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# Ex.2.5 Q10 Linear Equations in One Variable Solution - NCERT Maths Class 8

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## Question

Simplify and solve the linear equation \begin{align}0.25(4f - 3) = 0.05(10f - 9)\end{align}

Video Solution
Linear Equations
Ex 2.5 | Question 10

## Text Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

First open the brackets then transpose variable to one side and constant to another side.

Steps:

\begin{align}0.25\left( {4f - 3} \right)&= 0.05\left( {10f - 9} \right)\\\frac{1}{4}\left( {4f - 3} \right) &= \frac{1}{{20}}\left( {10f - 9} \right)\end{align}

Multiplying both sides by $$20$$ we obtain

\begin{align}5\left( {4f - 3} \right) &= 10f - 9\\20f - 15 &= 10f - 9\,\\\qquad\qquad\qquad\text{(Opening }& \text{the brackets)}\\\\20f - 10f &= - 9 + 15\\\quad \,\,\,\,\,\,\,10f &= 6\\\quad \,\,\,\,\,\,\,\,\,\,\,\,f &= \frac{3}{5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 0.6\end{align}

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