# 8. Equations Involving Fractions

In this section, we can find the solution easily by multiplying throughout by the **lowest common denominator**** (LCD)****.** This will simplify our equation and make it easier to solve.

We need to remember to **multiply all terms** in the equation (both sides of the equal sign) by the LCD, otherwise the final answer won't be correct.

We start with some algebraic examples, then follow with some word problems involving fractions.

## a. Algebraic types

### Example 1

Solve for *x*:

`x/5+3/10 = 1/2`

Answer

We first look at the **denominators** of the fractions and determine the lowest common denominator. In this case, it will be `10.`

We multiply throughout by `10` and the result is:

`10xx(x/5+3/10) = 10xx(1/2)`

`2x+3=5`

Then, we just solve the simpler equation we've found.

Subtract `3` from both sides:

`2x=2`

Divide boths sides by `2`, giving us the final answer:

`x=1`

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### Example 2

Solve for *x*:

`(2x)/3+2/5 = 8+x/2`

Answer

Once again, we look at the **denominators** of the fractions and determine the lowest common denominator. In this case, it will be `30.`

We multiply throughout by `30` and the result is:

`30xx((2x)/3+2/5) = 30xx(8+x/2)`

`20x+12=240+15x`

Subtracting `12` from both sides:

`20x=228+15x`

Substract `15x` from both sides:

`5x=228`

Divide both sides by `5`, giving us the final answer:

`x=228/5=45.6`

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### Example 3

This next one has the variable in the **denominator**. We'll still need to find the LCD as before.

Solve for *x*:

`4/x+1/3 = 7/(5x)`

Answer

In this case, the lowest common denominator is `15x.`

We multiply throughout by `15x` and the result is:

`15x xx(4/x+1/3) = 15x xx(7/(5x))`

`60+5x=21`

Subtracting `60` from both sides:

`5x=-39`

Divide both sides by `5`, giving us the final answer:

`x=-39/5=-7.8`

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## b. Word problems

### Example 4

An aquarium can be filled by one hose in 7 minutes and a second thinner hose in 10 minutes.

How long will it take to fill the tank if both hoses operate together?

Answer

Since each hose makes the filling time **less**, we have
to add the reciprocals together and take the reciprocal of the result.

We need to use:

`1/T=1/T_1+1/T_2`

So we have:

`1/T=1/7+1/10`

`1/T=(10+7)/70=17/70`

So

`T=70/17=4.1176`

So it will take 4.1 minutes to fill the tank with both hoses operating together.

### Example 5

### Related pages

You may find these useful:

For 2 resistors with resistances
*R*_{1}and
*R*_{2} in parallel, the combined resistance *R* is given by:

`1/R=1/R_1+1/R_2`

For a particular circuit, the combined resistance
*R* was found to be 4 ohms (Ω), and *R*_{1}= 10 Ω. Find *R*_{2}.

Answer

We have:

`1/4=1/10+1/R_2`

We need to multiply throughout by the **lowest common denominator: 20***R*_{2}

`1/4=1/10+1/R_2`

`(20R_2)/4=(20R_2)/10+(20R_2)/(R_2)`

`5R_2=2R_2+20`

`3R_2=20`

`R_2=20/3=6 2/3Omega`

"`Omega`" is the symbol for "ohms", the unit of resistance.

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### Example 6

A car averaged 30 km/h going from home to work and 40 km/h on the return journey. If the total time for the two journeys is 50 minutes, how far is it from home to work?

Answer

Let the length of the journey from home to work be *x* km.

Recall that

`text(speed) = text(distance)/text(time)`

So

`text(time) = text(distance)/text(speed)`

We must use the same time units throughout. We will use hours.

Now

` 50\ text(minutes)=50/60=5/6text(hours)`

In the forward journey, the car's time was `x/30` hours.

For the return journey, the time was `x/40` hours.

The total time was `x/30+x/40=5/6\ text(hours)`

So `(4x+3x)/120=(7x)/120=5/6`

This gives us `7x=(5xx120)/6=100`

That is `x=100/7=14.286`

So the distance from home to work is 14.3 km.

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