How do you solve $\frac{6 y}{y^{2} - 25} + \frac{9}{y - 5} = \frac{5}{y + 5}$ $\frac { 6y } { y ^ { 2} - 25} + \frac { 9} { y - 5} = \frac { 5} { y + 5}$ ?

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How do you solve $\frac{6 y}{y^{2} - 25} + \frac{9}{y - 5} = \frac{5}{y + 5}$ $\frac { 6y } { y ^ { 2} - 25} + \frac { 9} { y - 5} = \frac { 5} { y + 5}$ ?

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Answer 1 · 2017-10-19T18:46:58

$y = - 7$ $y=-7$

Explanation:

First, we should try to get the common denominator for all of the fractions.

We should know that $(y - 5) (y + 5) = y^{2} - 25$ $(y-5)(y+5)=y^2-25$ . Therefore, we multiply the second fraction $(\frac{9}{y - 5})$ $(9/(y-5))$ by $(y + 5)$ $(y+5)$ to get $(\frac{9 y + 45}{y^{2} - 25})$ $((9y+45)/(y^2-25))$ . The third fraction should be multiplied by $y - 5$ $y-5$ to make the denominator all the same, so we get: $5 y - 25$ $5y- 25$ .

Since all of the denominators are the same, we can ignore them for now to get....

$6 y + 9 y + 45 = 5 y - 25$ $6y+9y+45=5y-25$ .

We subtract $5 y$ $5y$ on both sides and subtract $45$ $45$ (in order to find what y is) and combine like terms...

$10 y = - 70$ $10y=-70$ .

Divide 10 on both sides to get...

$y = - 7$ $y=-7$

You can plug it in if you want to, and you should get...

$- \frac{30}{12} = - \frac{5}{2}$ $-30/12=-5/2$ .

If you need me to go through some details, just ask me in the comments.

-Sakuya

Answer 2 · 2017-10-19T19:45:56

$y = - 7$ $y=-7$

Explanation:

Notice that the first denominator is $y^{2} - 25$ $y^2-25$

This is the same as $y^{2} - 5^{2} \to (y - 5) (y + 5)$ $y^2-5^2->(y-5)(y+5)$ so this is our starting point.

$\frac{6 y}{(y - 5) (y + 5)} + \frac{9}{y - 5} = \frac{5}{y + 5}$ $(6y)/((y-5)(y+5))+9/(y-5)=5/(y+5)$

$[\frac{6 y}{(y - 5) (y + 5)}] + [\frac{9}{y - 5}] = [\frac{5}{y + 5}]$ $color(green)([(6y)/((y-5)(y+5))]+[9/(y-5)]=[5/(y+5)]$

Multiply by 1 and you do not change the value. However, 1 comes in many forms.

$[\frac{6 y}{(y - 5) (y + 5)}] + dd [\frac{9}{y - 5} \times 1] dd = dd [\frac{5}{y + 5} \times 1]$ $color(green)([(6y)/((y-5)(y+5))]+color(white)("dd")[9/(y-5)color(red)(xx1)]color(white)("dd")=color(white)("dd")[5/(y+5)color(red)(xx1)]$

$[\frac{6 y}{(y - 5) (y + 5)}] + [\frac{9}{y - 5} \times \frac{y + 5}{y + 5}] = [\frac{5}{y + 5} \times \frac{y - 5}{y - 5}]$ $color(green)([(6y)/((y-5)(y+5))]+[9/(y-5)color(red)(xx(y+5)/(y+5))]=[5/(y+5)color(red)(xx(y-5)/(y-5))]$

$\frac{6 y}{(y - 5) (y + 5)} dd + d \frac{9 y + 45}{(y - 5) (y + 5)} d = d \frac{5 y - 25}{(y - 5) (y + 5)}$ $(6y)/((y-5)(y+5))color(white)("dd")+color(white)("d") (9y+45)/((y-5)(y+5) )color(white)("d")=color(white)("d") (5y-25)/((y-5)(y+5)$

Now that the denominators are all the same we can just forget about them. Or, if you wish to be a purist you can say: multiply all of both sides by $(y - 5) (y + 5)$ $(y-5)(y+5)$

$6 y + 9 y + 45 = 5 y - 25$ $6y+9y+45=5y-25$

Collecting like terms we have:

$6 y + 9 y - 5 y = - 25 - 45$ $6y+9y-5y=-25-45$

$10 y = - 70$ $10y=-70$

divide both sides by 10

$y = - 7$ $y=-7$

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How do you solve $\frac{6 y}{y^{2} - 25} + \frac{9}{y - 5} = \frac{5}{y + 5}$ $\frac { 6y } { y ^ { 2} - 25} + \frac { 9} { y - 5} = \frac { 5} { y + 5}$ ?

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How do you solve 6yy2−25+9y−5=5y+5\frac { 6y } { y ^ { 2} - 25} + \frac { 9} { y - 5} = \frac { 5} { y + 5}?

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How do you solve $\frac{6 y}{y^{2} - 25} + \frac{9}{y - 5} = \frac{5}{y + 5}$ $\frac { 6y } { y ^ { 2} - 25} + \frac { 9} { y - 5} = \frac { 5} { y + 5}$ ?