Question

solve:
$\frac{2x - 3}{3} = 1 - \frac{2}{3}$
$\frac{y}{9} - \frac{y}{12} = \frac{1}{103}$
$2y + 3 = 5y + 7$

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Answer 1

Solutions :-

1)

Given that (2x-3)/3 = 1 - (2/3)

=> (2x-3)/3 = (3-2)/3

=> (2x-3)/3 = 1/3

On cancelling 3 both sides then

=> 2x-3 = 1

=> 2x = 1+3

=> 2x = 4

=> x = 4/2

=> x = 2

Therefore, The value of x = 2

2)

Given that (y/9)-(y/12) = 1/103

LCM of 9 and 12 = 36

=> [(4×y)-(3×y)]/36 = 1/103

=> (4y-3y)/36 = 1/103

=> y/36 = 1/103

=> y = 36/103

Therefore, The value of y = 36/103

3)

Given that 2y+3 = 5y+7

=> 2y-5y = 7-3

=> -3y = 4

=> y = 4/-3

=> y = -4/3

Therefore, y = -4/3

Answers :-

1) x = 2

2) y = 36/103

3) y = -4/3

Check :-

1)If x = 2 then LHS becomes

[2(2)-3]/3

= (4-3)/3

= 1/3

and RHS = 1-(2/3) = (3-2)/3 = 1/3

LHS = RHS is true for x = 2

2)If y = 36/103 then LHS becomes

(y/9)-(y/12)

= [(36/103)/9]-[(36/103)/12)]

= [36-(103×9)]-[36-(103×12)]

= [(36×4)-(36×3)]/(36×103)

= [36(4-3)]/(36×103)

= 36/(36×103)

=1/103)

= RHS

LHS = RHS is true for y = 36/103

3)If y = -4/3 then LHS becomes 2y+3

2(-4/3)+3

= (-8/3)+3

= (-8+9)/3

= 1/3

and RHS becomes 5y+7

= 5(-4/3)+7

= (-20/3)+7

= (-20+21)/3

= 1/3

LHS = RHS is true for y = -4/3

Verified the given relations in the given problems.

Answer 2

SOLUTIONS :-

$\bf 1. \: \frac{2x - 3}{3} = 1 - \frac{2}{3}$

$\rm⇢ \: \: \frac{2x - 3}{3} = \frac{1}{3}$

$\rm⇢ \: \: 2x - 3 = 1$

$\rm⇢ \: \: 2x = 4$

$\rm⇢ \: \: x = 4 \div 2 = 2$

value of x is 2 .

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$\bf2. \: \: \frac{y}{9} - \frac{y}{12} = \frac{1}{103}$

$\rm⇢ \: \: \frac{12y - 9y}{9 \times 12} = \frac{1}{103}$

$\rm⇢ \: \: \frac{3y}{9 \times 12} = \frac{1}{103}$

$\rm⇢ \: \: \frac{y}{36} = \frac{1}{103}$

$\rm⇢ \: \: 103y = 36$

$\rm⇢ \: \: y = \frac{36}{103}$

value of y is 36/103 .

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$\bf3. \: \: 2y + 3 = 5y + 7$

$\rm⇢ \: \: 5y - 2y = 3 - 7$

$\rm⇢ \: \: 3y = - 4$

$\rm⇢ \: \: y = \frac{ - 4}{3}$

value of y is -4/3 .