Question

(i) Solve for the value of x from the following equation $0.12x+\left( \dfrac{0.5+x}{2} \right)=\dfrac{x}{3}+1.5$.
(ii) Solve for the value of z from the following equation $z+6.1=\dfrac{0.5\left( z-0.4 \right)}{0.35}-\dfrac{0.6\left( z-6.63 \right)}{0.42}$.

Answer 1

Hint: We start solving the problem (i) by applying the result $\dfrac{\left( a+b \right)}{2}=\dfrac{a}{2}+\dfrac{b}{2}$ in the given equation. We then make necessary arrangements to bring the x terms on one side and constants on another side. We then make the necessary calculations to get the desired value of x. We start solving the problem (ii) by applying the result $\dfrac{a}{c}-\dfrac{b}{c}=\dfrac{\left( a-b \right)}{c}$ in the given equation. We then make necessary arrangements to bring the z terms on one side and constants on another side. We then make the necessary calculations to get the desired value of z.

Complete step-by-step answer:
(i) According to the problem, we need to find the value of x from the equation $0.12x+\left( \dfrac{0.5+x}{2} \right)=\dfrac{x}{3}+1.5$. Let us make calculations and bring x terms on one side and constants on the other side.
So, we have $0.12x+\left( \dfrac{0.5+x}{2} \right)=\dfrac{x}{3}+1.5$ ---(1).
We know that $\dfrac{\left( a+b \right)}{2}=\dfrac{a}{2}+\dfrac{b}{2}$. We use this result in equation (1).
$\Rightarrow 0.12x+\left( \dfrac{0.5}{2}+\dfrac{x}{2} \right)=\dfrac{x}{3}+1.5$.
$\Rightarrow 0.12x+0.25+0.5x=0.33x+1.5$.
$\Rightarrow 0.12x+0.5x-0.33x=1.5-0.25$.
$\Rightarrow 0.29x=1.25$.
$\Rightarrow x=\dfrac{1.25}{0.29}$.
$\Rightarrow x=\dfrac{125}{29}$.
So, we have found the value of x as $\dfrac{125}{29}$.
∴ The value of x is $\dfrac{125}{29}$.
(ii) According to the problem, we need to find the value of z from the equation $z+6.1=\dfrac{0.5\left( z-0.4 \right)}{0.35}-\dfrac{0.6\left( z-6.63 \right)}{0.42}$. Let us make calculations and bring z terms on one side and constants on the other side.
So, we have $z+6.1=\dfrac{0.5\left( z-0.4 \right)}{0.35}-\dfrac{0.6\left( z-6.63 \right)}{0.42}$.
$\Rightarrow z+6.1=\dfrac{\left( z-0.4 \right)}{0.7}-\dfrac{\left( z-6.63 \right)}{0.7}$ ---(2).
We know that $\dfrac{a}{c}-\dfrac{b}{c}=\dfrac{\left( a-b \right)}{c}$. We use this result in equation (2).
$\Rightarrow z+6.1=\dfrac{\left( z-0.4 \right)-\left( z-6.63 \right)}{0.7}$.
$\Rightarrow z+6.1=\dfrac{z-0.4-z+6.63}{0.7}$.
$\Rightarrow z+6.1=\dfrac{6.33}{0.7}$.
$\Rightarrow z+6.1=9.04$.
$\Rightarrow z=9.04-6.1$.
$\Rightarrow z=2.94$.
So, we have found the value of z as 2.94.
∴ The value of z is 2.94.

Note: We can also use trial and error methods in order to solve these problems. In this problem, we have taken the values up to two decimal places as the constants given in the problem are with two decimal places. We should not make mistakes while doing calculations and approximations. We can verify the obtained answer by substituting it in the original equation. We can also expect problems to find the value of variables which contain quadratic equations.