solve each equation. choose a method

Question

5/7 (x+4)+1/3x= x+3

Answer 1

I know most of the students are afraid of fractions so lets first get rid of the first fraction (5/7) by multiplying EACH term in this equation by 7 {because 7*5/7=5)

so 7*[5/7(x+4)]+7*[1/3x]=7*x+7*3

 

5(x+4)+7/3x=7x+21

 

now lets get rid of the second fraction 7/3 by multiplying EACH term in this equation by 3

 

3*5(x+4) + 3*[7/3x]=3*7x+21*3

 

15(x+4)+ 7x=21x+63 =================(1)

 

distribute 15 inside the parenthesis  (x+4)

 15x+60+7x=21x+63

 

now

 

move all of the terms that have x to one side of the equation and the constants to the other side of the equation ( if you move a term beyond = you change its sign )

 

15x+7x-21x=63-60

(notice how I changed the sign of 21x and 60 as the moved from one side of the = to the other side)

 

simplify

 

22x-21x=3

 

simplify

 

 x=3

 

and that is the answer

 

 (the answer based on 1/3x=(1/3)x i.e x is in the numerator )

 

if x is in the denominator then equation (1) will become

 

15(x+4)+ 7/x=21x+63

 

distribute 15 first

 

15x+60+7/x=21x+63

 

multiply each term in the equation by x to eliminate it from the denominator.

 

15x^2+60x+7=21x^2+63x

 

combine like terms

 

15x^2-21x^2+60x-63x+7=0

 

-6x^2-3x+7=0

multiply by -1 just to make it look nice

 

6x^2+3x-7=0

solve this quadratic equation using the formula and you'll get this answer

 

x=(1/12)(3-sqr(177))

x=(1/12)(3+sqr(177))