Question

Solve the given pair of linear equations and find the value of x and z:
$\sf{\longrightarrow\:\dfrac{4}{16x+24z}+\dfrac{12}{21x-14z}=\dfrac{1}{2}}$
and
$\sf{\longrightarrow\:\dfrac{14}{4z+6z}+\dfrac{4}{3x-2z}=2}$
Please provide well-explained answers :)

Answer 1

Correct equations :-

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

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

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Solution :-

Given first equation:

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

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

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

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Assume that :-

• 4x + 6z = u
• 3x - 2z = v

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

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

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

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Given second equation:

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

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

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

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

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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

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Now we get :-

• u = 14 = 4x + 6z
• v = 4 = 3x - 2z ---(3)

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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.

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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.

Answer 2

Answer :-

Required values are x = 2 and z = 1 .

Explanation :-

Firstly, let's put :-

• 1/(4x + 6z) = a

• 1/(3x - 2z) = b

Now, we have the two equations as :-

⇒ a + 12b/7 = 1/2

⇒ (7a + 12b)/7 = 1/2

⇒ 2(7a + 12b) = 7

⇒ 14a + 24b = 7 ---(1)

⇒ 14a + 4b = 2 ----(2)

Subtracting eq.2 from eq.1, we get :-

⇒ 14a + 24b - 14a - 4b = 7 - 2

⇒ 20b = 5

⇒ b = 1/4

Putting value of "b" in eq.2 :-

⇒ 14u + 1 = 2

⇒ a = 1/14

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Now, let's put the value of "a" and "b" .

⇒ 1/(4x + 6z) = 1/14

⇒ 4x + 6z = 14 ---(3)

⇒ 1/(3x - 2z) = 1/4

⇒ 3x - 2z = 4 ----(4)

Multiplying eq.4 with 3 and then adding it with eq.3, we get :-

⇒ 4x + 6z + 9x - 6z = 14 + 12

⇒ 13x = 26

⇒ x = 26/13

⇒ x = 2

Putting value of "x" in eq.4 :-

⇒ 3(2) - 2z = 4

⇒ 6 - 2z = 4

⇒ -2z = -2

⇒ z = 1