Question

How do you solve $\dfrac{2}{5}\left( {z + 1} \right) = y$ for $z$?

Answer 1

Hint: Here in this given equation is a linear equation with two variables. Here we have to solve for one variable. To solve this equation for $z$ by using arithmetic operation we can shift the $y$ variable to the right-hand side of the equation then solve the equation for $z$ and on further simplification we get the required solution for the above equation.

Complete step by step solution:
Given: $\dfrac{2}{5}\left( {z + 1} \right) = y$
We first multiply the terms in the left hand side of the equation with $\dfrac{2}{5}$.
$ \Rightarrow \left( {\dfrac{2}{5} \times z} \right) + \left( {\dfrac{2}{5} \times 1} \right) = y$
Calculate the product of each term in left hand side of the equation
$ \Rightarrow \dfrac{2}{5}z + \dfrac{2}{5} = y$
We need to transpose ‘$\dfrac{2}{5}$’ to the right-hand side of the equation by subtracting $\dfrac{2}{5}$ on the right-hand side of the equation.
$ \Rightarrow \dfrac{2}{5}z = y - \dfrac{2}{5}$
Now, divide both sides of the equation by $\dfrac{2}{5}$.
$ \Rightarrow z = \dfrac{{y - \dfrac{2}{5}}}{{\dfrac{2}{5}}}$
$ \Rightarrow z = \dfrac{5}{2}y - 1$
This is the required solution.
If we observe the obtained solution, we notice that it is in the form of the equation slope intercept form. That is $y = mx + c$, where ‘$m$’ is slope and ‘$c$’ is $y$-intercept.
It is in the exact slope intercept form no need to rearrange the equation,
$z = \dfrac{5}{2}y - 1$, where slope is $\dfrac{5}{2}$ and the intercept is $ - 1$.

$z = \dfrac{5}{2}y - 1$ is the required solution of the given equation.

Note: By putting different values of $y$ and then solving the equation, we can find the values of $z$. The algebraic equation or an expression is a combination of variables and constants, it also contains the coefficient. Generally, we denote the variables with the alphabets. Here both ‘$z$’ and ‘$y$’ are variables. The numerals are known as constants and here $\dfrac{2}{5}$ is constant. The numeral of a variable is known as co-efficient and here $1$ is coefficient of ‘$y$’.