All categories

3 years ago

How would I solve this Rational Equation? Solve for all values of x.

Mathematics

1 answer:

Nesterboy [21]3 years ago

4 0

9514 1404 393

Answer:

x = 4

Step-by-step explanation:

I like to put these in the form f(x) = 0. We can do that by subtracting the right side. Common factors can be cancelled from numerator and denominator, provided they are not zero.

$\dfrac{7}{x+3}+\dfrac{3}{x-3}-\dfrac{x}{x-3}=0\\\\ \dfrac{7(x-3)+(x+3)(3-x)}{(x+3)(x-3)}=0\\\\\dfrac{(x-3)(4-x)}{(x-3)(x+3)}=0\\\\ \dfrac{4-x}{x+3}=0\qquad x\ne3\\\\x=4 \qquad\text{equate the numerator to zero, add $x$}$

_____

If you leave the numerator as (x-3)(4-x), then there are two values of x that make it zero. Because x=3 makes the equation "undefined", it cannot be considered to be a solution.

You might be interested in

20 subtracted from three times a number is 6 more than the number

kompoz [17]

<h3>20 subtracted from three times a number is 6 more than the number. Then the number is 13</h3>

<em><u>Solution:</u></em>

<em><u>Given statement is:</u></em>

20 subtracted from three times a number is 6 more than the number

We can translate the sentence into algebraic expression

Let "a" be the unknown number

Then,

20 subtracted from three times "a" is 6 more than "a"

Which means,

3a - 20 = 6 + a

3a - a = 6 + 20

2a = 26

Divide both sides by 2

a = 13

Thus the number is 13

3 0

3 years ago

Which of the following is the simplest form of this expression?

Softa [21]

Step-by-step explanation:

which expression there is no expression here

4 0

3 years ago

A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a

Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\$

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)$ [as x+y=14]

$\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\$

$\Rightarrow x =(25 \times 14)/32=10.9375$ grams

and $y = 14-x= 14-10.9375=3.0625$ grams.

Hence, 10.9375 grams of 82% gold and 3.0625 grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\$