Word Problem?

Question

The sum of two numbers is 15 and the difference of their reciprocals is 1/10. Find the numbers.

Please show me the steps please! Thanks!

Answer 1

Equations: 1. x + y = 15 and 2. (1/x) - (1/y) = 1/10
*Take either equation and solve for "x" or "y" > let's solve for "x" in the 1st equation.

First: solve "y" by isolating it - subtract "x" from both sides...

x - x + y = 15 - x
y = 15 - x

Sec: replace "15 - x" with the y-variable in the 2nd equation...

(1/x) - (1/15 - x) = 1/10

*Eliminate parenthesis-multiply the denominators by everything...

(x)(15-x)(10)(1/x) - (x)(15-x)(10)(1/15 - x) = (x)(15-x)(10)(1/10)

*Cross cancel "like" terms-combine the remaining terms...

(15 - x)(10)(1) - (x)(10)(1) = (x)(15 - x)(1)
150 - 10x - 10x = 15x - x^2
150 - 20x = 15x - x^2

*Set the equation to "0" - subtract 150 from both sides...

150 - 150 - 20x = 15x - x^2 - 150
- 20x = 15x - x^2 - 150

*Add 20x to both sides...

- 20x + 20x = 15x - x^2 - 150+ 20x
0 = 35x - x^2 - 150

*Write the equation in alph descending order-"x" variables come first & factor the equation...

0 = -x^2 + 35x - 150
0 = -1(x^2 - 35x + 150)
0 = -1(x - 30)(x - 5)

*Solve for both x-variables by setting the parenthesis to equal "0"

x - 30 = 0
x - 30 + 30 = 0 + 30
x = 30

x - 5 = 0
x - 5 + 5 = 0 + 5
x = 5

Third: replace 5 with "x" in the 1st equation and solve for "y"

5 + y = 15
5 - 5 + y = 15 - 5
y = 10

*Replace 30 with "x" in the 1st equation and solve for "y"

30 - 30 + y = 15 - 30
y = - 15

(5, 10) and (30, -15)

Answer 2

Sum of 2 numbers is 15:

a + b = 15

The difference of the reciprocals is 1/10

1/a - 1/b = 1/10

So you have 2 equations and 2 unknowns. In the first equation
a+b = 15 or
a = 15-b

Now use that to substitute in the 2nd

1/(15-b) - 1/b = 1/10 Now to solve: multiply both sides by 10b*(15-b)

10b - 150 + 10b = 15b - b^2 Combine like terms

b^2 + 5b - 150 = 0 and factor

(b+15)(b-10) = 0
b = 10, b = -15

Now sub back in the first equation

a + b = 15
a + 10 = 15
a = 5
-OR-
a + b = 15
a -15 = 15
a = 30

So the solutions are (30, -15) and (5, 10)

Answer 3

Let's express it in "math" terms.

The sum of two numbers is 15.
a + b = 15 (i pulled "a" and "b" out of thin air, it doesn't matter what you call the two numbers.)

The difference of their reciprocals is 1/10.
(1/a) - (1/b) = 1/10
It doesn't matter whether a or b is first. you could just as easily write
(1/b) - (1/a) = 1/10
and it wouldn't matter, you'd just switch around which number is a, and which number is b. But for the sake of consistency, let's use the first one.
A nice trick to remember when you have a bunch of denominators that are ALL different, multiply every single term by the product of all the denominators to get rid of the fractions. In this case, it's 10ab.

(1/a)(10ab) - (1/b)(10ab) = (1/10)(10ab)
10b - 10a = ab

So! Our two equations are
a + b = 15
10b - 10a = ab

I hope you remember how to solve systems of equations, yes?

Answer 4

Let x = one number
Then 15 - x = other number
1/x - 1/(15-x) = 1/10
[(15-x) -x)] /x(15-x)= 1/10
(15-2x)/(x(15-x) = 1/10
x(15-x) = 10(15-2x) [cross multiply]
-x^2 +15x = 150 - 20x
-x^2 + 35x - 150 = 0
x^2 - 35x +150 = 0
(x-5)(x-30)= 0
x=5 or x=30
If x=5, then 15-x =10
If x = 30, then 15-x = -15

Both solutions work.

Answer 5

x+y=15

1/x-1/y=1/10

Solve for x or y in the first equation and substitute into the second equation.

If both numbers are positive, you should get x=5, y=10