Question

HELP me with above attachment​

Answer 1

Question :

$\bullet\ \; \sf \red{\dfrac{32}{x}+\dfrac{33}{y}=31}$

$\bullet\ \; \sf \red{\dfrac{33}{x}+\dfrac{32}{y}=34}$

Solve for 'x' and 'y'

Solution :

Let ,

$\bullet\ \; \sf p= \dfrac{1}{x}\ and\ q=\dfrac{1}{y}$

Equations become ,

32 p + 33 q = 31  ... (1)

33 p + 32 q = 34 ... (2)

Note : This kind of equations can easily solved by addition or subtraction of equations .

On adding (1) & (2) ,

⇒ 65 p + 65 q = 65

⇒ p + q = 1  ... (3)

On subtracting (1) from (2) ,

⇒ p - q = 3 ... (4)

On adding (3) & (4) ,

➠ 2p = 4

➠ p = 2  $\pink{\bigstar}$

On subtracting (4) from (3) ,

➠ 2q = - 2

➠ q = - 1  $\green{\bigstar}$

So ,

$\bullet\ \; \sf \pink{p=\dfrac{1}{x}=2}$

$\longrightarrow\ \sf \pink{ x= \dfrac{1}{2} }$

$\bullet\ \; \sf \green{q=\dfrac{1}{y}=-1}$

$\longrightarrow\ \sf \green{y=-1}$

Answer 2

Question: To solve:

$\sf \dfrac{32}{x} + \dfrac{33}{y} = 31 \ \ \ \ ; \ \ \ \dfrac{33}{x} + \dfrac{32}{y} = 34$

Solution:

Let the first equation be Eq(1), and let the second equation be Eq(2).

$\sf \Longrightarrow \dfrac{32}{x} + \dfrac{33}{y} = 31 \ \ \dashrightarrow Eq(1)$

$\sf \Longrightarrow \dfrac{33}{x} + \dfrac{32}{y} = 34 \ \ \dashrightarrow Eq(2)$

To make solving easier, let's consider that:

$\Longrightarrow \sf {\dfrac{1}{x} = a \ \ and \ \ \dfrac{1}{y} = b}$

Substitute these values in the two equations given in the question.

For equation 1:

$\sf \Longrightarrow \dfrac{32}{x} + \dfrac{33}{y} = 31$

$\sf \Longrightarrow 32 \times \Bigg[\dfrac{1}{x}\Bigg] + 33 \times \Bigg[\dfrac{1}{y}\Bigg] = 31$

$\sf \Longrightarrow 32 \times \Bigg[a\Bigg] + 33 \times \Bigg[b\Bigg] = 31$

$\sf \Longrightarrow 32a + 33b = 31 \ \ \dashrightarrow Eq(3)$

Let the above equation be named Eq(3).

For equation 2:

$\sf \Longrightarrow \dfrac{33}{x} + \dfrac{32}{y} = 34$

$\sf \Longrightarrow 33 \times \Bigg[\dfrac{1}{x}\Bigg] + 32 \times \Bigg[\dfrac{1}{y}\Bigg] = 34$

$\sf \Longrightarrow 33 \times \Bigg[a\Bigg] + 32 \times \Bigg[b\Bigg] = 34$

$\sf \Longrightarrow 33a + 32b = 34 \ \ \dashrightarrow Eq(4)$

Let the above equation be named Eq(4).

On observing both the equations 3 and 4, you can tell that they're of the form;

• ax + by = c₁
• bx + ay = c₂

When equations are of this form, we can use the Add & Subtract Method to solve them.

[Any other method is fine too, this one's quicker for these kind of questions]

Adding equations (3) and (4) we get:

⇒ 32a + 33b + [33a + 32b] = 31 + [34]

⇒ 65a + 65b = 65

⇒ 65[a + b] = 65

⇒ a + b = 1 → Let this be Eq(5)

Subtracting equations (3) and (4) we get:

⇒ 32a + 33b - [33a + 32b] = 31 - 34

⇒ 32a + 33b - 33a - 32b = -3

⇒ -a + b = -3 → Let this be Eq(6)

Now, on adding Eq(5) and Eq(6) we get:

⇒ a + b + [-a + b] = 1 + [-3]

⇒ a + b - a + b = 1 - 3

⇒ 2b = -2

⇒ b = -1

Substitute the value of "b" in Eq(6).

⇒ -a + b = -3

⇒ -a + [-1] = -3

⇒ -a - 1 = -3

⇒ -a = -3 + 1

⇒ -a = -2

⇒ a = 2

Now we've got the values of both "a" and "b". Now substitute both "a" and "b" in 1/x = a and 1/y = b respectively,

For 'x':

$\Longrightarrow \sf \dfrac{1}{x} = a$

$\Longrightarrow \sf \dfrac{1}{x} = 2$

$\Longrightarrow \boxed{\sf x = \dfrac{1}{2}}$

For 'y':

$\Longrightarrow \sf \dfrac{1}{y} = b$

$\Longrightarrow \sf \dfrac{1}{y} = -1$

$\Longrightarrow \boxed{\sf y = -1}$

Therefore:

x = 1/2

y = -1

Hence solved.