Anonymous

# need some help solving 0= (K/s+1)-(s/s+1)?

im blanking on some basic algebra...someone help please?

Update:

sorry...didnt use my brakets well at all...to make it easier to show what i meant we can use

0=(k-s)/(s+1)

Relevance

0=((K)/(s)+1)-((s)/(s)+1)

Since K is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.

((K)/(s)+1)-((s)/(s)+1)=0

To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is s. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.

((K)/(s)+1*(s)/(s))-((s)/(s)+1)=0

Complete the multiplication to produce a denominator of s in each expression.

((K)/(s)+(s)/(s))-((s)/(s)+1)=0

Combine the numerators of all expressions that have common denominators.

((K+s)/(s))-((s)/(s)+1)=0

Remove the common factors that were cancelled out.

((K+s)/(s))-(1+1)=0

Add 1 to 1 to get 2.

((K+s)/(s))-(2)=0

Multiply -1 by the 2 inside the parentheses.

(K+s)/(s)-2=0

Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of s.

(K+s)/(s)-2*(s)/(s)=0

Multiply the expression by a factor of 1 to create the least common denominator (LCD) of s.

(K+s)/(s)-(2*s)/(s)=0

Multiply 2 by s to get 2s.

(K+s)/(s)-(2s)/(s)=0

The numerators of expressions that have equal denominators can be combined. In this case, ((K+s))/(s) and -((2s))/(s) have the same denominator of s, so the numerators can be combined.

((K+s)-(2s))/(s)=0

Simplify the numerator of the expression.

(K+s-2s)/(s)=0

Since s and -2s are like terms, add -2s to s to get -s.

(K-s)/(s)=0

Multiply each term in the equation by s.

(K-s)/(s)*s=0*s

Simplify the left-hand side of the equation by canceling the common factors.

K-s=0*s

Multiply 0 by s to get 0.

K-s=0

Since -s does not contain the variable to solve for, move it to the right-hand side of the equation by adding s to both sides.

K=s