Question

solve the the pair of linear equations by the elemination method 3y/2-5x/3 =-2 and y/3+x/3 = 13 / 16​

Answer 1

$\huge\purple{\boxed{\underline{ANSWER}}}$

GIVEN THAT:

$➨$ There are two liner equation

$➞ \: \: \frac{3y}{2} - \frac{5x}{3} = - 2 \: \: \: \: \: \: \: \: (1) \\ \\ ➞ \: \: \frac{y}{3} + \frac{x}{3} = \frac{13}{16} \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$

SOLUTIONS:

$➨$ Equation (1)

$➞ \: \: \frac{3y}{2} - \frac{5x}{3} = - 2 \\ \\ &#10142 \: \: \frac{9y - 10x}{6} = - 2 \\ \\ &#10142 \: \: 9y - 10x = - 12$

$➨$ Equation (2)

$➞ \: \: \frac{y}{3} + \frac{x}{3} = \frac{13}{16} \\ \\ &#10142 \: \: \frac{y + x}{3} = \frac{13}{16} \\ \\ &#10142 \: \: x + y = \frac{39}{16} \: \:$

$➨$ From Equation (2)

$➞ \: \: x = \frac{39}{16} - y$

$➨$ Now putting the value of x in equaton (1)

$&#10142 \: \: 9y - 10( \frac{39}{16} - y) = - 12 \\ \\ &#10142 \: \: 9y - \frac{390}{16} + 10y = - 12 \\ \\ &#10142 \: \: 19y = \frac{390}{16} - 12 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10142 \: \: 19y = \frac{390 - 192}{16} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10142 \: \: 19y = \cancel\frac{198}{16} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10142 \: \: \: y = \frac{99}{8 \times 19} = \frac{99}{152} \: \: \: \: \: \: \: \:$

$➨$ Now putting the value of y in equaton (2)

$&#10142 \: \: x = \frac{39}{16} - \frac{99}{152} \\ \\ &#10142 \: \: x = \frac{741 - 198}{304} \\ \\ &#10142 \: \: x = \frac{543}{304} \: \: \: \: \: \: \: \: \: \: \:$