Question

3(2x-9) - 4(3x + 4) = 2(x − 12) + 7/3
pls answer the question​

Answer 1

Answer:

$\normalsize \textsf{ x = -\dfrac{8}{3} }$

Step-by-step explanation:

Given: $\normalsize \textsf{ 3(2x - 9) - 4(3x + 4) = 2(x - 12) + \dfrac{7}{3} }$

1. Distribute

For 3:

$\normalsize \textsf{ 3(2x - 9) }\\\normalsize \textsf{ 3(2x) + 3(- 9) }\\\normalsize \boxed{\textsf{ 6x -27 }}$

For -4:

$\normalsize \textsf{ -4(3x + 4) }\\\normalsize \textsf{ -4(3x) + -4(4) }\\\normalsize \boxed{\textsf{ -12x -16 }}$

For 2:

$\normalsize \textsf{ 2(x - 12) }\\\normalsize \textsf{ 2(x) + 2(-12) }\\\normalsize \boxed{\textsf{ 2x -24 }}$

Now we have: $\normalsize \textsf{ 6x - 27 - 12x -16 = 2x-24 + \dfrac{7}{3} }$

2. Combine like terms:

$\normalsize \textsf{ -6x -43 = 2x -\dfrac{72}{3} + \dfrac{7}{3} }\\\\\normalsize \textsf{ -6x -43 = 2x -\dfrac{65}{3} }$

3. Subtract 2x from both sides:

$\normalsize \textsf{ -6x-2x -43 = 2x -2x -\dfrac{65}{3} }\\\\\normalsize \textsf{ -8x -43 = -\dfrac{65}{3} }$

4. Add 43 to both sides:

$\normalsize \textsf{ -8x -43 +43 = -\dfrac{65}{3} + 43 }\\\\\normalsize \textsf{ -8x = -\dfrac{65}{3} + \dfrac{129}{3} }\\\\\normalsize \textsf{ -8x = \dfrac{64}{3} }$

5. Divide both sides by -8

$\normalsize \textsf{ -8x = \dfrac{64}{3} }\\\\\normalsize \textsf{ -8x \div 8 = \dfrac{64}{3} \div -\dfrac{8}{1} }\\\\\normalsize \boxed{\textsf{ x = -\dfrac{8}{3} }}$

Learn more about equations and the distributive property here:

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