Answer:
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:
a. For equation [tex]x+5=\frac{6}{x}[/tex] value of x are 1 and -6.
b. For equation [tex]x=\frac{24}{x}-5[/tex] value of x are 3 and -8
c. For equation [tex]x=\frac{1}{x}[/tex] value of x are -1 and 1.
Solution:
a)Given equation is [tex]x+5=\frac{6}{x}[/tex]
On cross multiplying we get,
x(x+5)=6
[tex]x^{2}+5 x-6=0[/tex]
On splitting the middle term we get,
[tex]\begin{array}{l}{=x^{2}+6 x-x-6=0} \\ {=x(x+6)-1(x+6)=0} \\ {=(x-1)(x+6)=0}\end{array}[/tex]
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 [tex]x=\frac{24}{x}-5[/tex]
[tex]\begin{array}{l}{=x \times x=24-5 x} \\ {=x^{2}+5 x-24=0}\end{array}[/tex]
On splitting the middle term we get
[tex]\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}[/tex]
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 [tex]x=\frac{1}{x}[/tex]
[tex]\begin{array}{l}{=x \times x=1} \\ {=x^{2}-1=0} \\ {=x^{2}-1^{2}=0}\end{array}[/tex]
Using algebraic identity [tex]a^{2}-b^{2}=(a-b)(a+b)[/tex] 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.