Question

Solve the given equation $4 - \dfrac{{2(z - 4)}}{3} = \dfrac{1}{2}(2z + 5)$

Answer 1

Hint: Here we are asked to solve the given equation which means that we need to find the value of the variable here which is $z$ by solving this above equation. So we just need to simplify it and get the required value.

Complete step-by-step answer:
Here we are given the equation which is $4 - \dfrac{{2(z - 4)}}{3} = \dfrac{1}{2}(2z + 5)$ and we need to solve it. Solving the equation means to find the value of the variable which is $z$ here in this equation. So let us start by simplifying it till we get the value of the variable $z$
We must know that whenever we make the terms transfer from one side to other side the signs of the terms get changed.
For example: if we have $x + 1 = 3$
To solve this when we will take $1$ to other side of the sign changes from
So we get $x = 3 - 1 = 2$
In this way we will solve this above equation and get the value of the variable.
$\Rightarrow$ $4 - \dfrac{{2(z - 4)}}{3} = \dfrac{1}{2}(2z + 5)$
First of all we need to open the brackets. So we need to multiply $2$ with $(z - 4)$ and $1$ with $(3z + 5)$ we get
$\Rightarrow$ $4 - \dfrac{{2z - 8}}{3} = \dfrac{{2z + 5}}{2}$
Now we will separate the denominator as we know that $\dfrac{{a + b}}{3} = \dfrac{a}{3} + \dfrac{b}{3}$
So we can write that $\dfrac{{2z - 8}}{3} = \dfrac{{2z}}{3} - \dfrac{8}{3}$
So we get $4 - \dfrac{{2z}}{3} + \dfrac{8}{3} = \dfrac{{2z}}{2} + \dfrac{5}{2}$
Now taking the terms which contain $z$ to one side of the equation because then it will be easier to solve the equation.
$\Rightarrow$ $4 + \dfrac{8}{3} - \dfrac{5}{2} = \dfrac{{2z}}{2} + \dfrac{{2z}}{3}$
Now we need to take the LCM of the denominators on both the sides which is the least common multiple of $2{\text{ and }}3$
If we have two prime numbers then their LCM is the multiplication of the two terms so here we get the LCM as $(3)(2) = 6$
So now we make the denominator as $6$ on both the sides.
$\Rightarrow$ $\dfrac{{4 \times 6}}{{1 \times 6}} + \dfrac{{8 \times 2}}{{3 \times 2}} - \dfrac{{5 \times 3}}{{2 \times 3}} = \dfrac{{2z \times 3}}{{2 \times 3}} + \dfrac{{2z \times 2}}{{3 \times 2}}$
Now we have made the denominator off the whole term as $6$
We have got
$\Rightarrow$ $\dfrac{{24 + 16 - 15}}{6} = \dfrac{{6z + 4z}}{6}$
So we know that if the denominators are same on both sides, then we can equate the numerators
So we get
$
  24 + 16 - 15 = 6z + 4z \\
\Rightarrow 25 = 10z \\
\Rightarrow 10z = 25 \\
\Rightarrow z = \dfrac{{25}}{{10}} = \dfrac{5}{2} \\
$
Hence we get that $z = \dfrac{5}{2}$

Note: Here a student can make mistakes in the calculations only because here only the terms with the variable need to be shifted from one side to the other. So one must be aware of it. We should know how we have to take LCM of the numbers. We must know if we are given that $\dfrac{{a + b}}{3} = \dfrac{{c + d}}{3}$ where the denominator are equal on both the sides then we can equate the numerators and we can say that $a + b = c + d$