Answer:
x+y=14⇒y=14−x
Substituting this value in the second equation, we get
x−(14−x)=4
2x=18⇒x=9
Substituting this value of x in the first equation, we get
9+y=14⇒y=5
(ii) s−t=3⇒s=t+3
Substituting in 2nd equation
3
t+3
+
2
t
=6
6
2t+6+3t
=6
5t+6=36⇒t=6
Substituting value of t in 1st equation
s=t+3=9
(iii) ∵
a
2
a
1
=
b
2
b
1
=
c
2
c
1
Hence all the points lying on the line y=3x−3 like x=2,y=3; are a solution.
(iv) 0.2x+0.3y=1.3⇒x=
0.2
1.3−0.3y
Substituting this value in the second equation
0.4×
0.2
1.3−0.3y
+0.5y=2.3
2.6−0.6y+0.5y=2.3
2.6−2.3=0.6y−0.5y
0.1y=0.3⇒y=3
Substituting this value of y,
x=
0.2
1.3−0.3y
=
0.2
1.3−0.3×3
=
0.2
0.4
x=2
(v)
2
x+
3
y=0⇒y=−
3
2
x
Substituting this in 2nd equation
3
x−
8
×(−
3
2
x)=0
3x+4x=0⇒x=0
Substituting value of x,
y=−
3
2
×0=0
(vi)
2
3x
−
3
5y
=−2
3
5y
=
2
3x
+2⇒y=
10
9x+12
Substituting this in 2nd equation
3
x
+
20
9x+12
=
6
13
60
47x+36
=
6
13
47x+36=130
47x=94⇒x=2
Substituting this value of x in 1st equation
y=
10
9x+12
=
10
9×2+12
=3