Answer

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**Hint:**To solve the given question we must first know and understand the standard form of a linear equation in two variables. We must also know the conditions on the coefficients of the x and y terms and the constant.

**Complete step by step solution:**

Let us first understand what is meant by a linear equation. A linear equation is an algebraic equation in which all the variables are raised to the power of 1. The given equation in the question is a linear equation in two variables. We usually write a given linear equation (in two variables) in a standard form so that we can generalise some formulae according to this standard form. The standard form of a linear equation in two variables is given as,

$Ax+By=C$ ….. (i),

where A, B and C are integers and A, B and C are all co prime numbers. This means that A, B and C do not have any common factors other than one.

The given linear equation in the question is $0.25x=0.1+0.2y$.Let us first club the x and y terms on the left side of the equation and the constant term on the right hand side of the equation. With this we get the equation as,

$0.25x-0.2y=0.1$ …. (ii)

We can see decimal coefficients in the equation. Therefore, we have to convert the coefficients in integer form.

Now, multiply equation (ii) by 100 as the highest decimal place is two.

Therefore,

$25x-20y=10$

We know that the numbers 25, 20 and 10 have 5 as a common factor. Therefore, divide the above equation by 5.

$\dfrac{25x-20y}{5}=\dfrac{10}{5}$

$\Rightarrow 5x-4y=2$

Now, this equation satisfies the condition for the standard form of linear equation.

Therefore, the standard form of the given equation is $5x-4y=2$ …. (iii)

**Now, if we compare equations (i) and (iii) then we get that $A=5,B=-4,C=2$.**

**Note:**Other than saying that we shall club the x and y terms on the left hand side and the constant term on the right side of the equation, we can subtract 0.2y on both sides of the equation so that the y term on the right hand side becomes zero. Then we can simplify further as above.

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