Solve for x:
(7 x)/3+1/5 = (5 (x+2/5))/(3)
Put each term in x+2/5 over the common denominator 5: x+2/5 = (5 x)/5+2/5:
(7 x)/3+1/5 = 5/3 (5 x)/5+2/5
(5 x)/5+2/5 = (5 x+2)/5:
(7 x)/3+1/5 = 5/3 (5 x+2)/5
(5 (5 x+2))/(5×3) = 5/5×(5 x+2)/3 = (5 x+2)/3:
(7 x)/3+1/5 = (5 x+2)/3
Put each term in (7 x)/3+1/5 over the common denominator 15: (7 x)/3+1/5 = (35 x)/15+3/15:
(35 x)/15+3/15 = (5 x+2)/3
(35 x)/15+3/15 = (35 x+3)/15:
(35 x+3)/15 = (5 x+2)/3
Multiply both sides by 15:
(15 (35 x+3))/15 = (15 (5 x+2))/3
(15 (35 x+3))/15 = 15/15×(35 x+3) = 35 x+3:
35 x+3 = (15 (5 x+2))/3
15/3 = (3×5)/3 = 5:
35 x+3 = 5 (5 x+2)
Expand out terms of the right hand side:
35 x+3 = 25 x+10
Subtract 25 x from both sides:
(35 x-25 x)+3 = (25 x-25 x)+10
35 x-25 x = 10 x:
10 x+3 = (25 x-25 x)+10
25 x-25 x = 0:
10 x+3 = 10
Subtract 3 from both sides:
10 x+(3-3) = 10-3
3-3 = 0:
10 x = 10-3
10-3 = 7:
10 x = 7
Divide both sides of 10 x = 7 by 10:
(10 x)/10 = 7/10
10/10 = 1:
Answer: |x = 7/10
