Question

y/4-1/3=y/5+2/7 solve the following equation​

Answer 1

Hey there!

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Answer:

→ $\sf{\dfrac{y}{4} - \dfrac{1}{3} = \dfrac{y}{5} + \dfrac{2}{7}}$

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Simplifying both sides, we get:

→ $\sf{\dfrac{1}{4}y + \dfrac{-1}{3} = \dfrac{1}{5}y + \dfrac{2}{7}}$

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Subtracting '1/5y' from both sides:

→ $\sf{\dfrac{1}{4}y + \dfrac{-1}{3} - \dfrac{1}{5}y = \dfrac{1}{5}y + \dfrac{2}{7} - \dfrac{1}{5}y}$

→ $\sf{\dfrac{1}{20}y + \dfrac{-1}{3} = \dfrac{2}{7}}$

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Adding '1/3' on both sides:

→ $\sf{\dfrac{1}{20}y + \dfrac{-1}{3} + \dfrac{1}{3} = \dfrac{2}{7} + \dfrac{1}{3}}$

→ $\sf{\dfrac{1}{20}y = \dfrac{13}{21}}$

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Multiplying '20' on both sides:

→ $\sf{20 \times \bigg(\dfrac{1}{20}y\bigg) = 20 \times \bigg(\dfrac{13}{21}\bigg)}$

→ ${\boxed{\bold{y = \dfrac{260}{21}}}}$

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Thanks!

Answer 2

Answer:

The solution of the equation $\frac{y}{4}-\frac{1}{3}=\frac{y}{5}+\frac{2}{7}$ is $y=\frac{260}{21}$.

Step-by-step explanation:

Consider the given equation as follows:

$\frac{y}{4}-\frac{1}{3}=\frac{y}{5}+\frac{2}{7}$ . . . . . (1)

Notice that the equation is in $y$ variable.

Step 1 of 4

Subtract $\frac{y}{5}$ from both sides in the equation (1) as follows:

$\frac{y}{4}-\frac{1}{3}-\frac{y}{5} =\frac{y}{5}+\frac{2}{7}-\frac{y}{5}$

Now, simplify as follows:

⇒ $\left(\frac{y}{4}-\frac{y}{5} \right) -\frac{1}{3}=\left(\frac{y}{5}-\frac{y}{5} \right) +\frac{2}{7}$

⇒ $\left(\frac{5y-4y}{20} \right) -\frac{1}{3}=\frac{2}{7}$ . . . . . (2)

Step 2 of 4

Add $\frac{1}{3}$ both the sides in the equation (2) as follows:

$\left(\frac{5y-4y}{20} \right) -\frac{1}{3}+\frac{1}{3}=\frac{2}{7}+\frac{1}{3}$

Now, simplify as follows:

⇒ $\left(\frac{5y-4y}{20} \right) =\frac{6+7}{21}$

⇒ $\frac{5y-4y}{20} =\frac{13}{21}$ . . . . . . (3)

Step 3 of 4

Cross multiply the equation (3) as follows:

⇒ $21(5y-4y)=20(13)$

⇒ $105y-84y=260$

Step 4 of 4

Simplify to find the value of $y$ as follows:

⇒ $21y=260$

⇒ $y=\frac{260}{21}$

Therefore, the solution of the equation $\frac{y}{4}-\frac{1}{3}=\frac{y}{5}+\frac{2}{7}$ is $y=\frac{260}{21}$.

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