Question 4990
#1 
y=5-x
3x-4y=-20 

sol:
let,
y=5-x      ..eq(i)
3x-4y=-20  ..eq(ii)

substitute the value of y from eq(i) in eq(ii),
then eq(ii) will be
3x-4(5-x)=-20
3x-20-4x=-20
-20-x=-20
x=-20+20
x=0


#2 
x+2y=6
2x+3y=8 

sol:
let,
x+2y=6  ..eq(i)
2x+3y=8 ..eq(ii)

from eq(i),
x=6-2y

substitute the value of x in eq(ii),
then eq(ii) will be
2(6-2y)+3y=8
12-4y+3y=8
12-y=8
y=12-8
y=4


#3 
3x+y=1
x=2y+5 

sol:
let,
3x+y=1  ..eq(i)
x=2y+5  ..eq(ii)

substitute the value of x from eq(ii) in eq(i),
then eq(i)will be
3(2y+5)+y=1
6y+15+y=1
7y+15=1
7y=1-15
7y=-14
{{{y=-14/7}}}
y=-2


#4 
x+y=6
y=3-2x

sol:
let,
x+y=6  ..eq(i)
y=3-2x ..eq(ii)

substitute the value of y from eq(ii) in eq(i)
then eq(i) will be
x+(3-2x)=6
x+3-2x=6
3-x=6
x=3-6
x=-3


#5 
s+t=5
s=13-3t

sol:
let,
s+t=5   ..eq(i)
s=13-3t .eq(ii)

substitute the value of s from eq(ii) in eq(i)
(13-3t)+t=5
13-2t=5
13-5=2t
8=2t
{{{8/2=t}}}
t=4


#6 
x-y=4
y=2-x

sol:
let,
x-y=4 ..eq(i)
y=2-x ..eq(ii)

substitute y in eq(i)
x-(2-x)=4
x-2+x=4
2x-2=4
2x=4+2
2x=6
{{{x=6/2}}}
x=3