Question

please help if you can and maybe explain ty :)​​

Answer 1

$\large\sf\underline{Given\::}$

• $\sf\:\frac{2}{3} + \frac{}{6}=1\frac{1}{6}$

‎

$\large\sf\underline{To\:find\::}$

• Find the value of missing numerator ?

‎

$\large\sf\underline{Solution\::}$

$\sf\:\frac{2}{3} + \frac{}{6}=1\frac{1}{6}$

Let the missing numerator be x

$\sf\implies\:\frac{2}{3} + \frac{x}{6}=1\frac{1}{6}$

• LCM of 3 and 6 = 6

$\sf\implies\: \frac{(2 \times 2) + (1 \times x) }{6}=\frac{7}{6}$

$\sf\implies\: \frac{4 + x}{6}=\frac{7}{6}$

• Cross multiplying

$\sf\implies\: 6(4 + x) =7 \times 6$

• Multiplying the numbers

$\sf\implies\: 24 +6x=42$

• Transposing +24 to RHS it becomes -24

$\sf\implies\:6x=42-24$

$\sf\implies\:6x=18$

• Transposing 6 to RHS it goes to the denominator

$\sf\implies\:x=\frac{18}{6}$

• Reducing it to lower terms

$\sf\implies\:x=\cancel{\frac{18}{6}}$

$\small{\underline{\boxed{\mathrm\red{\implies\:x\:=\:3}}}}$ ★

‎

‎$\large\sf\underline{Verifying\::}$

Let check if our answers are correct.

We would check it by substituting the value of x as 3 in the given equation. Doing so if we get LHS = RHS, our answer would be correct.

$\sf\:\frac{2}{3} + \frac{x}{6}=1\frac{1}{6}$

• Substituting x as 6

$\sf\implies\:\frac{2}{3} + \frac{3}{6}=\frac{7}{6}$

• LCM of 3 and 6 = 6

$\sf\implies\: \frac{(2 \times 2) + (1 \times 3)}{6}=\frac{7}{6}$

$\sf\implies\: \frac{4 + 3}{6}=\frac{7}{6}$

$\sf\implies\: \frac{7}{6}=\frac{7}{6}$

$\bf\implies\:LHS~=~RHS$

$\small\fbox\green{Hence~Verified~!! }$

========================

$\dag\:\underline{\sf So\:the\:required\:value\:of\:numerator\:is\:3.}$‎

!! Hope it helps !!‎

Answer 2

Answer:

\large\sf\underline{Given\::}

Given:

\sf\:\frac{2}{3} + \frac{}{6}=1\frac{1}{6}

3

2

+

6

=1

6

1

‎

\large\sf\underline{To\:find\::}

Tofind:

Find the value of missing numerator ?

‎

\large\sf\underline{Solution\::}

Solution:

\sf\:\frac{2}{3} + \frac{}{6}=1\frac{1}{6}

3

2

+

6

=1

6

1

Let the missing numerator be x

\sf\implies\:\frac{2}{3} + \frac{x}{6}=1\frac{1}{6}⟹

3

2

+

6

x

=1

6

1

LCM of 3 and 6 = 6

\sf\implies\: \frac{(2 \times 2) + (1 \times x) }{6}=\frac{7}{6}⟹

6

(2×2)+(1×x)

=

6

7

\sf\implies\: \frac{4 + x}{6}=\frac{7}{6}⟹

6

4+x

=

6

7

Cross multiplying

\sf\implies\: 6(4 + x) =7 \times 6⟹6(4+x)=7×6

Multiplying the numbers

\sf\implies\: 24 +6x=42⟹24+6x=42

Transposing +24 to RHS it becomes -24

\sf\implies\:6x=42-24⟹6x=42−24

\sf\implies\:6x=18⟹6x=18

Transposing 6 to RHS it goes to the denominator

\sf\implies\:x=\frac{18}{6}⟹x=

6

18

Reducing it to lower terms

\sf\implies\:x=\cancel{\frac{18}{6}}⟹x=

6

18

\small{\underline{\boxed{\mathrm\red{\implies\:x\:=\:3}}}}

⟹x=3

★

‎

‎\large\sf\underline{Verifying\::}

Verifying:

Let check if our answers are correct.

We would check it by substituting the value of x as 3 in the given equation. Doing so if we get LHS = RHS, our answer would be correct.

\sf\:\frac{2}{3} + \frac{x}{6}=1\frac{1}{6}

3

2

+

6

x

=1

6

1

Substituting x as 6

\sf\implies\:\frac{2}{3} + \frac{3}{6}=\frac{7}{6}⟹

3

2

+

6

3

=

6

7

LCM of 3 and 6 = 6

\sf\implies\: \frac{(2 \times 2) + (1 \times 3)}{6}=\frac{7}{6}⟹

6

(2×2)+(1×3)

=

6

7

\sf\implies\: \frac{4 + 3}{6}=\frac{7}{6}⟹

6

4+3

=

6

7

\sf\implies\: \frac{7}{6}=\frac{7}{6}⟹

6

7

=

6

7

\bf\implies\:LHS~=~RHS⟹LHS = RHS

\small\fbox\green{Hence~Verified~!! }

Hence Verified !!

========================

\dag\:\underline{\sf So\:the\:required\:value\:of\:numerator\:is\:3.}†

Sotherequiredvalueofnumeratoris3.

‎

!! Hope it helps !!‎