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$\dfrac{3}{4}(x - 1) = x - 3$

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Hint: The given equation is a degree one polynomial equation. The solution of these equations is given by using simple arithmetic operations.

Given equation is

$ \Rightarrow \dfrac{3}{4}(x - 1) = x - 3$

Multiplying both sides of the equation by 4, we get

$

\Rightarrow 4 \times \dfrac{3}{4}(x - 1) = 4 \times (x - 3) \\

\Rightarrow 3(x - 1) = 4(x - 3) \\

$

Solving the above equation for the value of x, we get

$

\Rightarrow 3x - 3 = 4x - 12 \\

\Rightarrow 4x - 3x = 12 - 3 \\

\Rightarrow x = 9 \\

$

Hence the value of x is $9$ .

Note: Polynomial equations of degree one can be easily solved by simple arithmetic operations while polynomials of degree greater than one are little bit difficult to solve. To solve polynomials of degree greater than one arithmetic operations and factorization methods are required. By factorising we break the higher degree polynomial in multiples of lower degree polynomial.

Given equation is

$ \Rightarrow \dfrac{3}{4}(x - 1) = x - 3$

Multiplying both sides of the equation by 4, we get

$

\Rightarrow 4 \times \dfrac{3}{4}(x - 1) = 4 \times (x - 3) \\

\Rightarrow 3(x - 1) = 4(x - 3) \\

$

Solving the above equation for the value of x, we get

$

\Rightarrow 3x - 3 = 4x - 12 \\

\Rightarrow 4x - 3x = 12 - 3 \\

\Rightarrow x = 9 \\

$

Hence the value of x is $9$ .

Note: Polynomial equations of degree one can be easily solved by simple arithmetic operations while polynomials of degree greater than one are little bit difficult to solve. To solve polynomials of degree greater than one arithmetic operations and factorization methods are required. By factorising we break the higher degree polynomial in multiples of lower degree polynomial.