Question

If $29x+37y=103,37x+29y=95$ then:

• A. $x=1,y=2$
• B. $x=2,y=1$
• C. $x=2,y=3$
• D. $x=3,y=2$

Answer 1

The correct option is A

$x=1,y=2$

Step 1: Consider the given system of equations

The given system of equations is as follows:

$$\begin{gathered} 29x+37y=103......\left(1\right) \\ 37x+29y=95.......\left(2\right) \end{gathered}$$

Now, add equations (1) and (2) to get:

$\begin{array}{rcl}29x+37x+29y+37y& =& 103+95\\ 66x+66y& =& 198\\ x+y& =& 3........\left(3\right)\end{array}$

Now, subtract equations (1) from (2):

$\begin{array}{rcl}37x-29x+29y-37x& =& 95-103\\ x-y& =& -1........\left(4\right)\end{array}$

Therefore, by adding equations (3) and (4), we get:

$\begin{array}{rcl}x+y+x-y& =& 3-1\\ 2x& =& 2\\ x& =& 1\end{array}$

Step 2: Find the value of y

Now, substitute the value $x=1$ into equation (3):

$\begin{array}{rcl}x+y& =& 3\\ 1+y& =& 3\\ y& =& 3-1\\ y& =& 2\end{array}$

Hence, the required value is $x=1,y=2$. Thus, option (a) is correct.