Question

x+y = 7xy, 2x-3y+xy=0

Answer 1

ek(1)×3
3x+3y=21xy.....eq(3)

eq(2)+(3)
2x-3y=-xy
3x+3y=21xy

;5x=20xy

;y=1/4

y=1/4 put eq (1)

x+1/4=7x × 1/4

;x=7x/4 - 1/4

;x=7x-1/4

;4x=7x-1

;-3x=-1

;x=1/3

x=1/3 and y=1/4

is your answer

Answer 2

The solution of equations $\bf x + y = 7xy$ and $\bf 2x - 3y + xy = 0$ are $\bf \red{x= \frac{1}{3} }$ and $\bf \red{y = \frac{1}{4}}$

Given:

• Two equations.
• $x + y = 7xy$
• $2x - 3y + xy = 0$

To find:

• Find the solution of equations.

Solution:

Concept to be used:

1. Divide the equations by xy.
2. Assume 1/x and 1/y as 'a' and 'b' respectively, hence solve the linear equations.

Step 1:

Divide the equations by xy.

$\frac{x}{xy} + \frac{y}{xy} = \frac{7xy}{xy}$

or

$\frac{1}{y} + \frac{1}{x} = 7$

or

$\bf \frac{1}{x} + \frac{1}{y} = 7 \: \: ...eq1$

and

$\frac{2x}{xy} - \frac{3y}{xy} + \frac{xy}{xy} = 0$

or

$\frac{2}{y} - \frac{3}{x} + 1 = 0$

or

$\frac{2}{y} - \frac{3}{x}= - 1$

or

$\bf \frac{3}{x} - \frac{2}{y} = 1 \: ... eq2$

Step 2:

Assume 1/x and 1/y as 'a' and 'b' respectively,

Let

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

and

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

so,

$\bf a + b = 7...eq3$

and

$\bf 3a - 2b = 1 \: ...eq4$

Step 3:

Solve eq3 and eq4 for 'a' and 'b'.

Multiply eq3 by 2 and add both equations.

$2a + 2b = 14 \\ 3a - 2b = 1 \\ - - - - - - \\ 5a = 15 \\ - - - - - -$

So,

$a = \frac{15}{5}$

or

$\bf a = 3$

put the value of a in eq3

$3 + b = 7$

or

$b = 7 - 3$

or

$\bf b =4$

Step 4:

Find the value of x and y.

As,

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

so,

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

or

$\bf x = \frac{1}{3}$

and

$\bf y = \frac{1}{4}$

Thus,

Values of x and y are 1/3 and 1/4 respectively.

Learn more:

1) 3x+14y=5xy

21y-x=2xy.

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