Question

How do you solve $\dfrac{2}{-3}=\dfrac{4v+4}{2v+14}$ ?

Answer 1

Hint: We are given a equation as $\dfrac{2}{-3}=\dfrac{4v+4}{2v+14}$ , we can see that we are given a term in fraction, so firstly we will simplify term to solve fractions out, after that we will separate the variables and constant on each side. After that we will use an algebraic tool to solve and reduce our equation, then we will look for the value of ‘v’ which will satisfy our equation.

Complete step by step answer:
We are given that our equation is $\dfrac{2}{-3}=\dfrac{4v+4}{2v+14}$ .
We can clearly see that our equation involves fractions, so we start our solution by reducing it to a normal equation without fraction, to do so we will multiply.
If we have a fraction as $\dfrac{a}{b}=\dfrac{c}{d}$ , then we multiply right side by ‘b’ and left side by ‘d’. Hence the equation becomes $a\times d=b\times c$ .
Now in our equation, we have $\dfrac{2}{-3}=\dfrac{4v+4}{2v+14}$
We cross multiply, we get –
$\Rightarrow 2\left( 2v+14 \right)=\left( -3 \right)\left( 4v+4 \right)$
Now, we open brackets, we get –
$\Rightarrow 2\times 2v+2\times 14=-3\times \left( 4v \right)+-3\times 4$
By solving, we get –
$\Rightarrow 4v+28=-12v-12$
Now, we separate the variable on one side (on the left side).
So, we add 12v on both sides, so,
$\Rightarrow 4v+12v+28=-12v+12v-12$
By simplifying, we get –
$\Rightarrow 16v+28=-12$ (as $-12v+12v=0$ )
Now we separate the constants on one side (on right side).
So, we subtract 28 on both sides.
$\Rightarrow 16v+28-28=-12-28$
By simplifying, we get –
$\Rightarrow 16v=-40$ (as $12-28=-40$ and $28-28=0$ )
Now, we divide above equation by 11, to separate 16 from ‘v’. so,
$\Rightarrow \dfrac{16v}{16}=\dfrac{-40}{16}$
By simplifying, we get –
$v=\dfrac{-5}{2}$
So, the value of ‘v’ is $\dfrac{-5}{2}$ .
Hence, the solution is $v=\dfrac{-5}{2}$ .
As the equation has power just 1, so it is a linear equation.
Means it can have just one solution.

Note:
When we try to solve a fraction, we need to be careful that we solve term correcting or not.
If we have $\left( -3 \right)\left( a+b \right)$ then -3 will be multiplied by both so, it will give $-3a-3b$ , if we do it as $-3a+b$ or $-3a+3b$ it will be incorrect as -3 is multiplied to ‘a’ as well as ‘b’.
Also we need to remember that we cannot add a constant with a variable, so if we have $x+2$ , we leave it as it is because we cannot add ‘x’ with 2.
Also $2x+4=6x$ is wrong as one cannot add 2x with 4.
We can check that our solution is correct or not.
We put $v=\dfrac{-5}{2}$ in $\dfrac{2}{-3}=\dfrac{4v+4}{2v+14}$.
We get –
$\Rightarrow \dfrac{2}{-3}=\dfrac{4\left( \dfrac{-5}{2} \right)+4}{2\left( \dfrac{-5}{2} \right)+14}$
By simplifying, we get –
$\Rightarrow \dfrac{2}{-3}=\dfrac{-10+4}{-5+14}$
By simplifying further, we get –
$\begin{align}
  & \Rightarrow \dfrac{2}{-3}=\dfrac{-6}{9} \\
 & \Rightarrow \dfrac{2}{-3}=\dfrac{2}{-3} \\
\end{align}$
So, it satisfies our equation.
Hence $v=\dfrac{-5}{2}$ is the correct solution.