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What is the value of x in both a & b?

What is the value of x in both a & b?-example-1
User EricWF
by
4.4k points

2 Answers

6 votes

Answer:

7 and
(16)/(5)

Explanation:

(a)

Δ ABC and Δ ADE are similar thus the ratios of corresponding sides are equal


(AB)/(AD) =
(AC)/(AE) , substitute values


(2)/(4) =
(x)/(x+7) ( cross- multiply )

4x = 2(x + 7) ← distribute

4x = 2x + 14 ( subtract 2x from both sides )

2x = 14 ( divide both sides by 2 )

x = 7

---------------------------------------------

(b)

Δ ABC and Δ CDF are similar thus ratios of corresponding sides are equal


(AB)/(CD) =
(BC)/(DF) , substitute values


(2)/(5) =
(x)/(8) ( cross- multiply )

5x = 16 ( divide both sides by 5 )

x =
(16)/(5)

User Cristy
by
5.5k points
1 vote

Answer:

I've pretty much already solved this, in my discussion above:

| 3 | = 3

| –3 | = 3

So then x must be equal to 3 or equal to –3.

But how am I supposed to solve this if I don't already know the answer? I will use the positive / negative property of the absolute value to split the equation into two cases, and I will use the fact that the "minus" sign in the negative case indicates "the opposite sign", not "a negative number".

For example, if I have x = –6, then "–x " indicates "the opposite of x" or, in this case, –(–6) = +6, a positive number. The "minus" sign in "–x" just indicates that I am changing the sign on x. It does not indicate a negative number. This distinction is crucial!

Whatever the value of x might be, taking the absolute value of x makes it positive. Since x might originally have been positive and might originally have been negative, I must acknowledge this fact when I remove the absolute-value bars. I do this by splitting the equation into two cases. For this exercise, these cases are as follows:

a. If the value of x was non-negative (that is, if it was positive or zero) to start with, then I can bring that value out of the absolute-value bars without changing its sign, giving me the equation x = 3.

b. If the value of x was negative to start with, then I can bring that value out of the absolute-value bars by changing the sign on x, giving me the equation –x = 3, which solves as x = –3.

Then my solution

User Lars Lind Nilsson
by
4.9k points