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 $x$}$

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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.

4 0

2 years ago

A student got 85% on a math test with 40 problems. How many did the student get correct?

dalvyx [7]

Answer:

The student got 34 of the 40 questions correct.

Step-by-step explanation:

85% of 40 is 34.

Hope this helps! :)

I would appreciate brainliest, if not that's okay!

5 0

2 years ago

Read 2 more answers

—7m<28 solve please hurry

mario62 [17]

Answer:

the answer is -4

Step-by-step explanation:

step 1: -7m<28

step 2: -7m÷ (-7)>28÷(-7)

step 3: m>28÷(-7)

step 4: m>-28÷7

step 5: m>-4

8 0

2 years ago