Question

solve equations 3x+2y=5 and 2x-3y=7
by substitution method are elimination method​

Answer 1

By substitution-----

3x+2y=5

3x=5-2y

x=(5-2y)/3----(i)

Now, 2x-3y=7-----(ii)

putting eqn (i) in eqn (ii)

[2*(5-2y)/3]-3y=7

[(10-4y)/3]-3y=7

(10-4y-9y)/3=7

10-13y=21

-13y=11

y=-11/13------(iii)

putting eqn (iii) in eqn (i)

x=[5-2*(-11/13)]/3

= (5+22/13)/3

= [(65+22)/13]/3

= 87/13*1/3

= 29/13

( elimination one I have given in photo)

this is substitution method

if you like it, mark me brainliest^o^

Answer 2

Answer:

elimination method,

3x+2y=5 (1)

2x-3y=7 (2)

2 - 5 3 2

-3 - 7 2 -3

x/2 -5 =y/-5 3 =1/3 2

-3 -7 -7 2 2 -3

cross multiply them,

x/-14-15 =y/-10-(-21) =1/-9-4

x/-29 =y/11 =1/13

equate them,

x/-29=1/13 y/11=1/13

cross multiply them,

13x=-29 13y=11

x=-29/13 y=11/13