ong>Answer:

The VALUES of x and y in the given equation by substitution method is x=20 and y=30 for x\neq 0x



=0 and y\neq 0y



=0

Step-by-step explanation:

Given equations are \frac{4}{x}+\frac{9}{y}=\frac{1}{2}\hfill (1)

x

4

+

y

9

=

2

1

\hfill(1)

and \frac{12}{x}+\frac{12}{y}=1\hfill (2)

x

12

+

y

12

=1\hfill(2) for x\neq 0x



=0 and y\neq 0y



=0

Now solving the given equations by substitution method :

Let a=\frac{1}{x}a=

x

1

and b=\frac{1}{y}b=

y

1

for x\neq 0x



=0 and y\neq 0y



=0

Equation (1) implies 4a+9b=\frac{1}{2}\hfill (3)4a+9b=

2

1

\hfill(3)

Equation (2) implies 12a+12b=1\hfill (4)12a+12b=1\hfill(4)

From equation (3) 4a+9b=\frac{1}{2}4a+9b=

2

1

4a=\frac{1}{2}-9b4a=

2

1

−9b

a=\frac{\frac{1}{2}-9b}{4}a=

4

2

1

−9b

a=\frac{1}{8}-\frac{9b}{4}a=

8

1

−

4

9b

Now substitute the value of "a" in equation (4)

12(\frac{1}{8}-\frac{9b}{4})+12b=112(

8

1

−

4

9b

)+12b=1

\frac{12}{8}-\frac{108b}{4}+12b=1

8

12

−

4

108b

+12b=1

\frac{3}{2}-27b+12b=1

2

3

−27b+12b=1

-15b=1-\frac{3}{2}−15b=1−

2

3

-15b=\frac{2-3}{2}−15b=

2

2−3

-15b=-\frac{1}{2}−15b=−

2

1

b=\frac{1}{2\times 15}b=

2×15

1

b=\frac{1}{30}b=

30

1

Since b=\frac{1}{y}b=

y

1

for y\neq 0y



=0

b=\frac{1}{30}=\frac{1}{y}b=

30

1

=

y

1

for y\neq 0y



=0

Therefore y=30

Substitute the value of b in equation a=\frac{1}{8}-\frac{9b}{4}a=

8

1

−

4

9b

we get

a=\frac{1}{8}-\frac{9(\frac{1}{30})}{4}a=

8

1

−

4

9(

30

1

)

=\frac{1}{8}-\frac{\frac{9}{30}}{4}=

8

1

−

4

30

9

=\frac{1}{8}-\frac{9}{120}=

8

1

−

120

9

=\frac{1}{8}-\frac{3}{40}=

8

1

−

40

3

=\frac{5-3}{40}=

40

5−3

=\frac{2}{40}=

40

2

a=\frac{1}{20}a=

20

1

Since a=\frac{1}{x}a=

x

1

for x\neq 0x



=0

a=\frac{1}{20}=\frac{1}{x}a=

20

1

=

x

1

for x\neq 0x



=0

Therefore x=20

Therefore x=20 and y=30