Question

Solve the given equation
Only stars/mod
$\large \sf \dfrac{4t}{t^{2} - 25} = \dfrac{1}{5 - t}$

Answer 1

Given

• $\large \sf \dfrac{4t}{t^{2} - 25} = \dfrac{1}{5 - t}$

Solution

$= > \frac{4t}{ {t}^{2} - 25} = \frac{1}{5 - t} \\ cross \: multiply \: \\ = > 4t(5 - t) = {t}^{2} - 25 \\ = > 20t - 4 {t}^{2} = {t}^{2} - 25 \\ = > 20t + 25 = {t}^{2} + 4 {t}^{2} \\ = > 5 {t}^{2} = 20t + 25 \: \: \: \: \: \: \: \\ = > 5 {t}^{2} - 20t - 25 = 0 \\ = > 5( {t}^{2 } - 4t - 5) = 0 \\= > {t}^{2} - 4t - 5 = 0 \: \: \: \: \: \: \: \\ = > {t}^{2} - 5t + t - 5 = 0 \\ = > t(t - 5) + 1(t - 5) = 0 \\ = > (t - 5)(t + 1) = 0 \: \\ = > t = 5 \: or \: t = - 1$

•°• t = 5 or -1

Answer 2

Step-by-step explanation:

Given:-

4t/(t^2-25) = 1/(5-t)

On applying cross multiplication then

=>4t(5-t)=1(t^2-25)

=>20t-4t^2 = t^2-25

=>t^2-25+4t^2-20t = 0

=>5t^2 -20t -25 = 0

=>5(t^2-4t-5)=0

=>t^2-4t-5 = 0×5

=>t^2-4t-5=0

=>t^2+t-5t -5=0

=>t(t+1)-5(t+1)=0

=>(t+1)(t-5)=0

=>t+1 = 0 or t-5 =0

=>t = -1 and t=5

The values of t are -1 and 5

(or)

Given that

4t/(t^2-25) = 1/(5-t)

It can be written as

4t/(t^2-25) = -1/(t-5)

=>4t/[t^2-5^2] = -1/(t-5)

=>4t/[(t+5)(t-5)] = -1/(t-5)

On cancelling (t-5) both sides then

=>4t/(t+5)=-1

==>4t=-1(t+5)

=>4t = -t-5

=>4t+t = -5

=>5t = -5

=>t= -5/5

=>t = -1

Answer :-

The value of t for the given problem is -1 and 5

Check:-

If t = 5 then

LHS: 4t/(t^2-25)

=>4(5)/(5^2-25)

=>20/(25-25)

=>20/0

=>not defined

RHS:

-1/(t-5)

=>-1/(5-5)

=>-1/0

Not defined

LHS = RHS

If t = -1 then

LHS :4t/(t^2-25)

=>4(-1)/[(-1)^2-25)

=>-4/(1-25)

=>-4/-24

=>1/6

RHS:

-1/(t-5)

=>-1/[(-1-5)

=>-1/-6

=>1/6

LHS = RHS is true for x= -1