Answer

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**Hint:**At first, we will convert both of the equations by removing the fraction coefficients. Then we put the value of the $x$ from the second equation in the first equation. We will get the constant value of the variable $y$ from this. Lastly, we will put the value $y$ in the second equation and get the solution.

**Complete step by step answer:**

We have given $\dfrac{1}{2}x + \dfrac{1}{3}y = 5$ ……. $\left( 1 \right)$

And $\dfrac{1}{4}x + y = 10$ …… $\left( 2 \right)$

Multiplying $6$ in both side of the equation $\left( 1 \right)$ we get;

$3x + 2y = 30$ ….. $\left( 3 \right)$

Multiplying $4$ in both side of the equation $\left( 2 \right)$ we get;

$x + 4y = 40$ ….. $\left( 4 \right)$

Subtracting $4\;y$ from both side of the equation $\left( 4 \right)$ we get;

$\Rightarrow x = 40 - 4y$

Putting this value $x$ in the equation $\left( 3 \right)$ we get;

$\Rightarrow 3\left( {40 - 4y} \right) + 2y = 30$

Simplifying the above equation we get;

$\Rightarrow 120 - 12y + 2y = 30$

Simplifying both sides of the equation we get;

$\Rightarrow 120 - 10y = 30$

Subtracting $\;120$ from both sides of the equation we get;

$\Rightarrow - 10y = - 90$

Dividing with $- 10$ both side of the equation we get;

$\Rightarrow y = 9$

Putting this value $y$ in the equation $\left( 4 \right)$ we get;

$x + 4 \cdot 9 = 40$

After multiplication on the left-hand side we get;

$\Rightarrow x + 36 = 40$

Subtracting $\;36$ from both sides of the equation we get;

$\Rightarrow x = 4$

**So our required solution by substitution is $x = 4$ And $y = 9$ .**

**Note:**To avoid the complicated addition and subtraction of fractions we will try to remove the fractions by multiplication with the least common multiple of the denominators of the fractions. The addition and subtraction will be easier in natural numbers than fractions.

Alternative Method:

We have given $\dfrac{1}{2}x + \dfrac{1}{3}y = 5$ ……. $\left( 1 \right)$

And $\dfrac{1}{4}x + y = 10$ …… $\left( 2 \right)$

From the equation $\left( 2 \right)$ we get;

$\Rightarrow y = 10 - \dfrac{1}{4}x$

Putting this value of the variable $y$ in the equation $\left( 1 \right)$ we get;

$\dfrac{1}{2}x + \dfrac{1}{3}\left( {10 - \dfrac{1}{4}x} \right) = 5$

After multiplication in the left-hand side of the equation we get;

$\Rightarrow \dfrac{1}{2}x + \dfrac{{10}}{3} - \dfrac{1}{{12}}x = 5$

Simplifying the above equation we get;

$\Rightarrow \dfrac{{10}}{3} + \dfrac{5}{{12}}x = 5$

Subtracting $\dfrac{{10}}{3}$ from both side of the equation we get;

$\Rightarrow \dfrac{5}{{12}}x = \dfrac{5}{3}$

Dividing both sides of the equation with $\dfrac{5}{{12}}$ we get;

$\Rightarrow x = 4$

Put this value in the equation $\left( 2 \right)$ we get;

$\Rightarrow y = 10 - \dfrac{1}{4} \cdot 4$

Simplifying the above equation we get;

$\Rightarrow y = 9$

So our required solution by substitution is $x = 4$

And $y = 9$ .

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