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Add terms

C = \frac{2 x}{\frac{x}{1 8} + \frac{x}{6 0}}

$\frac{a}{b} = \frac{a \cdot 1 0}{b \cdot 1 0}$

C = \frac{2 x}{\frac{1 0 x}{1 8 0} + \frac{x}{6 0}}

$\frac{a}{b} = \frac{a \cdot 3}{b \cdot 3}$

C = \frac{2 x}{\frac{1 0 x}{1 8 0} + \frac{3 x}{1 8 0}}

AddFrac

Add fractions

C = \frac{2 x}{\frac{1 3 x}{1 8 0}}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

C = \frac{3 6 0 x}{1 3 x}

ReduceFrac

$\frac{a}{b} = \frac{a / x}{b / x}$

C = \frac{3 6 0}{1 3}

Calculate quotient

C = 27.692307 \dots

RoundDec

Round to $1$ decimal place(s)

C \approx 27.7

In this option, the combined mpg is

C \approx 27.7 .

Option II: Buy a New Hybrid, Keep SUV as It Is

In this option, it is suggested to buy a new hybrid with a mpg of $75$ and keep the SUV. As in the procedure followed in the previous option, the information at hand can be organized in a table.

	miles	mpg	$gallons = \frac{miles}{mpg}$
SUV	$x$	$15$	$\frac{x}{1 5}$
Hybrid	$x$	$75$	$\frac{x}{7 5}$

Substitute these expressions into the formula.

C = \frac{SUV miles + Hybrid miles}{SUV gallons + Hybrid gallons}

SubstituteExpressions

Substitute expressions

C = \frac{x + x}{\frac{x}{1 5} + \frac{x}{7 5}}

▼

Simplify right-hand side

Add terms

C = \frac{2 x}{\frac{x}{1 5} + \frac{x}{7 5}}

$\frac{a}{b} = \frac{a \cdot 5}{b \cdot 5}$

C = \frac{2 x}{\frac{5 x}{7 5} + \frac{x}{7 5}}

AddFrac

Add fractions

C = \frac{2 x}{\frac{6 x}{7 5}}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

C = \frac{1 5 0 x}{6 x}

ReduceFrac

$\frac{a}{b} = \frac{a / x}{b / x}$

C = \frac{1 5 0}{6}

Calculate quotient

C = 25

In this option, the combined mpg is

C = 25 .

Conclusion

If Option I is chosen, the combined mpg will be about $27.7 .$ If Option II is chosen, it will be $25 .$ Therefore, the first option gives a better combined mpg.

Example

Adjusting Estimated Time of Arrival

Tiffaniqua's parents want to thank her for solving their car problem and drove to a store $40$ miles away in the next city over to buy a new $3 D$ printer that she has dreamed of for years. On the way back, there was a road closure due to a landslide. This caused them to drive $12$ mph slower.

A map that shows lake, forrest, some buildings, and road construction

This resulted in the return trip taking

3

hours longer. How many hours did it take them to get home from the store?

Hint

The distance traveled is equal to the product of the speed and the time.

Solution

Only the distance to the store is known. The time it takes to go to the store and the speed are unknown. Then, let $t$ and $r$ be the time elapsed and the speed, respectively.

	Speed	Time	Distance
To the store	$r$	$t$	$40$

On the way home, Tiffaniqua's mother drove $12 mph$ slower and the trip took $3$ hours longer.

	Speed	Time	Distance
To the store	$r$	$t$	$40$
From the store	$r - 12$	$t + 3$	$40$

Since the distance traveled is equal to the product of the speed and the time, a system of equations can be written.

{r t = 40 (r - 12) (t + 3) = 40 (I) (II)

By manipulating the equations in the system, rational equations can be formed.

{r t = 40 (r - 12) (t + 3) = 40

$(I):$ $LHS / t = RHS / t$

{r = \frac{4 0}{t} (r - 12) (t + 3) = 40

$(II):$ $LHS / (t + 3) = RHS / (t + 3)$

{r = \frac{4 0}{t} r - 12 = \frac{4 0}{t + 3}

Next, since

r

is isolated in Equation (I), its equivalent expression can be substituted into Equation (II).

{r = \frac{4 0}{t} r - 12 = \frac{4 0}{t + 3} (I) (II)

$(II):$ $r = \frac{4 0}{t}$

{r = \frac{4 0}{t} \frac{4 0}{t} - 12 = \frac{4 0}{t + 3}

Now, the second equation is written only in terms of

t .

Since only the time is required, this equation will be solved for the time

t .

\frac{4 0}{t} - 12 = \frac{4 0}{t + 3}

▼

Simplify

NumberToFrac

$a = \frac{t \cdot a}{t}$

\frac{4 0}{t} - \frac{1 2 t}{t} = \frac{4 0}{t + 3}

SubFrac

Subtract fractions

\frac{4 0 - 1 2 t}{t} = \frac{4 0}{t + 3}

$LHS \cdot t = RHS \cdot t$

40 - 12 t = \frac{4 0}{t + 3} t

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

40 - 12 t = \frac{4 0 t}{t + 3}

$LHS \cdot (t + 3) = RHS \cdot (t + 3)$

(40 - 12 t) (t + 3) = 40 t

Distribute $(t + 3)$

40 (t + 3) - 12 t (t + 3) = 40 t

Distribute $40 & - 12 t$

40 t + 120 - 12 t^{2} - 36 t = 40 t

$LHS - 40 t = RHS - 40 t$

120 - 12 t^{2} - 36 t = 0

CommutativePropAdd

Commutative Property of Addition

- 12 t^{2} - 36 t + 120 = 0

Factor out $- 12$

- 12 (t^{2} + 3 t - 10) = 0

$LHS / (- 12) = RHS / (- 12)$

t^{2} + 3 t - 10 = 0

This equation is now a quadratic equation in terms of

t .

It can be solved by factoring.

t^{2} + 3 t - 10 = 0

▼

Factor

WriteSum

Write as a sum

t^{2} - 2 t + 5 t - 10 = 0

Factor out $t$

t (t - 2) + 5 t - 10 = 0

Factor out $5$

t (t - 2) + 5 (t - 2) = 0

Factor out $(t - 2)$

(t - 2) (t + 5) = 0

▼

Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

t - 2 = 0 t + 5 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 2 = RHS + 2$

t = 2 t + 5 = 0

$(II):$ $LHS - 5 = RHS - 5$

t = 2 t = - 5

Since time cannot be negative, the solution

t = - 5

is disregarded. Therefore,

t = 2

is the only solution in this context. Recall that

t + 3

represents the time to get home from the store. This means that it took Tiffaniqua

2 + 3 = 5

hours to get home from the store.

Example

Concentration of a Mixture

Tiffaniqua and Mark are excited to print parts to make a drone using Tiffaniqua's new $3 D$ printer. They want to make a special mixture for the raw material, called the filament. Tiffaniqua filled a flask with $500$ milliliters of water. She then adds $60$ grams of water soluble polyester resin.

Mark then adds more water at a rate of $10$ milliliters per minute and simultaneously adds more resin at a rate of $6$ grams per minute.

a Write a function

C (t)

for the concentration of resin in the flask after

t

minutes.

b After how many minutes will the concentration be

0.2 ?

Hint

a How can the amount of water after

t

minutes be represented? How can the amount of resin after

t

minutes be represented?

b Substitute the given value into the function found in Part A.

Solution

a Since water increases at a rate of

10

milliliters per minute and resin at

6

grams per minute, these are constant rates of change. Let

W (t)

be the amount of water and

R (t)

the amount of resin in the flask after

t

minutes. Considering that the flask initially has

500

milliliters of water and

60

grams of resin, two functions can be written.

Water : Resin : W (t) = 500 + 10 t R (t) = 60 + 6 t

The concentration

C (t)

is the ratio of grams of resin to milliliters of water.

C (t) = \frac{R ( t )}{W ( t )} ⇕ C (t) = \frac{6 0 + 6 t}{5 0 0 + 1 0 t}

b To find after how many minutes the concentration will be

0.2,

substitute

0.2

for

C (t)

into the equation found in Part A. Then, solve it for

t .

C (t) = \frac{6 0 + 6 t}{5 0 0 + 1 0 t}

$C (t) = 0.2$

0.2 = \frac{6 0 + 6 t}{5 0 0 + 1 0 t}

▼

Solve for

t

$LHS \cdot (500 + 10 t) = RHS \cdot (500 + 10 t)$

0.2 (500 + 10 t) = 60 + 6 t

Distribute $0.2$

100 + 2 t = 60 + 6 t

$LHS - 100 = RHS - 100$

2 t = - 40 + 6 t

$LHS - 6 t = RHS - 6 t$

- 4 t = - 40

$LHS / (- 4) = RHS / (- 4)$

t = 10

After

10

minutes, the concentration in the flask will be

0.2 .

Using constant rates of change, Tiffaniqua and Mark have successfully made the special mixture for the filament that they hope will become an awesome

3 D

printed drone — a drone that may have more purpose than what they expect.

Example

Finding the Printing Time of the New $3$ D Printer

Tiffaniqua and Mark realize they can use their filament with Mark's old $3 D$ printer in addition to the new printer. There is one major difference between the printers, however, it would take the old printer $50$ minutes longer than the new printer to create the same parts.

External credits: macrovector_official, macrovector

Working together — some parts are printed with the new printer and some with the old printer — they can complete the task in

3

hours and

20

minutes. How long would it take the new printer to create the parts alone?

Hint

If $t$ is the printing time of the new printer, then $t + 50$ is the printing time of the old printer.

Solution

Let

t

be the time in minutes it takes the new printer to print the parts Tiffaniqua and Mark need. Then,

t + 50

will be the printing time for the old printer. In this situation, the work is to print the drone parts, which can be considered as

1 .

Each machine's rate of work is then

1

over the printing time.

New Printer ’ s Rate \frac{1}{t} Old Printer ’ s Rate \frac{1}{t + 5 0}

When both printers work at the same time, the job is completed in

3

hours and

20

minutes. Since

t

is defined in minutes, this time should also be expressed in minutes.

3 h and 20 min = 200 min

The combined rate of the printers is

1

over the combined printing time

200 .

Therefore, the combined rate is

\frac{1}{2 0 0} .

This rate should be equal to the sum of the individual rates of the printers.

\frac{1}{t} + \frac{1}{t + 5 0} = \frac{1}{2 0 0}

The above rational equation can be solved for

t .

To eliminate the denominators, the equation can be multiplied by the least common denominator, which is the product of all the denominators.

\frac{1}{t} + \frac{1}{t + 5 0} = \frac{1}{2 0 0}

$LHS \cdot 200 t (t + 50) = RHS \cdot 200 t (t + 50)$

(\frac{1}{t} + \frac{1}{t + 5 0}) (200 t (t + 50)) = \frac{1}{2 0 0} (200 t (t + 50))

▼

Simplify

Distribute $200 t (t + 50)$

\frac{1}{t} (200 t (t + 50)) + \frac{1}{t + 5 0} (200 t (t + 50)) = \frac{1}{2 0 0} (200 t (t + 50))

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

\frac{2 0 0 t ( t + 5 0 )}{t} + \frac{2 0 0 t ( t + 5 0 )}{t + 5 0} = \frac{2 0 0 t ( t + 5 0 )}{2 0 0}

CancelCommonFac

Cancel out common factors

\frac{2 0 0 t ( t + 5 0 )}{t} + \frac{2 0 0 t ( t + 5 0 )}{t + 5 0} = \frac{2 0 0 t ( t + 5 0 )}{2 0 0}

SimpQuot

Simplify quotient

200 (t + 50) + 200 t = t (t + 50)

▼

Rewrite

Distribute $200 & t$

200 t + 10000 + 200 t = t^{2} + 50 t

Add terms

400 t + 10000 = t^{2} + 50 t

$LHS - 400 t = RHS - 400 t$

10000 = t^{2} - 350 t

$LHS - 10000 = RHS - 10000$

0 = t^{2} - 350 t - 10000

RearrangeEqn

Rearrange equation

t^{2} - 350 t - 10000 = 0

To solve this quadratic equation, the Quadratic Formula can be used.

t^{2} - 350 t - 10000 = 0

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a = 1, b = - 350, c = - 10000$

t = \frac{- ( - 3 5 0 ) \pm ( - 3 5 0 ) ^{2} - 4 ( 1 ) ( - 1 0 0 0 0 )}{2 ( 1 )}

NegNeg

$- (- a) = a$

t = \frac{3 5 0 \pm ( - 3 5 0 ) ^{2} - 4 ( 1 ) ( - 1 0 0 0 0 )}{2 ( 1 )}

CalcPowProd

Calculate power and product

t = \frac{3 5 0 \pm 1 2 2 5 0 0 + 4 0 0 0 0}{2}

Add terms

t = \frac{3 5 0 \pm 1 6 2 5 0 0}{2}

Next, the solutions can be individualized by using the positive and negative signs.

$t = \frac{3 5 0 \pm 1 6 2 5 0 0}{2}$
$t = \frac{3 5 0 + 1 6 2 5 0 0}{2}$	$t = \frac{3 5 0 - 1 6 2 5 0 0}{2}$
$t \approx 377$	$t \approx - 27$

Since time cannot be negative,

t \approx - 27

is ignored. The solution

t \approx 377

needs to be checked.

\frac{1}{t} + \frac{1}{t + 5 0} = \frac{1}{2 0 0}

$t \approx 377$

\frac{1}{3 7 7} + \frac{1}{3 7 7 + 5 0} \approx ? \frac{1}{2 0 0}

▼

Evaluate left-hand side

Add terms

\frac{1}{3 7 7} + \frac{1}{4 2 7} \approx ? \frac{1}{2 0 0}

UseCalc

Use a calculator

0.002652 \dots + 0.002341 \dots \approx ? 0.005

Add terms

0.004994 \dots \approx 0.005 ✓

Since a true statement was obtained,

t \approx 377

is a solution to the equation. The new printer prints the parts needed in about

377

minutes. The final step is to write this in hours.

377 min

▼

Rewrite

WriteSum

Write as a sum

360 min + 17 min

SplitIntoFactors

Split into factors

6 (60 min) + 17 min

HourToMin

$1 h = 60 min$

6 (1 h) + 17 min

6 h + 17 min

The answer is

6

hours and

17

minutes.

Example

How Far Can the Drone Go?

Drones communicate using radio waves on specific radio frequencies. The drone operates at a frequency of $2.4$ gigahertz (GHz). How far a drone can travel depends on a number of factors such as the power of the signal transmitted by the controller.

External credits: macrovector

Let $P_{t}$ be the power of the radio signal transmitted by the controller, $P_{r}$ the power of the radio signal received by the drone, $λ$ the wavelength, and $d$ the distance between the drone and its controller. It is known that $P_{r}$ varies directly with $P_{t}$ and the square of $λ,$ and inversely with the square of $d .$

a The controller transmits

300

milliwatts (mW) of power. When the distance between the drone and the controller is

240

meters, the power received by the drone is

4.5 \times 1 0^{- 7}

milliwatts. The wavelength of a

2.4 GHz

signal is about

0.12

meters. Find the constant of variation.

b If the power received is less than or equal to

1 0^{- 9}

milliwatts, data exchange is not possible with the drone. How far can the drone go? Round the answer to the nearest integer.

Hint

a Use the equation of the combined variation

z = \frac{k x}{y},

where

k

is the constant of variation.

b Substitute the given value into the equation.

Solution

a It is given that

P_{r}

varies directly with

P_{t}

and

λ^{2},

and inversely with

d^{2} .

Start by labeling the quantities.

P_{r} : P_{t} : λ : d : Power received Power transmitted Wavelength of signal Distance between the drone and the controller

A combined variation equation can then be written as follows.

P_{r} = \frac{k P _{t} λ ^{2}}{d ^{2}}

Here

k

is the constant of variation and cannot be

0 .

To find the value of

k,

review the given information.

P_{r} : P_{t} : λ : d : 4.5 \times 1 0^{- 7} mW 300 mW 0.12 m 240 m

Substitute these values into the equation and solve for

k .

P_{r} = \frac{k P _{t} λ ^{2}}{d ^{2}}

SubstituteValues

Substitute values

4.5 \times 1 0^{- 7} = \frac{k ( 3 0 0 ) ( 0 . 1 2 ^{2} )}{2 4 0 ^{2}}

▼

Solve for

k

MovePartNumLeft

$\frac{a \cdot b}{c} = a \cdot \frac{b}{c}$

4.5 \times 1 0^{- 7} = k \frac{( 3 0 0 ) ( 0 . 1 2 ^{2} )}{2 4 0 ^{2}}

CalcPow

Calculate power

4.5 \times 1 0^{- 7} = k \frac{( 3 0 0 ) ( 0 . 0 1 4 4 )}{5 7 6 0 0}

4.5 \times 1 0^{- 7} = k \frac{4 . 3 2}{5 7 6 0 0}

Calculate quotient

4.5 \times 1 0^{- 7} = k \cdot 0.000075

Write in scientific notation

4.5 \times 1 0^{- 7} = k \cdot 7.5 \times 1 0^{- 5}

$LHS / (7.5 \times 1 0^{- 5}) = RHS / (7.5 \times 1 0^{- 5})$

\frac{4 . 5 \times 1 0 ^{- 7}}{7 . 5 \times 1 0 ^{- 5}} = k

WriteProdFrac

Write as a product of fractions

\frac{4 . 5}{7 . 5} \cdot \frac{1 0 ^{- 7}}{1 0 ^{- 5}} = k

Calculate quotient

0.6 \cdot \frac{1 0 ^{- 7}}{1 0 ^{- 5}} = k

DivPow

$\frac{a ^{m}}{a ^{n}} = a^{m - n}$

0.6 \cdot 1 0^{- 7 - (- 5)} = k

SubNeg

$a - (- b) = a + b$

0.6 \cdot 1 0^{- 2} = k

CalcPow

Calculate power

0.6 \cdot 0.01 = k

0.006 = k

RearrangeEqn

Rearrange equation

k = 0.006

The constant of variation is

0.006 .

b Consider the equation written in Part A.

P_{r} = \frac{0 . 0 0 6 P _{t} λ ^{2}}{d ^{2}}

When

P_{r}

is less than or equal to

1 0^{- 9}

milliwatts, the connection is lost. To keep the drone under control, the drone must be within a certain distance. This distance can be found by solving the following equation.

1 0^{- 9} = \frac{0 . 0 0 6 P _{t} λ ^{2}}{d ^{2}}

The same values from Part A can be used for

P_{t}

and

λ,

as they are the same for this case as well.

1 0^{- 9} = \frac{0 . 0 0 6 P _{t} λ ^{2}}{d ^{2}}

SubstituteValues

Substitute values

1 0^{- 9} = \frac{0 . 0 0 6 ( 3 0 0 ) ( 0 . 1 2 ^{2} )}{d ^{2}}

▼

Solve for

d

CalcPow

Calculate power

1 0^{- 9} = \frac{0 . 0 0 6 ( 3 0 0 ) ( 0 . 0 1 4 4 )}{d ^{2}}

1 0^{- 9} = \frac{0 . 0 2 5 9 2}{d ^{2}}

$LHS \cdot d^{2} = RHS \cdot d^{2}$

1 0^{- 9} d^{2} = 0.02592

$LHS / 1 0^{- 9} = RHS / 1 0^{- 9}$

d^{2} = \frac{0 . 0 2 5 9 2}{1 0 ^{- 9}}

FracToMultNegExponent

$\frac{a}{b ^{m}} = a b^{- m}$

d^{2} = 0.02592 \cdot 1 0^{9}

d^{2} = 25920000

SqrtEqn

$LHS = RHS$

d^{2} = 25920000

$a^{2} = \pm a$

d = \pm 25920000

UseCalc

Use a calculator

d = \pm 5091.168824 \dots

RoundInt

Round to nearest integer

d \approx \pm 5091

The negative solution is ignored. The drone can be operated within a distance of about

5091

meters.

Closure

Calculating the Speed of the Wind

From solving her parents' car issues, to successfully printing a drone, Tiffaniqua and Mark are now trying to locate a missing dog, Quipu! The owner tells them that Quipu was last seen running along the road that leads to the next city where Tiffaniqua bought the printer. They desperately want to find Quipo quickly. The drone can fly at a max speed of $25$ kilometers per hour.

External credits: macrovector

Before setting out to find Quipo, they test flew the drone $5$ kilometers against the wind before needing to bring it back to the starting point. Suppose that the drone flew at its maximum rate of $25$ km/h throughout the trip.

a Write a rational function for the time of the flight

T (w),

where

w

is the speed of the wind.

b What is the speed of the wind if the total time of the trip is

25

minutes?

Hint

a Write two rational expressions for the times it takes to go upwind and downwind.

b Convert the given time into hours.

Solution

a Recall that the distance is the product of the rate and the time.

d = r t \Leftrightarrow t = \frac{d}{r}

Since

w

is the speed of the wind in kilometers per hour,

25 - w

will be the speed of the drone while going upwind and

25 + w

will be the speed of the drone while going downwind. The information about the distance, speed, and time can be organized in a table.

	Upwind	Downwind
Distance (km)	$5$	$5$
Speed (km/h)	$25 - w$	$25 + w$
Time (h)	$\frac{5}{2 5 - w}$	$\frac{5}{2 5 + w}$

Therefore, the flight time of the drone

T (w)

is the sum of the times.

T (w) = \frac{5}{2 5 - w} + \frac{5}{2 5 + w}

b In the previous part, the unit of

T (w)

is hours. For this reason, the given time must be converted into hours. To do so,

25 min

is multiplied by the conversion factor

\frac{1 h}{6 0 min} .

25 min \cdot \frac{1 h}{6 0 min} = \frac{5}{1 2} h

Now substitute this value for

T (w)

and solve the equation for

w .

T (w) = \frac{5}{2 5 - w} + \frac{5}{2 5 + w}

$T (w) = \frac{5}{1 2}$

\frac{5}{1 2} = \frac{5}{2 5 - w} + \frac{5}{2 5 + w}

▼

Solve for

w

$LHS \cdot \frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5} = RHS \cdot \frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5}$

\frac{5}{1 2} (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5}) = (\frac{5}{2 5 - w} + \frac{5}{2 5 + w}) (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5})

Distribute $(\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5})$

\frac{5}{1 2} (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5}) = \frac{5}{2 5 - w} (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5}) + \frac{5}{2 5 + w} (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5})

CancelCommonFac

Cancel out common factors

\frac{5}{1 2} (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5}) = \frac{5}{2 5 - w} (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5}) + \frac{5}{2 5 + w} (\frac{1 2 ( 2 5 - w ) ( 2 5 + w )}{5})

SimpQuotProd

Simplify quotient and product

(25 - w) (25 + w) = 12 (25 + w) + 12 (25 - w)

ExpandDiffSquares

$(a + b) (a - b) = a^{2} - b^{2}$

625 - w^{2} = 12 (25 + w) + 12 (25 - w)

Distribute $12$

625 - w^{2} = 300 + 12 w + 300 - 12 w

AddSubTerms

Add and subtract terms

625 - w^{2} = 600

$LHS - 625 = RHS - 625$

- w^{2} = - 25

$LHS \cdot (- 1) = RHS \cdot (- 1)$

w^{2} = 25

SqrtEqn

$LHS = RHS$

w^{2} = 25

$a^{2} = \pm a$

w = \pm 25

CalcRoot

Calculate root

w = \pm 5

The negative solution is ignored because speed cannot be negative. Therefore, the speed of the wind is

5

kilometers per hour. Do not forget to verify the solution in the equation.

\frac{5}{1 2} = \frac{5}{2 5 - w} + \frac{5}{2 5 + w}

$w = 5$

\frac{5}{1 2} = ? \frac{5}{2 5 - 5} + \frac{5}{2 5 + 5}

▼

Evaluate right-hand side

AddSubTerms

Add and subtract terms

\frac{5}{1 2} = ? \frac{5}{2 0} + \frac{5}{3 0}

$\frac{a}{b} = \frac{a \cdot 3}{b \cdot 3}$

\frac{5}{1 2} = ? \frac{1 5}{6 0} + \frac{5}{3 0}