Question

How do you solve\[\dfrac{5}{12}=\dfrac{x+1}{4}\] ?

Answer 1

Hint: By observing the question we can understand that we have to solve the given equation i.e. we have to find the value of ‘x’. We can see that ‘x’ is a numerator for solving the question we have pulled the ‘x’ out of the equation. For that we have to do cross multiplication and then by solving the equation we can find the value of ‘x’.

Complete step by step answer:
For the given question we have to solve the equation\[\dfrac{5}{12}=\dfrac{x+1}{4}\]. For that we have to consider the above equation as equation (1).
Let us consider the given equation as equation (1).
\[\dfrac{5}{12}=\dfrac{x+1}{4}...............\left( 1 \right)\]
By observing equation (1) we can see that ‘x’ is in the numerator. So for solving the given equation we have to pull the ‘x’ out from the numerator for that we have to do cross multiplication for the equation (1).
As we know cross multiplication means we multiply the numerator of the first fraction with the denominator of the second fraction and the numerator of the second fraction with the denominator of the first fraction i.e.
\[\begin{align}
  & \Rightarrow \dfrac{a}{b}=\dfrac{c}{d} \\
 & \Rightarrow ad=bc \\
\end{align}\]
Let us do cross multiplication for the equation (1), we get
\[\Rightarrow 5\times 4=12\left( x+1 \right)\]
Let us consider the above equation as equation (2).
\[5\times 4=12\left( x+1 \right)...........\left( 2 \right)\]
Simplifying a bit in RHS (right hand side) of equation (2), we get
\[\Rightarrow 5\times 4=12x+12\]
Simplifying a bit in LHS (left hand side) of equation (2), we get
\[\Rightarrow 20=12x+12\]
Subtracting with 12 on both sides, we get
\[\Rightarrow 20-12=12x+12-12\]
Cleaning a bit, we get
\[\Rightarrow 8=12x\]
Rewriting the above equation by changing LHS and RHS, we get
\[\Rightarrow 12x=8\]
Dividing the both sides of equation with 12, we get
\[\Rightarrow x=\dfrac{8}{12}\]
Lowering the terms in above equation, we get
\[\Rightarrow x=\dfrac{2}{3}\]
Let us consider the above equation as equation (3).
\[x=\dfrac{2}{3}..........\left( 3 \right)\]

Note: We should note a point that cross multiplication of an equation \[\dfrac{a}{b}=\dfrac{c}{d}\]is \[ad=bc\] but not like \[ac=bd\] which throws the whole problem into error. Examiners may include some quadratic expression in the equation in that type of questions we have to find roots of the given quadratic equation.