Question

solve the following pairs of equations by reducing them to a pair of linear equations.. by elimination method​

Answer 1

Question :

$\longmapsto \: \bf{\dfrac{4}{x} + 3y = 14 \: }, \\ \\ \longmapsto \bf{ \: \: \dfrac{3}{x} - 4y = 23}$

Answer:

• x = 1/5 , y = -2

Step-by-step explanation:

Given information

• 4/x + 3y = 14 & 3/x - 4y = 23

Need to find out

• solve the following pairs of equations by reducing them to a pair of linear equations.. by elimination method

Solution

Let 1/x be k

$\rightarrow \frac{4}{x} + 3y = 14 - - - (1) \\ \rightarrow \frac{3}{x} - 4y = 23 - - - (2)$

So , our equations became

• ⇒ 4k + 3y = 14 ----(3)
• ⇒ 3k - 4y = 23 ---(4)

Now , solving equation (3)

⇒ 4k + 3y = 14

⇒ 4k = 14 - 3y

⇒ k = 14 - 3y/4

Putting value of k in equations (4) we get

$\\ \tt \: ⇒3k - 4y = 23$

$\\ ⇒ \tt3 \bigg( \frac{14 - 3y}{4} \bigg) - 4y = 23$

Multiplying both side by 4 we get

$\\ \tt ⇒3 \times 4 \bigg( \frac{14 - 3y}{4} \bigg) - 4 \times 4y = 23 \times 4$

$\\ \tt⇒ \cancel{12} \times \dfrac{14 - 3y}{ \cancel4} - 16y = 92$

$\\ \tt⇒3 \times (14 - 3y) - 16y = 92$

$\\ \tt⇒4 - 9y - 16y = 92$

$\\ \tt⇒ - 9 - 16y = 92 - 42$

$\\ \tt⇒ - 25y = 50$

$\\ \tt ⇒y = \cancel \frac{ 50}{ - 25}$

$\\ \large\pink{ \tt⇒y = - 2 }$

Now ,Putting value of yn= -2 in equations (3) we get

$\\ \tt⇒4k + 3y = 14$

$\\ \tt⇒4k + 3( - 2) = 14$

$\\ \tt⇒4k - 6 = 14$

$\\ \tt⇒4k = 14 + 6$

$\\ \tt⇒4k = 20$

$\\ \tt⇒ k = \cancel \frac{20}{4}$

$\\ { \text⇒k = 5\ \: \: \: b ut \: \: \red{k = \frac{1}{x} } \:} \\ \\ \\ \bf \: therefour \: \: \\ \\ \longmapsto \large \pink{x= \frac{1}{5} }$

$\\ \text{ Hence } \bf{\red{x = \dfrac{1}{5} , y = -2}}\text{ is the Solution of the given equation}$