-8y+5 & = 49 \\ 6x-8y & = 44 \\ 3x-4y & = 22 && (\cdots 4) \end{aligned}$$Now, using the elimination method, we obtain
$$\begin{aligned} \! \begin{aligned} 4x+9y & = -28 \\ 3x-4y & = 22 \end{aligned} \left| \! \begin{aligned} \times 3 \\ \times 4 \end{aligned} \right| & \! \begin{aligned}~12x+27y & = -84 \\ 12x-16y & = 88 \end{aligned} \\ & \rule{3.4cm}{0.6pt} – \\ & \! \begin{aligned} 43y & = -172 \\ y & = -4 \end{aligned} \end{aligned}$$Substitute $y = -4$ into one of the equations, for example Equation $(4)$.
$\begin{aligned} 3x-4\color{red}{y}& = 22 \\ 3x-4(-4) & = 22 \\ 3x+16 & = 22 \\ 3x & = 6 \\ x & = 2 \end{aligned}$
Therefore, the solution of the system of linear equations is $\boxed{(2, -4)}.$
Answer b)
Known
$\begin{cases} \dfrac{2}{x}+\dfrac{1}{y} & = 1\dfrac15 \\ \dfrac{1}{x}-\dfrac{3}{y} & = -\dfrac{1}{10} \end{cases}$
Suppose $a = \dfrac{1}{x}$ and $b = \dfrac{1}{y}$ so that we obtain the following system of linear equations.
$\begin{aligned} 2a + b & = \dfrac65 && (\cdots 1) \\ a-3b & = -\dfrac{1}{10} && (\cdots 2) \end{aligned}$
Now, using the elimination method, we obtain
$\begin{aligned} \! \begin{aligned} 2a+b & = \frac65 \\ a-3b & = -\frac{1}{10} \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 2 \end{aligned} \right| & \! \begin{aligned}~2a+b & = \frac65 \\ 2a-6b & = -\frac15 \end{aligned} \\ & \rule{3.2cm}{0.6pt} – \\ & \! \begin{aligned} 7b & = \frac75 \\ b & = \frac15 \end{aligned} \end{aligned}$
Since $b = \dfrac{1}{y}$, it means that $y = 5$.
Substitute $y = 5$ into one of the equations $\dfrac{2}{x}+\dfrac{1}{y} = \dfrac65$.
$\begin{aligned} \dfrac{2}{x} + \dfrac{1}{5} & = \dfrac65 \\ \dfrac{2}{x} & = 1 \\ x & = 2 \end{aligned}$
Therefore, the solution of the system of equations is $\boxed{(2, 5)}.$

Suppose $x$ and $y$ represent the weight of the bottle and the weight of the water that completely fills the bottle, respectively. Thus, the following system of linear equations can be formed.
$$\begin{cases} x + y & = 1.500 && (1) \\ x + \dfrac12y & = 900 && (2) \end{cases}$$Eliminating $x$ by subtracting Equation $(2)$ from Equation $(1)$ gives
$$\begin{aligned} x-\dfrac12y & = 1.500-900 \\ \dfrac12y & = 600 \\ y & = 1.200. \end{aligned}$$This means that the weight of the water that completely fills the bottle is $1.200$ grams. Consequently, the weight of the bottle itself is $1.500-1.200 = 300$ grams.
Therefore, the weight of the bottle itself (without water) is $\boxed{300~\text{grams}}.$