4 years ago

Solve for X. 5+4/7(21+ 3X)= 41

Mathematics

2 answers:

Igoryamba4 years ago

8 0

Consider the equation $5+\dfrac{4}{7}\left(21+3x\right)=41.$

1. Multiply this equation by 7:

$5\cdot 7+\dfrac{4}{7}\cdot 7 (21+3x)=41\cdot 7,\\\\35+4(21+3x)=287.$

2. Eliminate brackets:

$35+4\cdot 21+4\cdot 3x=287,\\\\35+84+12x=287.$

3. Separate terms with x into the left side and without x into the right side:

$12x=287-35-84,\\\\12x=168.$

4. Find x:

$x=168:12,\\ \\x=14.$

Answer: x=14.

oksano4ka [1.4K]4 years ago

4 0

5+4/7(21+ 3X)= 41

-5

4/7(21+3x)=36

*7/4

21+3x=63

-21

3x=42

x=14

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