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Answer 1

Question :

$\bf{\dfrac{1}{2(x + 2y)} + \dfrac{5}{3(3x - 2y)} = -\dfrac{3}{2}}$

$\bf{\dfrac{5}{4(x + 2y)} - \dfrac{3}{5(3x - 2y)} = \dfrac{61}{60}}$

Solution :

From the given equation , we notice that $\bf{\dfrac{1}{(x + 2y)}}$ and $\bf{\dfrac{1}{(3x - 2y)}}$ are common in both the equations.

So let us denote them by a single variable :-

• $\bf{\dfrac{1}{(x + 2y)} = p}$

• $\bf{\dfrac{1}{(3x - 2y)} = q}$

Now substituting them in the equation , we get :

$\bf{\dfrac{p}{2} + \dfrac{5q}{3} = -\dfrac{3}{2}}$⠀⠀⠀⠀⠀...Eq.(1)

$\bf{\dfrac{5p}{4} - \dfrac{3q}{5} = \dfrac{61}{60}}$⠀⠀⠀⠀⠀⠀...Eq.(2)

Now by solving the Eq.(1) , we get :

$:\implies \bf{\dfrac{1p}{2} + \dfrac{5q}{3} = -\dfrac{3}{2}}$

$:\implies \bf{\dfrac{3p + 10q}{6} = -\dfrac{3}{2}}$

$:\implies \bf{3p + 10q = -\dfrac{3}{2} \times 6}$

$:\implies \bf{3p + 10q = -9}$

$\boxed{\therefore \bf{3p + 10q = - 9}}$⠀⠀⠀⠀⠀⠀...Eq.(3)

Now by solving the Eq.(2) , we get :

$:\implies \bf{\dfrac{5p}{4} - \dfrac{3q}{5} = \dfrac{61}{60}}$

$:\implies \bf{\dfrac{25p - 12q}{20} = \dfrac{61}{60}}$

$:\implies \bf{25p - 12q = \dfrac{61}{60} \times 20}$

$:\implies \bf{25p - 12q = \dfrac{61}{3}}$

$:\implies \bf{3(25p - 12q) = 61}$

$:\implies \bf{75p - 36q = 61}$

$\boxed{\therefore \bf{75p - 36q = 61}}$⠀⠀⠀⠀⠀⠀...Eq.(4)

Now , we have to make a common in both the equations.

So by multiplying Eq.(3) by 25 , we get :

$:\implies \bf{(3p + 10q = -9) \times 25}$

$:\implies \bf{75p + 250q = -225}$⠀⠀⠀⠀⠀⠀...Eq.(5)

By subtracting Equation.(4) from Equation.(5) , we get :

$:\implies \bf{(75p + 250q) - (75p - 36q) = -225 - 61}$

$:\implies \bf{75p + 250q - 75p + 36q = -225 - 61}$

$:\implies \bf{250q + 36q = -225 - 61}$

$:\implies \bf{286q = -286}$

$:\implies \bf{q = \dfrac{-286}{286}}$

$:\implies \bf{q = -1}$

$\boxed{\therefore \bf{q = (-1)}}$

Hence the value of q is (-1).

Now by substituting the value of q in Eq.(3) , we get :

$:\implies \bf{3p + 10q = -9}$

$:\implies \bf{3p + 10(-1) = -9}$

$:\implies \bf{3p - 10 = -9}$

$:\implies \bf{3p = 10 - 9}$

$:\implies \bf{3p = 1}$

$:\implies \bf{p = \dfrac{1}{3}}$

$\boxed{\therefore \bf{p = \dfrac{1}{3}}}$

Hence the value of p is 1/3.

Since we have taken the value of $\bf{\dfrac{1}{(x + 2y)}}$ as p and the value of $\bf{\dfrac{1}{(3x - 2y)}}$ as q , we can say that :-

• $\bf{\dfrac{1}{(x + 2y)} = \dfrac{1}{3}}$⠀⠀⠀⠀⠀⠀...Eq.(6)

• $\bf{\dfrac{1}{(3x - 2y)} = (-1)}$⠀⠀⠀⠀⠀⠀...Eq.(7)

By solving Eq.(6) , we get :

$:\implies \bf{\dfrac{1}{(x + 2y)} = \dfrac{1}{3}}$

$:\implies \bf{3 = x + 2y}$

$\boxed{\therefore \bf{x + 2y = 3}}$⠀⠀⠀⠀⠀⠀...Eq.(8)

By solving Eq.(7) , we get :

$:\implies \bf{\dfrac{1}{(3x - 2y)} = (-1)}$

$:\implies \bf{\dfrac{1}{(-1)} = 3x - 2y}$

$:\implies \bf{-1 = 3x - 2y}$

$:\implies \bf{3x - 2y = -1}$

$\boxed{\therefore \bf{3x - 2y = -1}}$⠀⠀⠀⠀⠀⠀...Eq.(9)

Now by adding Eq.(8) and Eq.(9) , we get :

$:\implies \bf{(x + 2y) + (3x - 2y) = 3 + (-1)}$

$:\implies \bf{x + 2y + 3x - 2y = 3 - 1}$

$:\implies \bf{x + 3x = 3 - 1}$

$:\implies \bf{4x = 2}$

$:\implies \bf{x = \dfrac{2}{4}}$

$:\implies \bf{x = \dfrac{1}{2}}$

$\boxed{\therefore \bf{x = \dfrac{1}{2}}}$

Hence the value of x is 1/2.

Now by substituting the value of x in the Eq.(8) , we get :

$:\implies \bf{x + 2y = 3}$

$:\implies \bf{\dfrac{1}{2} + 2y = 3}$

$:\implies \bf{\dfrac{1 + 4y}{2} = 3}$

$:\implies \bf{1 + 4y = 3 \times 2}$

$:\implies \bf{1 + 4y = 6}$

$:\implies \bf{4y = 6 + 1}$

$:\implies \bf{4y = 5}$

$:\implies \bf{y = \dfrac{5}{4}}$

$\boxed{\therefore \bf{y = \dfrac{5}{4}}}$

Hence the value of y is 5/4.

Answer 2

Step-by-step explanation:

The given equations are

12(x+2y)+53(3x−2y)=−32.....(1)

and 54(x+2y)−35(3x−2y)=6160....(2)

Putting 1x+2y=u and 13x−2y= v in equation (1) & equation (2) so that we may get two linear equations in the variables u & v as following:-

12u+53v=−32.....................(1)

54u−35v=6160.................(2)

Multiplying (1) by 36 and (2) by 100, we get

18u+60v=−54...............(3)

125u−60v=3053.............(4)

Adding (3) and (4),we get

143u=3053−54=305−1622=1433

∴u=13=1x+2y

∴x+2y=3...........(5)

Putting value of u in (3), we get

1+10v=−9 (after dividing by 3)

∴10v=−10 or v=−1

⇒−1=13x−2y

⇒3x−2y=−1......(6)

Adding (5) and (6), we get

4x=2

∴x=12

Putting value of x in (5),

12+2y=3

or 2y=3−12=52

∴y=54

The required solution is x=12,y=54