How to do 3/2b+1/5b=17 ?

Mathematics

2 answers:

guajiro [1.7K]3 years ago

7 0

Answer:

b = 10

Explanation:

Lets start by combining like terms (b):

3/2b + 1/5b

to combine these fractions you must get them to both have the same denominator:

3/2 · 5/5 = 15/10 multiply top and bottom by other fractions denominator

1/5 · 2/2 = 2/10 multiply top and bottom by other fractions denominator

15/10 + 2/10 = 17/10

17/10b = 17

Now we want to get the b to be by itself, so lets first multiply both sides by 10:

17b = 170

Now lets divide both sides by 17:

b = 10

Hope this helped!

zalisa [80]3 years ago

3 0

3/2b + 1/5b = 17

Multiply both sides by the LCD. The LCD is 2 * 5 = 10.

10 * 3/2b + 10 * 1/5b = 10 * 17

15b + 2b = 170

17b = 170

b = 10

You might be interested in

5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is

Dvinal [7]

Answer:

x = -2 or -9

Step-by-step explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

<h3>Slope</h3>

The slope of a line is given by the slope formula:

m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

<h3>Perpendicular lines</h3>

The slopes of perpendicular lines have product of -1:

$\dfrac{x}{2x+3}\cdot\dfrac{2+3x}{6-x}=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}$