Question

Solve the following quadratic equations by factorization:
3(7x+1/5x-3)-4(5x-3/7x+1)=11;x≠3/5,-(1/7)

Answer 1

x = 1   or   x = 0

Step-by-step explanation:

Given;

$3(\frac{7x+1}{5x-3})-4(\frac{5x-3}{7x+1})=11$

We have to solve for x.

Solution,

Let $\frac{7x+1}{5x-3}=y$

So we can say that ;

$3y-\frac{4}{y}=11$

now we take the L.C.M. and get;

$\frac{3y^2-4}{y}=11$

On using cross multiplication, we get;

$3y^2-4=11y$

Or we can rewrite the equation as;

$3y^2-11y-4=0$

Now we have to factorize the equation.

For this we have to split the second term to get the product of first term and third term.

$3y^2-12y+y-4=0\\\\3y(y-4)+1(y-4)=0\\\\(3y+1)(y-4)=0$

Now we can say that;

$3y+1=0\\\\3y=-1\\\\y=-\frac{1}{3}$

Again,

$y-4=0\\\\y=4$

Now we substitute the value of 'y' to get the value of 'x'.

$\frac{7x+1}{5x-3}=-\frac{1}{3}$

On using cross multiplication, we get;

$3(7x+1)=-1(5x-3)\\\\21x+3=-5x+3$

On combining the like terms, we get;

$21x+5x=3-3\\\\26x=0\\\\\therefore x=0$

Again we substitute the value of 'y' to get the value of 'x'.

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

On using cross multiplication, we get;

$7x+1=4(5x-3)\\\\7x+1=20x-12$

On combining the like terms, we get;

$20x-7x=12+1\\\\13x=13\\\\x=1$

Hence The value of x is 0 and 1.