Question

Solve By Elimination Method.<br />$\large{ \tt{and \: the \: given \: equation \: are =}} \\ \\ \boxed{\large \star \red{ \: \dfrac{1}{2} x + \dfrac{1}{3}y = 2} }\\ \\ \boxed{\large \star \red{ \: \dfrac{1}{3} x + \dfrac{1}{2}y = \dfrac{13}{6} }}$<br />$\underline{\large \star \: \{\tt \purple{Note:-} }$<br />● Class 10.<br />● Need Correct Answer.<br /><br />​

Answer 1

Given to solve the equations by elimination method :- ㅤㅤㅤㅤㅤ

$\tt \dfrac{1}{2} x \: + \dfrac{1}{3} y = 2$ -- eq 1

$\tt \dfrac{1}{3} x + \dfrac{1}{2} y = \dfrac{13}{6}$ -- eq2

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$\ \red \maltese \ \bf\red {R equired \: E xplanation...}$

Let's multiply the equation 1 with ½ on both sides and 2nd equation with ⅓ on both sides

$\tt \dfrac{1}{2} x \: + \dfrac{1}{3} y = 2$ -- eq 1

$\tt \dfrac{1}{2} x \times \dfrac{1}{2} \: + \dfrac{1}{3} y \times \dfrac{1}{2} = 2 \times \dfrac{1}{2}$

$\tt \dfrac{x}{2 \times 2} \: + \dfrac{y}{3 \times 2} = 1$

$\red{\tt \dfrac{x}{4} \: + \dfrac{y}{6} = 1} - - 3$

$\tt \dfrac{1}{3} x + \dfrac{1}{2} y = \dfrac{13}{6}$ -- eq2

Now multiplying the 2nd equation with ⅓ on both sides

$</p><p>\tt \dfrac{1}{3} x + \dfrac{1}{2} y = \dfrac{13}{6} </p><p>$

$</p><p>\tt \dfrac{1}{3} x \times \dfrac{1}{3} + \dfrac{1}{2} y \times \dfrac{1}{3} = \dfrac{13}{6} \times \dfrac{1}{3} </p><p>$

$</p><p>\tt \dfrac{x}{3 \times 3} + \dfrac{y}{2 \times 3} = \dfrac{13}{6 \times 3} </p><p>$

$</p><p> \tt \red{ \dfrac{x}{9} + \dfrac{y}{6} = \dfrac{13}{18} - - - 4}</p><p>$

Now we got the new equations We shall subtract the both equations.

eq 3 - eq 4

$\green{\tt \dfrac{x}{4} \: + \dfrac{y}{6} - \dfrac{x}{9} - \dfrac{y}{6} = 1 - \dfrac{13}{18} }$

$\green{\tt \dfrac{x}{4} \: - \dfrac{x}{9} = \dfrac{18- 13}{18} }$

$\green{\tt \dfrac{9x - 4x}{36} = \dfrac{5}{18} }$

$\green{\tt \dfrac{5x}{2} = 5}$

$\green{\tt {5x} = {2} \times 5}$

$\green{\tt {5x} = 10}$

$\green{\tt {x} = \dfrac{ 10}{5} }$

$\purple{\large\boxed{\star{x} = 2}}$

Substitute value of x in equation 3

$\red{\tt \dfrac{x}{4} \: + \dfrac{y}{6} = 1} - - 3$

$\green{\tt \dfrac{2}{4} \: + \dfrac{y}{6} = 1}$

$\green{\tt \dfrac{1}{2} \: + \dfrac{y}{6} = 1}$

$\green{\tt \: \dfrac{y}{6} = 1 - \dfrac{1}{2} }$

$\green{\tt \: \dfrac{y}{6} = \dfrac{1}{2} }$

$\green{\tt \: \dfrac{y}{3} = 1 }$

$\purple{\large\boxed { \star\: {y} = {3} }}$

So, the values of x, y are 2, 3 respectively.

Answer 2

Answer:

andthegivenequationare=

⋆

2

1

x+

3

1

y=2

⋆

3

1

x+

2

1

y=

6

13

<br />\underline{\large \star \: \{\tt \purple{Note:-} }

⋆{Note:−

<br />● Class 10.<br />● Need Correct Answer.<br /><br />

Step-by-step explanation:

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