Solve the following system of equations:

$21x+47y=110$$47x+21y=162$

Given:

The given system of equations is:

$21x+47y=110$

$47x+21y=162$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as:

$21x+47y=110$......(i)

$47x+21y=162$.....(ii)

Adding equations (i) and (ii), we get,

$21x+47y+47x+21y=110+162$

$68x+68y=272$

$68(x+y)=68\times4$

$x+y=4$.....(iii)

Subtracting equation (ii) from equation (i), we get, 

$21x+47y-(47x+21y)=110-162$

$-26x+26y=-52$

$-26(x-y)=-26\times2$

$x-y=2$.....(iv)

Adding equations (iii) and (iv), we get,

$x+y+x-y=4+2$

$2x=6$

$x=\frac{6}{2}$

$x=3$

Using $x=3$ in equation (iii), we get,

$3+y=4$

$y=4-3$

$y=1$

Therefore, the solution of the given system of equations is $x=3$ and $y=1$.   

Tutorialspoint

Updated on: 10-Oct-2022

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