Question

The sum of the reciprocal of a positive number and the reciprocal of 3 more than the number is
7/10. Find the number.
Write an equation with the following characteristics:
1. It uses x to represent the number.
2. It uses the information as it is given above.
3. It can be solved to answer the question.
Equation:
The number is:

Answer 1

since the reciprocal is when the numerator and denominator are switched, we can figure out the first part would be 1/x and 1/3. knowing this, we can set up the equation as:

1/x + 1/3 = 7/10
1/x = 7/10 - 1/3
1/x = 21/30 - 10/30
1/x = 11/30
here you can cross multiply:
30 = 11x
x = 30/11

i apologize if the format looks bad, i did this on mobile lol

Answer 2

Answer:

$\sf Equation: \quad \dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{7}{10}$

The number is 2.

Step-by-step explanation:

The reciprocal of a number is simply 1 divided by the number.

The reciprocal of a fraction is where the numerator and denominator are swapped (so the fraction is turned upside down).

Let x be the positive number.

The reciprocal of the positive number x:

$\sf \implies \dfrac{1}{x}$

The reciprocal of 3 more than the number x:

$\sf \implies \dfrac{1}{x+3}$

The sum of the reciprocal of a positive number and the reciprocal of 3 more than the number is 7/10:

$\implies \sf \dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{7}{10}$

To solve, make the denominators of the two fractions on the left side the same by multiplying them.

Multiply the numerator of the first fraction by the denominator of the second fraction to get the new numerator of the first fraction.

Multiply the numerator of the second fraction by the denominator of the first fraction to get the new numerator of the second fraction.

$\implies \sf \dfrac{1(x+3)}{x(x+3)}+\dfrac{1(x)}{(x+3)(x)}=\dfrac{7}{10}$

$\implies \sf \dfrac{x+3}{x(x+3)}+\dfrac{x}{x(x+3)}=\dfrac{7}{10}$

Add the numerators and keep the denominator so that there is now one fraction:

$\implies \sf \dfrac{2x+3}{x(x+3)}=\dfrac{7}{10}$

Cross multiply:

$\implies \sf 10(2x+3)=7x(x+3)$

Expand:

$\implies \sf 20x+30=7x^2+21x$

Rearrange so the equation is equal to zero:

$\implies \sf 7x^2+x-30=0$

Split the middle term:

$\implies \sf 7x^2+15x-14x-30=0$

Factor the first two terms and last two terms separately:

$\implies \sf x(7x+15)-2(7x+15)=0$

Factor out the common term (7x + 15):

$\implies \sf (x-2)(7x+15)=0$

Apply the zero product property:

$\sf (x-2)=0 \implies x=2$

$\sf (7x+15)=0 \implies x=-\dfrac{15}{7}$

As x is a positive number, x = 2.

Learn more about fractions here:

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