Question

Question :

Solve...

$\longrightarrow\tt{\frac{3y + 4}{2-6y} = \frac{-2}{\:5}}$

Answer 1

Solution :-

We have,

$\longrightarrow\sf{\dfrac{3y + 4}{2-6y} = \dfrac{-2}{\:5}}$

• Cross multiply,

$\longrightarrow\sf{(3y+4) 5 = (2-6y)-2}$

$\sf\longrightarrow\sf{3y \times 5 +4 \times 5= 2 \times -2 -6y \times -2}$

• In case of -6y × -2 (-) (-) will get cancelled and will result to 12y.

$\longrightarrow\sf{15y + 20 = -4 + 12y}$

• Take 12y to L.H.S. and 20 to R.H.S.

$\longrightarrow\sf{15y - 12y = -4-20}$

$\longrightarrow\sf{3y = -24}$

• Take 3 to R.H.S.,

$\longrightarrow\sf{y = \dfrac{-24}{\:3}}$

• Cancel the terms,

$\longrightarrow\sf{y =\cancel{ \dfrac{-24}{\:3}}}$

$\red{\longrightarrow \boxed{\bf{ \pink{y = \pink{-8}}}} \: \bigstar}$

$\red{\sf{\underline{\therefore \: Value \: of \: y \: is \:\bf{-8.}}}}$

Answer 2

Given equation:

$\mathsf{\to \ \frac{3y+4}{2-6y} =\frac{-2}{5} }$

What to do:

We have to solve the given equation

Solution:

$\mathsf{\frac{3y+4}{2-6y} =\frac{-2}{5} }\\\\\mathsf{Or,5(3y+4)=-2(2-6y)}\\\\\mathsf{Or,15y+20=-4+12y}\\\\\mathsf{Or,15y-12y=-4-20}\\\\\mathsf{Or,3y=-24}\\\\\mathsf{Or,y=\frac{-24}{3} }\\\\\mathsf{Or,y=-8}$

The value of y=-8