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\[\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}\]

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Hint: First cross multiply the given equation and take a variable that is ‘y’ in the question on one side and constant on the other side. Then cross check by putting the value of "y” back in the given equation.

Here we have to solve the following equation

\[\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}\]

First of all, we will cross multiply the given equation. So, we will get,

\[\left( 3y+4 \right)\left( 5 \right)=\left( -2 \right)\left( 2-6y \right)\]

By simplifying the equation, we get,

\[15y+20=-4+12y\]

Now, we will take the terms containing ‘y’ to the left hand side (LHS) and constant terms to the right hand side (RHS).

So, we will get,

\[15y-12y=-4-20\]

By simplifying the above equation, we get,

\[\Rightarrow 3y=-24\]

By dividing by 3 on both sides, we get,

\[y=\dfrac{-24}{3}\]

Therefore we get y = - 8

Hence, y = -8 is the correct answer.

Note: Here, students must take special care of signs while cross multiplying. In questions involving one variable, students should always try to take variables on one side and constants to the other side. After obtaining the answer, the student can cross check the same by substituting y = - 8 in the question and checking if it satisfies the same.

We can cross check as follows:

We are given the equation

\[\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}\]

Here, we have to prove that LHS = RHS.

Since, we have got y = - 8, therefore, to cross check the answer, we will put y = - 8 in left hand side (LHS) of the given equation, we get LHS as,

\[LHS=\dfrac{3\left( -8 \right)+4}{2-6\left( -8 \right)}\]

By simplifying the above expression, we get,

\[LHS=\dfrac{-24+4}{2+48}=-\dfrac{20}{50}\]

Therefore, we get

\[LHS=-\dfrac{2}{5}\]

which is equal to RHS.

So our answer is correct.

Here we have to solve the following equation

\[\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}\]

First of all, we will cross multiply the given equation. So, we will get,

\[\left( 3y+4 \right)\left( 5 \right)=\left( -2 \right)\left( 2-6y \right)\]

By simplifying the equation, we get,

\[15y+20=-4+12y\]

Now, we will take the terms containing ‘y’ to the left hand side (LHS) and constant terms to the right hand side (RHS).

So, we will get,

\[15y-12y=-4-20\]

By simplifying the above equation, we get,

\[\Rightarrow 3y=-24\]

By dividing by 3 on both sides, we get,

\[y=\dfrac{-24}{3}\]

Therefore we get y = - 8

Hence, y = -8 is the correct answer.

Note: Here, students must take special care of signs while cross multiplying. In questions involving one variable, students should always try to take variables on one side and constants to the other side. After obtaining the answer, the student can cross check the same by substituting y = - 8 in the question and checking if it satisfies the same.

We can cross check as follows:

We are given the equation

\[\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}\]

Here, we have to prove that LHS = RHS.

Since, we have got y = - 8, therefore, to cross check the answer, we will put y = - 8 in left hand side (LHS) of the given equation, we get LHS as,

\[LHS=\dfrac{3\left( -8 \right)+4}{2-6\left( -8 \right)}\]

By simplifying the above expression, we get,

\[LHS=\dfrac{-24+4}{2+48}=-\dfrac{20}{50}\]

Therefore, we get

\[LHS=-\dfrac{2}{5}\]

which is equal to RHS.

So our answer is correct.