Question

How do you solve \[\dfrac{2}{y} = \dfrac{{10}}{{60}}\]?

Answer 1

Hint: The equation is an algebraic equation, where the algebraic equation is the combination of constants and variables and it is in the form of fraction. To solve the above algebraic equation, we use the tables of multiplication and we can find the value of y.

Complete step-by-step solution:
The algebraic expression is an expression which consists of variables and is consistent with the arithmetic operations. The above equation is a linear equation where the linear equation is defined as the equations are of the first order. These equations are defined for lines in the coordinate system. To solve this linear equation, we apply simple methods. Since by solving these types of equations we get only one value.

Now we solve the given equation, let us consider the equation
\[\dfrac{2}{y} = \dfrac{{10}}{{60}}\]
Now divide the equation by 2 we get
\[ \Rightarrow \dfrac{1}{y} = \dfrac{5}{{60}}\]
Taking reciprocal, we get
\[ \Rightarrow y = \dfrac{{60}}{5}\]
On simplifying we get
\[ \Rightarrow y = 12\]
Therefore, the value of y is 12.
We can also solve this by another method. Now consider the equation
\[\dfrac{2}{y} = \dfrac{{10}}{{60}}\]
Take 60 to the LHS and y to the RHS we get
\[ \Rightarrow 2(60) = 10(y)\]
On simplifying we get
\[ \Rightarrow 120 = 10y\]
Divide the above equation by 10 we get
\[ \Rightarrow y = 12\]
Therefore, we have \[y = 12\]
Hence we have solved the given equation
Therefore, the value of y is 12.
If we solve the equation by the different methods the result obtained will be the same.

Note: The algebraic equation or an expression is a combination of variables and constants, it also contains the coefficient. The alphabets are known as variables. The x, y, z etc., are called as variables. The numerals are known as constants. The numeral of a variable is known as co-efficient. We have 3 types of algebraic expressions namely monomial expression, binomial expression and polynomial expression. By using the tables of multiplication, we can solve the equation.