Question

48. Solve: 7x + 13y=27
13x + 7y=33​

Answer 1

Answer:

7x +13y =27 .......... (1)

13x +7y = 33 ............(2)

by elemination method......

we multiply 13 to eq. 1 & then 7 to eq. 2

91x + 169y = 351

91x + 49y= 231

120 y = 120

y= 1

put value y in eq. (1)

7x + 13= 27

7x= 14

x= 2

[ x = 2 , y = 1 ]

Answer 2

Let,

7x + 13y = 27     →     (1)

13x + 7y = 33​     →     (2)

First we add both (1) and (2).

    7x + 13y + 13x + 7y = 27 + 33

⇒  20x + 20y = 60

⇒  20(x + y) = 60

Now we divide both sides by 20.

⇒  x + y = 3

And multiply both sides by 7 and 13 each.

→→  13x + 13y = 39     →     (3)

    (3) - (2)

⇒  13x + 13y - 13x - 7y = 39 - 33

⇒  6y = 6

⇒  y = 1

    (3) - (1)

⇒  13x + 13y - 7x - 13y = 39 - 27

⇒  6x = 12

⇒  x = 2

Hence values are found!

Solution by matrices is given below.

By the two systems given, we get a matrix $\left[\begin{array}{cc|c}7&13&27\\ 13&7&33\end{array}\right].$

So,

$\left[\begin{array}{cc|c}7&13&27\\ 13&7&33\end{array}\right]\ \xrightarrow{R_1+R_2\to R_2}\ \left[\begin{array}{cc|c}7&13&27\\ 20&20&60\end{array}\right]\ \xrightarrow{\frac{1}{20}R_2\to R_2}\\ \\ \\ \left[\begin{array}{cc|c}7&13&27\\ 1&1&3\end{array}\right]\ \xrightarrow{R_1-7R_2\to R_1}\ \left[\begin{array}{cc|c}0&6&6\\ 1&1&3\end{array}\right]\ \xrightarrow{\frac{1}{6}R_1\to R_1}$

$\left[\begin{array}{cc|c}0&1&1\\ 1&1&3\end{array}\right]\ \xrightarrow{R_2-R_1\to R_2}\ \left[\begin{array}{cc|c}0&1&1\\ 1&0&2\end{array}\right]\ \xrightarrow{R_1\leftrightarrow R_2}\ \left[\begin{array}{cc|c}1&0&2\\ 0&1&1\end{array}\right]$

So we got the identity matrix. From this, we get,

x = 2     ;     y = 1