Solve the following system of equations:

$23x-29y=98$$29x-23y=110$

Given:

The given system of equations is:

$23x-29y=98$

$29x-23y=110$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$23x-29y=98$......(i)

$29x-23y=110$.....(ii)

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

$23x-29y+29x-23y=98+110$

$52x-52y=208$

$52(x-y)=52\times4$

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

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

$23x-29y-(29x-23y)=98-110$

$-6x-6y=-12$

$-6(x+y)=-6\times2$

$x+y=2$.....(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,

$3-y=4$

$y=3-4$

$y=-1$

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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