(3(x+5))/4+(-7(x+3))/(10)=4 — step-by-step solution

1) Let's solve in

ℝ

\frac{3 \cdot (x + 5)}{4} + \frac{-7 \cdot (x + 3)}{10} = 4

\frac{3 \cdot (x + 5)}{4} - \frac{7 \cdot (x + 3)}{10} = 4

2) Let's add fractions:

\frac{3 \cdot (x + 5)}{4} - \frac{7 \cdot (x + 3)}{10} - 4 = 0

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{3 \cdot (x + 5)}{4} - \frac{7 \cdot (x + 3)}{10}

\frac{3 \cdot (x + 5) \cdot 5 - 7 \cdot (x + 3) \cdot 2}{4 \cdot 5}

\frac{3 \cdot (x + 5) \cdot 5 - 7 \cdot (x + 3) \cdot 2}{4 \cdot 5} - 4 = 0

ⓘ Multiplications are done before additions and subtractions:

3 \cdot 5

15

7 \cdot 2

14

■To add fractions, put them over the same denominator, then add the numerators:

\frac{15 \cdot (x + 5) - 14 \cdot (x + 3)}{20} - 4

\frac{(15 \cdot (x + 5) - 14 \cdot (x + 3)) - 4 \cdot 20}{20}

ⅰ

4 \cdot 5

20

\frac{(15 \cdot (x + 5) - 14 \cdot (x + 3)) - 4 \cdot 20}{20} = 0

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 20

80

\frac{15 \cdot (x + 5) - 14 \cdot (x + 3) - 80}{20} = 0

Let's solve in

ℝ

15 \cdot (x + 5) - 14 \cdot (x + 3) - 80 = 0

ⓘ Expand by multiplying each term inside the parentheses:

15 \cdot (x + 5)

15 x + 75

- 14 \cdot (x + 3)

- (14 x + 42)

15 x

+ 75

- (14 x + 42)

- 80 = 0

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (14 x + 42)

- 14 x - 42

15 x

+ 75

- 14 x

- 42

- 80 = 0

ⓘ Start from

75

and move back

42

on the number line:

75 - 42

33

■Like terms are grouped by adding their coefficients:

15 x - 14 x

x

x + 33 - 80 = 0

ⓘ Start from

33

and move back

80

on the number line, going below zero:

33 - 80

-47

x - 47 = 0

▶Solution obtained:

x = 47