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# The price of two tables and three chairs is Rs 340. The cost of a table of Rs 20 more than a chair. Find the cost of the table and the chair.  Hint: Assume the cost of the table is Rs x, and that of the chair is Rs y. Form two linear equations in terms of x and y using the statement of the problem. Solve the system of equations using the substitution method or elimination method or matrix method. Hence find the value of x and y satisfying both the questions. The value of x will be the cost of the table, and the value of y will be the cost of the chair.

Let x be the cost of the table and y be the cost of the chair.
Hence, we have the cost of two tables = 2x and the cost of three chairs = 3y.
Hence, the cost of two tables and three chairs = 2x+3y.
Also, given that the cost of two tables and three chairs = 340.
Hence, we have 2x+3y=340 (i)
Also, it is given that the cost of a table is Rs 20 more than the cost of a chair.
Hence, we have x = y+20 (ii).
Solving the system using the substitution method:
Substituting the value of x from equation (ii) in equation (i), we get
2(y+20)+3y=340
Applying distributive law, a(b+c) = ab+ac , we get,
2y+40+3y = 340
i.e. 5y+40 = 340
Subtracting 40 from both sides of the equation we get
5y+40-40 = 340-40
i.e. 5y = 300
Dividing both sides of the equation by 5, we get
\begin{align} & \dfrac{5y}{5}=\dfrac{300}{5} \\ & \Rightarrow y=60 \\ \end{align}
Hence, we have y = 60.
Substituting the value of y in equation (ii), we get
x = 60+20
i.e. x = 80.
Hence the cost of a table is Rs 80, and that of a chair is Rs 60.

Note: In the above solution, we used the substitution method to find the value of x and y.
We can also solve the system by using Matrices.
Here augmented matrix A is given by
$A=\left[ \begin{matrix} 2 & 3 & 340 \\ 1 & -1 & 20 \\ \end{matrix} \right]$
We use row transformations to convert the given matrix in row reduced form.
Apply ${{R}_{2}}\to 2{{R}_{2}}-{{R}_{1}}$, we get
\begin{align} & R=\left[ \begin{matrix} 2 & 3 & 340 \\ 2-2 & -2-3 & 40-340 \\ \end{matrix} \right] \\ & \Rightarrow R=\left[ \begin{matrix} 2 & 3 & 340 \\ 0 & -5 & -300 \\ \end{matrix} \right] \\ \end{align}
where R is the row reduced form of A.
Hence, we have
-5y = -300
Dividing both sides by -5, we get
y =60
and 2x+3y=340
i.e. 2x+180 = 340
Subtracting 340 from both sides, we get
2x+180-180 = 340-180
i.e. 2x = 160
Dividing both sides by 2 we get
x = 80
Hence x = 80 and y = 60 is the solution of the given system.

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