Question

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Answer 1

Answer:

8x+5y=9

3x+2y=4

8x=9-5y

x=9-5y/8

sub x in 2

3( 9-5y/8) + 2y=4

27-15y/8+2y=4

27-15y+16y/8=4

27+y=32

y=5

sub in 1 and get the value of x

8x+5(5) =9

8x=-16

x=-2

b) 4x-7y=-28 ×7

7x-5y= 9 ×4

28x -49y= -196

28x -20y= 36

(-). -29y= -160

y= 5.8

x= 38.5

Answer 2

Answer:

a.) 8x+5y=9

x=(9-5y)/8 .......(1)

3x+2y=4

x=(4-2y)/3 .......(2)

Equating both the values of x from (1) and (2) :-

(9-5y)/8 = (4-2y)/3

3(9-5y) = 8(4-2y)

27-15y = 32-16y

16y-15y = 32-27

y = 5

Substituting the value of y in (1) :-

x = [4-2(5)]/3

x = (4-10)/3

x = -6/3

x = -2

So, x=-2 & y=5

(b.) 4x-7y+28=0 ....(1)

7x-5y-9=0 ....(2)

Since L.H.S of eq. (1) & (2) is equal.

Hence, R.H.S would also be equal.

Equating (1) & (2) :-

4x-7y+28 = 7x-5y-9

4x-7x-7y+5y = -9-28

-3x-2y = -37

3x+2y = 37

x = (37-2y)/3 .....(3)

Putting the value of x in (1) :-

4[(37-2y)/3] -7y+28=0

4[(37-2y)/3] = 7y-28

4(37-2y) = 3(7y-28)

148-8y = 21y-84

-8y-21y = -84-148

-29y = -232

29y = 232

y = 8

Putting the value of y in (3) :-

x = [37-2(8)]/3

x = (37-16)/3

x = 21/3

x = 7

So, x=7 & y=8