Question

Solve the following systems by Elmination by Substitution method.
23x-29y=98 ; 29x-23y=110
Answer x= 3 ; y= -1
(class 10th)

Answer 1

Hiii friend,

23X -29Y = 98......(1)

29X -23Y = 110......(2)

Using substitution method

From equation 1 we get,

23X -29Y = 98

23X = 98+29Y

X = 98+29Y/23.......(3)


Putting the value of X in equation 2


29X -23Y = 110

29 × (98+29Y/23) -23Y = 110

29 × 98 + 29 × 29Y/23 -23Y = 110

2842 + 841Y/23 -23Y = 110


2842+841Y - 529Y /23 = 110

2842+841Y-529Y = 110 × 23

2842+841-529Y = 2530

841-529Y = 2530-2842

312Y = -321

Y = -321/321 => -1

Putting the value of Y in equation (3)

X = 98+29Y/23 => 98 + 29 × -1/23


X = 98-29/23 => 69/23 => 3


Again by elmination method,


23X -29Y = 98.....(1)


29X-23Y = 110.......(2)

Multiply equation (1) and (2) by 29 and 23 we get,

667X - 841Y = 2842......(3)
667X -529Y = 2530.....(4)


Adding equation (4) and equation (3) we get


667X-841Y = 2842
667X-529Y = 2530
_______________

-321Y = 321

Y = 321/-321 => -1


Putting the value of Y in equation (1)

23X - 29Y = 98

23X -29 × -1 = 98

23X +29 = 98

23X = 98-29

23X = 69

X = 69/23 => 3


Hence,

X = 3 and Y = -1



HOPE IT WILL HELP YOU...... :-)