Question

Solve for x.
a. x + 5 = 6/x
b. x = 24/x - 5
c. x = 1/x

Answer 1

A.x+5=6x

Step 1:subtract 6x from both sides.

x+5-6x=6x-6x

-5x+5=0

Step 2:subtract 5 from both sides.

-5x+5-5=0-5

-5x=-5

Step 3:divide both by -5

-5x/5=-5/5

X=1

Answer 2

a. For equation
`x+5=(6)/(x)` value of x are 1 and -6.

b. For equation
`x=(24)/(x)-5` value of x are 3 and -8

c. For equation
`x=(1)/(x)` value of x are -1 and 1.

Solution:

a)Given equation is
`x+5=(6)/(x)`

On cross multiplying we get,

x(x+5)=6

`x^(2)+5 x-6=0`

On splitting the middle term we get,

`\begin{array}{l}{=x^(2)+6 x-x-6=0} \\ {=x(x+6)-1(x+6)=0} \\ {=(x-1)(x+6)=0}\end{array}`

When x - 1 = 0 , x = 1

When x + 6 = 0 , x = -6

So two values of x which satisfies the given equation are 1 and -6.

b)Given equation is
`x=(24)/(x)-5`

`\begin{array}{l}{=x * x=24-5 x} \\ {=x^(2)+5 x-24=0}\end{array}`

On splitting the middle term we get

`\begin{array}{l}{=x^(2)+8 x-3 x-24=0} \\ {=x(x+8)-3(x+8)=0} \\ {=(x-3)(x+8)=0}\end{array}`

When x - 3 = 0 , x = 3

When x + 8 = 0 , x = -8

So two values of x which satisfies the given equation are 3 and -8.

c) Given equation is
`x=(1)/(x)`

`\begin{array}{l}{=x * x=1} \\ {=x^(2)-1=0} \\ {=x^(2)-1^(2)=0}\end{array}`

Using algebraic identity
`a^(2)-b^(2)=(a-b)(a+b)` we get

=(x-1)(x+1)=0

When x - 1 = 0 , x =1

When x + 1 = 0 , x = -1

So two values of x which satisfies the given equation are 1 and -1.