(x+4)/3-x/7 =(x+7)/5
1) Let's solve in ℝ: x + 43 -x7 = x + 75
2) Let's add fractions: x + 43 -x7 -x + 75 = 0
ⓘ To add fractions, put them over the same denominator, then add the numerators: x + 43 -x7 = (x + 4)·7 -x·33·7
(x + 4)·7 -x·33·7 -x + 75 = 0
ⓘ To add fractions, put them over the same denominator, then add the numerators: (x + 4)·7 -x·321 -x + 75 = ((x + 4)·7 -x·3)·5 -(x + 7)·2121·5
ⅰ 3·7 = 21
((x + 4)·7 -x·3)·5 -(x + 7)·2121·5 = 0
ⅰ 21·5 = (20 + 1)·5 = 20·5 + 5 = 100 + 5 = 105
((x + 4)·7 -x·3)·5 -(x + 7)·21105 = 0
Let's solve in ℝ: ((x + 4)·7 -x·3)·5 -(x + 7)·21 = 0
ⓘ Expand by multiplying each term inside the parentheses: (x + 4)·7 = x·7 + 28, ((x·7 + 28) -x·3)·5 = (x·7 + 28)·5 + 15·(-x), -(x + 7)·21 = -(x·21 + 147)
(x·7 + 28)·5 + 15·(-x) -(x·21 + 147) = 0
ⓘ The minus sign is placed in front of the product: 15·(-x) = -15x■A minus sign before a parenthesis changes the sign of each of its terms: -(x·21 + 147) = -x·21 -147
(x·7 + 28)·5 -15x -x·21 -147 = 0
ⓘ Like terms are grouped by adding their coefficients: (x·7 + 28)·5 -15x -x·21 = 5·(x·7 + 28) -36x
5·(x·7 + 28) -36x -147 = 0
ⓘ Expand by multiplying each term inside the parentheses: 5·(x·7 + 28) = 35x + 140
35x + 140 -36x -147 = 0
ⓘ Start from 140 and move back 147 on the number line, going below zero: 140 -147 = -7■Like terms are grouped by adding their coefficients: 35x -36x = -x
-x -7 = 0
ⓘ We add 7 to both sides
-x = 7
ⓘ The sign of both sides is changed
▶Solution found: x = -7