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How do you solve #6+2(3j-2)=4(1+j)#?

Answer 1

Avery Alexander

#j=1#

#6 + 2 (3j-2) = 4 (1+j)#

#6 + 2 xx (3j) + 2 xx (-2) = 4 xx (1)+ 4 xx (j)#

#6 + 6j - 4 = 4 + 4j#

#2 + 6j = 4 + 4j#

#6j - 4j = 4-2 #

#2j = 2 #

#j=2/2#

#j=1#

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Answer 2

Savannah Anderson

To solve the equation 6 + 2(3j - 2) = 4(1 + j), follow these steps:

Distribute the 2 on the left side: 6 + 6j - 4 = 4(1 + j)
Combine like terms: 6j + 2 = 4 + 4j
Move all terms involving the variable j to one side and constants to the other side: 6j - 4j = 4 - 2
Simplify: 2j = 2
Solve for j: j = 1

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Answer 3

Julia Adams

To solve the equation (6 + 2(3j - 2) = 4(1 + j)), you would follow these steps:

First, distribute the terms inside the parentheses: [6 + 6j - 4 = 4 + 4j]
Combine like terms on both sides of the equation: [6 - 4 + 6j = 4 + 4j] [2 + 6j = 4 + 4j]
Move all the terms involving the variable (j) to one side of the equation and the constants to the other side: [6j - 4j = 4 - 2]
Combine like terms: [2j = 2]
Finally, solve for (j) by dividing both sides of the equation by 2: [j = \frac{2}{2}]

So, the solution is: [j = 1]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.