Question

solve the equation and check your result :. (a). 8x - 5 / 2x + 3 =1/2. (b). 9y -5 = 2y +3.​

Answer 1

1. $\dfrac{8x - 5}{2x + 3} = \dfrac{1}{2}$

Cross multiply the terms.

➾ 2(8x - 5) = 1(2x + 3)

Multiply the number outside (the bracket) with the numbers inside (the bracket)

➾ 16x - 10 = 2x + 3

Shift the terms

➾ 16x - 2x = 3 + 10

Calculate

➾ 14x = 13

Divide both the sides by 14, we get :

➾ x = $\boxed{\boxed{\dfrac{\tt{13}}{\tt{14}}}}$

Let's check the solution :-

$\dfrac{8x - 5}{2x + 3} = \dfrac{1}{2}$

By putting the value of x as $\dfrac{13}{14}$

➾ $\dfrac{8 \times {\dfrac{13}{14}} - 5}{2 \times {\dfrac{13}{14}} + 3} = \dfrac{1}{2}$

Reduce the numbers

➾ $\dfrac{4 \times {\dfrac{13}{7}} - 5}{\dfrac{13}{7} + 3}} = \dfrac{1}{2}$

Multiply and calculate

➾ $\dfrac{\dfrac{52}{7}-5}{\dfrac{34}{7}} = \dfrac{1}{2}$

Calculate

➾ $\dfrac{\dfrac{17}{7}}{\dfrac{34}{7}} = \dfrac{1}{2}$

Reduce the fraction by 7, we get

➾ $\dfrac{17}{34} = \dfrac{1}{2}$

Reduce the fraction by 17

➾ $\dfrac{1}{2} = \dfrac{1}{2}$

∴ LHS = RHS

2. 9y - 5 = 2y + 3

Shift the terms

➾ 9y - 2y = 3 + 5

Calculate

➾ 7y = 8

Divide both the sudes by 7, we get :

➾ $\boxed{\boxed{\tt{y\:=\:{\dfrac{8}{7}}}}}$

Let's check the solution :-

9y - 5 = 2y + 3

By putting the value of y as $\dfrac{8}{7}$

➾ 9 × $\dfrac{8}{7}$ - 5 = 2 × $\dfrac{8}{7}$ + 3

Multiply and Calculate

➾ $\dfrac{72}{7}$ - 5 = $\dfrac{16}{7}$ + 3

Multiply and calculate

➾ $\dfrac{37}{7} = \dfrac{37}{7}$

∴ LHS = RHS

Points to remember :-

1. While shifting the numbers from LHS to RHS and vice versa, :-

+ changes to -

- changes to +

× changes to ÷

÷ changes to ×

2. While cross multiplying, we must multiply numerator (in LHS) to the denominator (in RHS) and denominator (in LHS) to the numerator (in RHS).

Answer 2

Cross multiply the terms.

➾ 2(8x - 5) = 1(2x + 3)

Multiply the number outside (the bracket) with the numbers inside (the bracket)

➾ 16x - 10 = 2x + 3

Shift the terms

➾ 16x - 2x = 3 + 10

Calculate

➾ 14x = 13

Divide both the sides by 14, we get :

➾ x = \boxed{\boxed{\dfrac{\tt{13}}{\tt{14}}}}

14

13

Let's check the solution :-

\dfrac{8x - 5}{2x + 3} = \dfrac{1}{2}

2x+3

8x−5

=

2

1

By putting the value of x as \dfrac{13}{14}

14

13

➾ \dfrac{8 \times {\dfrac{13}{14}} - 5}{2 \times {\dfrac{13}{14}} + 3} = \dfrac{1}{2}

2×

14

13

+3

8×

14

13

−5

=

2

1

Reduce the numbers

➾ \dfrac{4 \times {\dfrac{13}{7}} - 5}{\dfrac{13}{7} + 3}} = \dfrac{1}{2}

Multiply and calculate

➾ \dfrac{\dfrac{52}{7}-5}{\dfrac{34}{7}} = \dfrac{1}{2}

7

34

7

52

−5

=

2

1

Calculate

➾ \dfrac{\dfrac{17}{7}}{\dfrac{34}{7}} = \dfrac{1}{2}

7

34

7

17

=

2

1

Reduce the fraction by 7, we get

➾ \dfrac{17}{34} = \dfrac{1}{2}

34

17

=

2

1

Reduce the fraction by 17

➾ \dfrac{1}{2} = \dfrac{1}{2}

2

1

=

2

1

∴ LHS = RHS

2. 9y - 5 = 2y + 3

Shift the terms

➾ 9y - 2y = 3 + 5

Calculate

➾ 7y = 8

Divide both the sudes by 7, we get :

➾ \boxed{\boxed{\tt{y\:=\:{\dfrac{8}{7}}}}}

y=

7

8

Let's check the solution :-

9y - 5 = 2y + 3

By putting the value of y as \dfrac{8}{7}

7

8

➾ 9 × \dfrac{8}{7}

7

8

- 5 = 2 × \dfrac{8}{7}

7

8

+ 3

Multiply and Calculate

➾ \dfrac{72}{7}

7

72

- 5 = \dfrac{16}{7}

7

16

+ 3

Multiply and calculate

➾ \dfrac{37}{7} = \dfrac{37}{7}

7

37

=

7

37

∴ LHS = RHS

Points to remember :-

1. While shifting the numbers from LHS to RHS and vice versa, :-

+ changes to -

- changes to +

× changes to ÷

÷ changes to ×

2. While cross multiplying, we must multiply numerator (in LHS) to the denominator (in RHS) and denominator (in LHS) to the numerator (in RHS).