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3 years ago

1/2 (8x-2) + 3/4 (8x-16) please help me with this

Mathematics

2 answers:

saul85 [17]3 years ago

6 0

The Answer is 10x−13

Hope this helps!

AnnyKZ [126]3 years ago

4 0

Answer:

10x - 13

Step-by-step explanation:

1/2 (8x - 2) + 3/4 (8x - 16) Distribute the fractions. ->

(to distribute multiply the number outside the parentheses to each one inside the parentheses. Ex: 1/2 x 8 = 4 so its 4x for the first one, then do that same for 1/2 and -2)

4x - 1 + 6x - 12 Add like terms. ->

(Add the like terms are numbers/variables that are similar. Ex: 4x and 6x, add them together and you get 10x then do the same for -1 and -12)

10x - 13 <- The Answer

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3 years ago

Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)=

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Answer:

Error: $lnx^2=ln 3x$ not $ln\frac{3x}{0}$

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Step-by-step explanation:

We have to find the error and correct answer

Given: $2ln x=ln(3x)-[ln9-2ln(3)]$

$lnx^2=ln(3x)-[ln9-ln3^2]$

Using the formula

$alog b=logb^a$

$lnx^2=ln(3x)-[ln9-ln9]$

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$lnx^2=ln(3x)$

$x^2=3x$

$x^2-3x=0$

$x(x-3)=0$

Therefore, x=0 and x=3

But last step in the given solution

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote $lnx^2=ln\frac{3x}{0}$ instead of $lnx^2=ln3x$ and solution is given by

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