Math, asked by baijayantiprusty2, 1 month ago

(2x-1/3x)²

( \sqrt{2x - 3y)} {}^{2}

Answers

Answered by patelmeshwa751
1

Answer:

Correct equations :-

\sf{\:\dfrac{4}{16x+24z}+\dfrac{12}{21x-14z}=\dfrac{1}{2}}

16x+24z

4

+

21x−14z

12

=

2

1

\sf{\:\dfrac{14}{4{\bf{x}}+6z}+\dfrac{4}{3x-2z}=2}

4x+6z

14

+

3x−2z

4

=2

‎ ‎ ‎

Solution :-

Given first equation:

\sf\implies\dfrac{4}{16x+24z}+\dfrac{12}{21x-14z}=1/2⟹

16x+24z

4

+

21x−14z

12

=1/2

\sf\implies\dfrac{4}{4(x+6z)}+\dfrac{12}{7(3x-2z)}=1/2⟹

4(x+6z)

4

+

7(3x−2z)

12

=1/2

\sf\implies\dfrac{1}{4x+6z}+\dfrac{12}{7(3x-2z)}=1/2⟹

4x+6z

1

+

7(3x−2z)

12

=1/2

‎ ‎ ‎

Assume that :-

4x + 6z = u

3x - 2z = v

‎ ‎ ‎

\sf\implies\dfrac{1}{u}+\dfrac{12}{7v}=1/2⟹

u

1

+

7v

12

=1/2

\sf\implies\dfrac{7v+12u}{7vu}=1/2⟹

7vu

7v+12u

=1/2

\sf\implies 14v + 24 u = 7vu---(1.)⟹14v+24u=7vu−−−(1.)

‎ ‎ ‎

Given second equation:

\sf\implies\dfrac{14}{4x+6z}+\dfrac{4}{3x-2z}=2⟹

4x+6z

14

+

3x−2z

4

=2

\sf\implies\dfrac{14}{u}+\dfrac{4}{v}=2⟹

u

14

+

v

4

=2

\sf\implies\dfrac{14v+4u}{vu}=2⟹

vu

14v+4u

=2

\sf\implies{14v+4u}=2vu---(2.)⟹14v+4u=2vu−−−(2.)

‎ ‎ ‎

Eq(1) - Eq(2)

\implies⟹ 14v+24u - (14 v + 4 u) =7vu-2vu

\implies⟹ 24u - 4u = 5vu

\implies⟹ 20 u = 5vu

\implies⟹ 4 = v

Put v in eq(2)

\implies⟹ 14(4) + 4u = 2(4)u

\implies⟹ 56 + 4u = 8u

\implies⟹ 56 = 8u - 4u

\implies⟹ 56 = 4u

\implies⟹ 56/4 = u

\implies⟹ 14 = u

‎ ‎ ‎

Now we get :-

u = 14 = 4x + 6z

v = 4 = 3x - 2z ---(3)

‎ ‎ ‎

Multiply equation (3) with 3

\implies⟹ 3(4 = 3x -2z)

\implies⟹ 12 = 9x - 6z ---(4)

Add u and eq(4)

\implies⟹ 14+12 = 4x + 6z + 9x - 6z

\implies⟹ 26 = 13x

\implies⟹ 26/13 = x

\implies⟹ 2 = x

So the value of x is 2.

‎ ‎ ‎

Put this value of x in u

\implies⟹ 14 = 4(2) + 6z

\implies⟹ 14 = 8 + 6z

\implies⟹ 14 - 8 = 6z

\implies⟹ 6 = 6z

\implies⟹ 1 = z

So the value of z is 1.

Answered by sagarhm15930
1

Answer:

44

Step-by-step explanation:

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