Question

4/x + 3y = 14 and 3/x — 4/y = 23 please solve it by elimination method Chapter – 3 linear equation in two variables​

Answer 1

Answer:

given,

4/x+3y=14..... eqn 1.....3

3/x-4/y=23..... eqn 2.....×4

we get

12/x+9/y=42

12x/-16/y=92

- + -

25/y=-50

25=-50y

y=-25/50

y=- 1/2

put value of y in eqn 1

4/x+3/(-1/2)=14

4/x-6=14

4/x=14+6

x/4=1/20

x=1/5

therefore ,

x=1/5and y=-1/2

Answer 2

Answer:

The solution of the given simultaneous equations is

$\boxed{\red{\sf\:\left(\:x\:,\:y\:\right)\:=\:\left(\:\dfrac{1}{5}\:,\:-\:2\:\right)}}$

Step-by-step-explanation:

The given simultaneous equations are

$\sf\:\dfrac{4}{x}\:+\:3y\:=\:14\:\:\:-\:-\:(\:1\:)\\\\\\\sf\:\dfrac{3}{x}\:-\:4y\:=\:23\:\:\:-\:-\:(\:2\:)$

By replacing $\sf\:\dfrac{1}{x}$ with 'a' in equation ( 1 ) & equation ( 2 ), we get,

$\sf\:4a\:+\:3y\:=\:14\:\:\:-\:-\:(\:3\:)\\\\\\\sf\:3a\:-\:4y\:=\:23\:\:\:-\:-\:(\:4\:)$

By multiplying equation ( 3 ) by 4 & equation ( 4 ) by 3, we get,

$\sf\:16a\:+\:12y\:=\:56\:\:\:-\:-\:(\:5\:)\\\\\\\sf\:9a\:-\:12y\:=\:69\:\:\:-\:-\:(\:6\:)$

By adding equation ( 5 ) & ( 6 ), we get,

$\sf\:16a\:+\:\cancel{12y}\:=\:56\:\:\:-\:-\:(\:5\:)\\\sf\:+\\\underline{\sf\:9a\:-\:\cancel{12y}\:=\:69}\sf\:\:\:-\:-\:(\:6\:)\\\\\\\implies\sf\:25a\:=\:125\\\\\\\implies\sf\:a\:=\:\cancel{\dfrac{125}{25}}\\\\\\\implies\boxed{\red{\sf\:a\:=\:5}}$

By substituting a = 5 in equation ( 3 ), we get,

$\sf\:4a\:+\:3y\:=\:14\:\:\:-\:-\:(\:3\:)\\\\\\\implies\sf\:4\:\times\:5\:+\:3y\:=\:14\\\\\\\implies\sf\:20\:+\:3y\:=\:14\\\\\\\implies\sf\:3y\:=\:14\:-\:20\\\\\\\implies\sf\:3y\:=\:-\:6\\\\\\\implies\sf\:y\:=\:-\:\cancel{\dfrac{6}{3}}\\\\\\\implies\boxed{\red{\sf\:y\:=\:-\:2}}$

Now, by replacing 'a' with $\sf\:\dfrac{1}{x}$, we get,

$\sf\:a\:=\:5\\\\\\\implies\sf\:\dfrac{1}{x}\:=\:5\\\\\\\implies\boxed{\red{\sf\:x\:=\:\dfrac{1}{5}}}$

The solution of the given simultaneous equations is

$\boxed{\red{\sf\:\left(\:x\:,\:y\:\right)\:=\:\left(\:\dfrac{1}{5}\:,\:-\:2\:\right)}}$