8. Using the substitution method, solve each of the following pairs of simultaneous equations. (d) 9x + 2y = 5 6x – 5y = 4 7x – 3y = 13
Answers
Answer:
(i) We have, 3x−5y−4=0
⇒3x−5y=4...(i)
Again 9x=2y+7
⇒9x−2y=7...(ii)
By Elimination Method:
Multiplying equation (i) by 3, we get
9x−15y=12...(iii)
Subtracting (ii) from (iii), we get
9x−15y=12
9x−2y=7
−13y=5
⇒y=−
13
5
Putting the value of equation (ii), we get
9x−2(−
13
5
)=7
⇒9x+
13
10
=7
⇒9x=7−
13
10
⇒9x=
13
91−10
⇒9x=
13
81
⇒x=
13
9
Hence, the required solution is x=
13
9
,y=−
13
5
By Substitution Method:
Expressing x in terms of y from equation (i), we have
x=
3
4+5y
Substituting the value of x in equation (ii), we get
9×(
3
4+5y
)−2y=7
⇒3×(4+5y)−2y=7
⇒12+15y−2y=7
⇒13y=7−12
∴y=−
13
5
Putting the value of y in equation (i), we have
3x−5×(−
13
5
)=4
⇒3x+
13
25
=4
⇒3x=4−
13
25
⇒3x=
13
27
∴x=
13
9
Hence, the required solution is x=
13
9
, y=−
13
5
.
(ii) We have,
2
x
+
2
2y
=−1
⇒
6
3x+4y
=−1
∴3x+4y=−6...(i)
And x−
2
y
=3⇒
3
3x−y
=3
∴3x−y=9...(ii)
By Elimination Method:
Subtracting (ii) from (i), we get
5y=−15 ⇒y=−
5
15
=−3
Putting the value of y in equation (i), we get
3x+r×(−3)=−6
⇒3x−12=−6
⇒3x=−6+12⇒3x=6
Hence, Solution is x=2 , y=−3
By Substitution Method:
Expressing x in terms of y from equation (i), we have
x=
3
−6−4y
Substituting the value of x in equation (ii) from equation (i), we get
3×(
3
−6−4y
)−y=9
⇒−6−4y−y=9
⇒−6−5y=9
⇒−5y=9+6=15
∴y=
−5
15
=−3
Putting the value of y in equation (i), we get
3x+×(−3)=−6
⇒3x−12=−6
⇒3x=12−6=6
∴x=
3
6
=2
Hence, the required solution is x=2,y=−3