Question

Solve the given equation $0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)$ \[\]

Answer 1

Hint: We open the bracket on both sides and use the distributive law of multiplication over subtraction to simplify. We add and subtract suitable values so that we can collect variable terms at one side of the equation and the constant terms at the other side. We divide the coefficient of the variable to find the value of the variable.

Complete step-by-step solution:
We know from algebra that the linear equation in one variable $x$ is given by
\[ax=b\]
Here $a$ and $b$ are real numbers called constant of the equation where $a$ cannot be zero. The side left to the sign of equality ${}^{'}{{=}^{'}}$ is called the left hand side and side right hand side of the equation. We also know that if we add, subtract, multiply or divide the same number on both sides of the equation, equality holds. It is called balancing the equation. It means for some real number $c$ we have,
\[\begin{align}
  & ax+c=b+c \\
 & ax-c=b-c \\
 & ax\times x=b\times c \\
 & \dfrac{ax}{c}=\dfrac{b}{c} \\
\end{align}\]
When we are asked to solve an equation we find the values of unknown variables $x$. If we are given variable terms on both sides for example $ax+b=cx+d$ then we collect the like terms (variable terms and constant terms ) at two different sides.
We are asked to solve both
\[0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)\]
We see that the variable is $f$ and the variable terms are at both sides . So we follow the BODMAS rule and open the bracket using the distributive law of multiplication over subtraction that is $a\left( b-c \right)=a\times b-a\times c$. We proceed to have
\[\begin{align}
  & 0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right) \\
 & \Rightarrow 0.25\times 4f-0.25\times 3=0.05\times 10f-0.05\times 9 \\
 & \Rightarrow 1\times f-0.75=0.5f+.45 \\
 & \Rightarrow f-0.75=0.5f+0.45 \\
\end{align}\]
We subtract $0.5f$both sides of above step to have;
\[\begin{align}
  & \Rightarrow f-0.5f-0.75=0.5f-0.5f-0.45 \\
 & \Rightarrow 0.5f-0.75=0.45 \\
\end{align}\]
We subtract $0.75$both sides of above step to have;
\[\begin{align}
  & 0.5f-0.75+0.75=0.75-.45 \\
 & \Rightarrow 0.5f=.30 \\
\end{align}\]
We divide both sides of above step by the coefficient $0.5$ to have;
\[\Rightarrow f=\dfrac{0.30}{0.5}=0.6\]
So the solution of the given equation is 6.

Note: We note that knowledge of decimal arithmetic operation is required to solve this equation. We should be careful during the decimal when placing the decimal point. If the dividend has $n$ and the divisor has $m$ significant digits after decimal point the quotient will have $n-m$ digits after decimal point. The constant that is multiplied with the variable is called the coefficient of that variable.