Question

Note: ● Not By Cross Multiplication Method
●Do By Elimination method or
Substitution method​

Answer 1

Step-by-step explanation:

a. $\color{\red}{6x+3y=6xy & 2x+4y= 5xy}$

6x + 3y = 6xy ....(1)

2x +4y = 5xy ......(2)

From Eq 1

Taking xy From Multiply To Devide

$\frac{6x + 3y }{xy} =6$

Breaking xy To Both Variable

$\frac{6x}{xy} + \frac{3x}{xy} = 6$

We Cut x With x and Y with y

so, These eq Will Be Written As

$\frac{6}{y} + \frac{3}{x} = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ......(3)$

From Eq 2

2x + 4y = 5xy

Same Process

$\frac{2x + 4y}{xy} = 5$

these Eq Will Be Written As

$\frac{2x}{xy} + \frac{4x}{xy} = 5$

Cut x With x And y with y

so,

$\frac{2}{y} + \frac{4}{x} = 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .....(4)$

From Equation 3 And 4

6/y + 3/x = 6

& 2/y + 4/x = 5

Multiply By 3 In Equation 4

3 (2/y + 4/x = 5 )

6/y + 12/x = 15 ...(5)

By Elmination Method

From Equation 3 And 5

6/y +3/x = 6

6/y + 12/x = 15

- - -

——————————

-9/x = - 9

——————————

- 9/x = -9

x = -9 / -9

$\bold{\boxed{x = 1}}$

Putting The Value Of x In Equation 1

6 × 1 + 3y = 6 × 1 ×y

6 + 3y = 6y

6 = 6y - 3y

6 = 3y

y = 6/3

$\bold{\boxed{y = 2}}$

$<marquee>x = 1 </marquee>$

$<marquee>y = 2</marquee>$

B). $\color{red}{10/x+y <strong>+</strong><strong> </strong>2/ x - y = 4}$

$\color{red}{15/x + y <strong>-</strong> 5/x-y =-2}$

Answer

$let \: \: \frac{1}{x + y} = a \:$

And

$let \: \frac{1}{x - y} \: = b$

So

10a + 2b =4 ......(1)

15a - 5b = -2 ......(2)

Multiply By 5 In Equation 1 And 2 In Eq 2

ie,

(10a + 2b = 4) × 5

(15a - 5b = -2 )×2

By Elmination Method

50a + 10b = 20

30b - 10b = -4

+

—————————

80b = 16

—————————

$b \: = \frac{16}{80}$

so

$b \: = \frac{1}{5}$

Putting The Value Of b in Equation 2

i•e

$15a + 5 \times \frac{1}{5} = - 2$

15a + 1 = -2

15a = -2-1

15a = -3

$a = \frac{ - 3}{15}$

i•e

$a = \frac{1}{3}$

But

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

$\frac{1}{x + y} = \frac{1}{3}$

So ,

x + y = 3 ......(3)

Again

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

$\frac{1}{x - y} = \frac{1}{5}$

so,

x - y = 5 ......(4)

NOW,

by Elmination Method

From Equation 3 And 4

x + y = 3

x - y = 5

+

——————

2x = 8

x = 8/2

$\bold{\boxed{x=4}}$

Putting The Value Of x in Equation 3

4+ y =3

y = 3-4

$\bold{\boxed{y= -1 }}$

C),

$let \frac{1}{3x + y} = a$

And

$let \: \frac{1}{3x - y} = b$

Putting These Value

ie,

$a + b \: = \frac{3}{4}$

Taking 4 From Devide To Multiply

ie,

4( a+b ) = 3

4a + 4b = 3 ....(1)

Taking Another eq

$\frac{a}{2} - \frac{b}{2} = \frac{ - 1}{8}$

these Eq Can Be Written As

$\frac{a - b}{2} = \frac{ - 1}{8}$

$a - b = \frac{ - 1}{8} \times 2$

$a - b = \frac{ - 1}{4}$

$4a - 4b = - 1 \: \: \: \: \: \: \: \: \: ....(2)$

By Elmination Method

From Equation 1 And 2

4a + 4b =3

4a - 4b = -1

+

————————

8a = 2

a = 2/8

a= 1/4

Putting The Value Of a In Eq 1

$4 \times \frac{1}{4} + 4b = 3$

1 + 4b = 3

4b = 3-1

4b = 2

b = 2/4

b = 1/2

Now,

$\frac{1}{3x + y} = a$

$\frac{1}{3x + y} = \frac{1}{4}$

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

Again

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

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

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

By Elmination Method

From Equation 3 And 4

3x + y = 4

3x - y = 2

+

————————

6x = 6

____________

x = 6/6

$\bold{\boxed{x=1}}$

Putting The Value Of x In Eq 3

3 × 1 + y = 4

3 + y = 4

y = 4-3

$\bold{\boxed{y = 1}$

$<marquee> Answer </marquee>$

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