Question

find (x+y) ÷ (x-y) if x=-6/5 y=8/9​

Answer 1

Answer:

47/7

Step-by-step explanation:

see the attached image for the detailed answer

Answer 2

★ How to do :-

Here, we are given with two variables namely x and y. We are also given with the value of those variables x and y. Also, we are given with a statement in which we should apply the values of those x and y and then simplify it. We can see that there are two brackets given to us. So, the first method we should always do is to solve the brackets. Later, we can divide the fraction. Note that the final answer is written in the lowest form always. So, let's solve!!

$\:$

➤ Solution :-

${\tt \leadsto (x + y) ÷ (x - y)}$

Substitute the given values.

${\tt \leadsto \bigg( \dfrac{(-6)}{5} + \dfrac{8}{9} \bigg) \div \bigg( \dfrac{(-6)}{5} - \dfrac{8}{9} \bigg) }$

Let's solve the first bracket.

${\tt \leadsto \dfrac{(-6)}{5} + \dfrac{8}{9}}$

LCM of 5 and 9 is 45.

${\tt \leadsto \dfrac{(-6) \times 9}{5 \times 9} + \dfrac{8 \times 5}{9 \times 5}}$

Multiply the numbers in numerator and denominator.

${\tt \leadsto \dfrac{(-54)}{45} + \dfrac{40}{45}}$

Add the values now.

${\tt \leadsto \dfrac{(-54) + 40}{45} + \dfrac{(-14)}{45}}$

$\:$

Let's solve the second bracket now.

${\tt \leadsto \dfrac{(-6)}{5} - \dfrac{8}{9}}$

LCM of 5 and 9 is 45.

${\tt \leadsto \dfrac{(-6) \times 9}{5 \times 9} - \dfrac{8 \times 5}{9 \times 5}}$

Multiply the numbers in numerator and denominator.

${\tt \leadsto \dfrac{(-54)}{45} - \dfrac{40}{45}}$

Subtract the values now.

${\tt \leadsto \dfrac{(-54) - 40}{45} + \dfrac{(-94)}{45}}$

$\:$

Now, insert the obtained values in their places where they were in first statement.

${\tt \leadsto \dfrac{(-14)}{45} \div \dfrac{(-64)}{45}}$

Take the reciprocal of second fraction and multiply both the fractions.

${\tt \leadsto \dfrac{(-14)}{45} \times \dfrac{45}{(-64)}}$

Multiply the numerator with numerator and the denominator with denominator.

${\tt \leadsto \dfrac{(-14) \times 45}{45\times (-64)} = \dfrac{(-630)}{(-2800)}}$

Qrite the obtained fraction in lowest form to get the answer.

${\tt \leadsto \cancel \dfrac{(-630)}{(-2800)} = \pink{\underline{\boxed{\tt \dfrac{63}{280}}}}}$

$\Huge\therefore$ The answer when it's simplified is ${\tt \dfrac{63}{280}}$