Subtract (In picture)

Mathematics

1 answer:

Sonbull [250]3 years ago

6 0

--First we have to simplfy

$\frac{2x-8}{x^2-x-12} -\frac{x-3}{x(x+1)}$

$\frac{2(x-4)}{(x-4)(x+3)} - \frac{x-3}{x(x+1)}$

--Cancel common factors

$\frac{2}{(x+3)} -\frac{x-3}{x(x+1)}$

--Here remember never cancel factors in a subtraction or addition problem

--Now Multiply each side until both denominators are equal to each other

$\frac{2[x(x+1)]}{x(x+3)(x+1)} -\frac{(x-3)(x+3)}{x(x+3)(x+1)}$

--Simplify

$\frac{2x^2+2x}{x(x+1)(x+3)} - \frac{x^2-9}{x(x+1)(x+3)}$

--Now that the denominators are the same: subtract!

$\frac{2x^2+2x-(x^2-9)}{x(x+1)(x+3)}$

$\frac{2x^2-x^2+2x+9}{x(x+1)(x+3)}$

--And LAST STEP! ......Simplify More.... To get your answer

$\frac{x^2+2x+9}{x(x+1)(x+3)}$

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Miki prepared two gallons of a beverage that contains tea and lemonade in the ratio tea:lemonade=3:1. If Miki then added one-hal

Advocard [28]

Ratio of tea to lemonade in the new mixture is 3 : 2

<h3>Solution:</h3>

Given that volume of a beverage prepared by Miki containing tea and lemonade = 2 gallons

Ratio of tea and lemonade in beverage prepared by Miki = 3:1

Then Miki added one half of gallon of lemonade to the mixture.

Need to determine ratio of tea and lemonade in the new mixture.

Lets first calculate volume of tea and volume of lemonade in old mixture that is the one before adding one half of gallon of lemonade.

Given ratio of tea and lemonade in old mixture = 3: 1 , means our of every four parts of mixture, 3 parts are of tea and 1 part is of lemonade.

$=>\text { In old mixture, } \frac{3}{4} \text { is tea and } \frac{1}{4} \text { is lemonade. }$

$\text { Volume of tea in old mixture }=\frac{3}{4} \times \text { volume of old mixture }=\frac{3}{4} \times 2 \text { gallons }=1.5 \text { gallons }$

$\text { Volume of lemonade in old mixture }=\frac{1}{4} \times \text { volume of old mixture }=\frac{1}{4} \times 2 \text { gallons }=0.5 \text { gallons }$