Question

Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x+y=5 and 2x−3y=4
(ii) 3x+4y=10 and 2x−2y=2
(iii) 3x−5y−4=0 and 9x=2y+7
(iv) x2+2y3=−1 and x−y3=3

Answer 1

(i) x+y=5 ...(i)

2x−3y=4 ...(ii) (elimination method)

Multiply eqn (i) by 2

2x+2y=10

2x−3y=4

−+−

____________

5y=6

⇒y=6/5

Substitute y=6/5in eqn (i)

x+ y=5

x+6/5=5

⇒x=5×5/5−6/5

=19/5

Substitution method

x+y=5 ....(i)

2x−3y=4 ....(ii)

y=5−x ...(iii)

Putting eqn. (3) in (2)

2x−3(5−x)=4

2x−15+3x=4

x=19/5

y=5×5/5−19/5

y=6/5

(ii)
Elimination method
3x+4y=10 ...(i)

2x−2y=2 ...(ii)

Multiply eqn (2) by 2

4x−4y=4

3x+4y=10

____________

7x=14

x=2

2×2−2y=2

⇒4−2=2y

y=1

Substitution method

3x+4y=10 ...(i)

2x−2y=2⇒x−y=1

x−1=y ...(ii)

Putting eqn (ii) in (i)

3x+4(x−1)=10

3x+4x−4=10

_________________

7x=14

x=2

y=x−1

y=2−1

⇒y=1

(iii) 3x−5y=4 ...(i)

9x−2y=7 ...(ii) (elimination method)

Multiply eqn (i) by 3
(ii)-(i)×3

9x−2y=7

9x−15y=12

_____________

13y=−5

⇒y=−5/13

3x−5(−5/13)=4

⇒3x=4−25/13

3x=(52−25)/13

x=27/(13×3)

⇒x=9/13

Substitution method

3x−5y=4 ...(i)

9x−2y=7

⇒9x−7=2y

(9x−7)/2=y

3x−5y=4

⇒3x−5(9x−7)/2=4

⇒3x−45/2x+35/2=4

⇒6x2−452x=4−352

⇒−39x2=8−352

x=2739

⇒x=913

Putting x=913 in 3x−5y=4

3×913−5y=4

2713−5y=4

⇒2713−4=5y

⇒27−5213=5y

−2513=5y

⇒y=−513

(iv)Elimination method

x2+2y3=−1

3x+4y6=−1

⇒3x+4y=−6 ...(i)

x−y3=3

⇒3x−y=9 ...(ii)

(i)−(ii)

3x+4y=−6

3x−y=−9

−+−

______________

5y=−15

⇒y=−3

Putting y=−3 in eqn (i)

3x−12=−6

⇒3x=−6+12

⇒3x=6

⇒x=2

Substitution method

3x+4y=−6 ...(i)

3x−y=9 ...(ii)

3x−9=y ....(iii)

Putting enq (iii) in (i)

3x+4(3x−9)=−6

⇒3x+12x−36=−6

⇒15x=−6+36

⇒15x=30

⇒x=2

Putting x=2 in eq (iii)

6−9=y

⇒y=−3