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Answer 1

Hey there!

Here's my solution over this linear equation:

$\boxed{\bf{\dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5 \\ \\ \dfrac{5y - 7}{2} + \dfrac{4x - 3}{6} = 18 - 5x}}$

Solve for variable "x" in this expression $\bf{\dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5}$.

(Least common multiplier is: 20, in 5 and 4); So:

$\bf{\dfrac{7 + x}{5} \times 20 - \dfrac{2x - y}{4} \times 20 = 3y \times 20 - 5 \times 20}$

$\bf{4(7 + x) - 5(2x - y) = 60y - 100}$

$\bf{28 + 4x - 10x + 5y = 60y - 100}$

Add same elements; subtract the variable value of "5y" from both the sides, So this becomes:

$\bf{28 - 6x = 55y - 100}$

Subtract the value of "28" from both sides:

$\bf{-6x = 55y - 128}$

Divide both of the sides by a negative value of "6":

$\bf{x = \dfrac{55y}{-6} - \dfrac{128}{-6}}$

$\bf{x = - \dfrac{55y - 128}{6}}$

Now, you need to substitute the current expression of variable "x" into the second equation or the second expression that is,

$\bf{\dfrac{5y - 7}{2} + \dfrac{4 \Bigg(- \dfrac{55y - 128}{6} \Bigg) - 3}{6} = 18 - 5 \Bigg(- \dfrac{55y - 128}{6} \Bigg)}$

Solve for variable "y":

Subtract (After removing parentheses and cancelling the common factors) by "5 (55y - 128)/6" from both the sides:

$\bf{\dfrac{5y - 7}{2} + \dfrac{- \dfrac{2 (55y - 128)}{3} - 3}{6} - \dfrac{5(55y - 128)}{6}}$

$\bf{= 18 + \dfrac{5(55y - 128)}{6} - \dfrac{5(55y - 128)}{6}}$

$\bf{- \dfrac{445y + 1052}{9} = 18}$

Multiply by the value of "9", from both the sides:

$\bf{-445y + 1052 = 162}$

Subtract by the value of "1052".

$\bf{-445y = - 890}$

Divide by the value of "-445":

$\boxed{\bf{\therefore \quad y = 2}}$

Substitute the value of variable "x" into our original expression of variable "x", in order to obtain the final solution for this set of linear equation via substitution method.

$\bf{\therefore \quad x = - \dfrac{55 \times 2 - 128}{6}}$

$\bf{x = - \dfrac{110 - 128}{6}}$

$\bf{x = - \dfrac{- 18}{6}}$

$\bf{x = -(-3)}$

$\boxed{\bf{\therefore \quad x = 3}}$

Therefore, the final solutions to this particular system of linear equation with the method of substitution is going to be:

$\boxed{\bf{\underline{\therefore Required \: Solutions \: to \: this \: set \: of \: Linear \: Equations \: is: \: x =3; \quad y = 2}}}$

Which is the required solution to this type of query.

Hope this helps you and clears your doubts for finding the solutions for a given system of linear equations!!!!!!