# Help solving a linear equation in one variable

15+1/2=0.60(20x)

#### MarkFL

Staff member
Re: pre-algebra help

Hello and welcome to MHB!

Do you have any thoughts on how you should begin?

#### kmooney

##### New member
Re: pre-algebra help

you need to put the 1/2x on both sides

#### Jameson

Staff member
Re: pre-algebra help

you need to put the 1/2x on both sides
Hi kmooney,

Just to confirm, is this the equation?

$$\displaystyle 15+\frac{1}{2}=0.60(20x)$$

Or is it $$\displaystyle 15+\frac{1}{2x}=0.60(20x)$$

#### kmooney

##### New member
Re: pre-algebra help

the first one

#### MarkFL

Staff member
Re: pre-algebra help

Do you mean $$\displaystyle \frac{1}{2}x$$ or $$\displaystyle \frac{1}{2x}$$?

And where in the original equation do you see anything like that? If it were me, I would combine terms on the left side and distribute the 0.60 on the right. What do you get after doing this?

#### kmooney

##### New member
Re: pre-algebra help

1/2

#### MarkFL

Staff member
Re: pre-algebra help

I would simply add the 1/2 to the 15, and multiply the 0.60 with the 20...what does the equation look like now?

edit: I have retitled this topic to indicate the type of problem for which help is being asked. A topic title ideally will indicate the nature of the problem being discussed.

#### kmooney

##### New member
Re: pre-algebra help

ok so I take 15+1/2=.60*20=15 1/2x=12x does this look right

#### MarkFL

Staff member
The right side of the equation is correct as $0.60\cdot20x=12x$, but how did you get an $x$ on the left side?

#### kmooney

##### New member
o sorry hit it by mistake

#### MarkFL

Staff member
Okay, no worries, so we now have:

$$\displaystyle 15+\frac{1}{2}=12x$$

What do we get when we perform the addition on the left side?

#### kmooney

##### New member
put the 12 over the same denominator

15/1+1/2=16/3

#### MarkFL

Staff member
No, we are only want to add the $15$ to the $$\displaystyle \frac{1}{2}$$, so how can we write $15$ so that it is some number over $2$? In other words, how can we get a common denominator on the left side?

#### MarkFL

Staff member
15/1+1/2=16/3
No, while we can mutiply across when multiplying fractions, we cannot add across when adding them. We need to get a common denominator, then put the sum of the numerators over this common denominator. For example:

$$\displaystyle 3+\frac{1}{3}=\frac{9}{3}+\frac{1}{3}=\frac{9+1}{3}=\frac{10}{3}$$

#### tutor

##### Banned
15+ 1/2 =0.60(20x)
multiply 0.60(20x)=12.00x
The answer be 12x
15+ 1/2=12x

15=12x-1/2 hence a+b=c then a=c-b

find LCM of 2 and 1

then [24x-1]/2

15=[24x-1]/2

15*2=24x-1 hence a/b=c then a=bc
30=24x-1
31=24x
x=31/24

#### MarkFL

Staff member
15+ 1/2 =0.60(20x)
multiply 0.60(20x)=12.00x
The answer be 12x
15+ 1/2=12x

15=12x-1/2 hence a+b=c then a=c-b

find LCM of 2 and 1

then [24x-1]/2

15=[24x-1]/2

15*2=24x-1 hence a/b=c then a=bc
30=24x-1
31=24x
x=31/24
Why would you first subtract through by $\dfrac{1}{2}$? It is okay, but seems overly complicated, to then have to add it back after multiplying through by 2.

I would proceed as follows:

$$\displaystyle 15+\frac{1}{2}=12x$$

Combine terms on the left:

$$\displaystyle \frac{31}{2}=12x$$

Multiply through by 2:

$$\displaystyle 31=24x$$

Divide through by 24, and arrange as:

$$\displaystyle x=\frac{31}{24}$$

#### alane1994

##### Active member
I realize that this problem has thoroughly been answered, but this stuff is fun for me!

$$\displaystyle 15+\frac{1}{2}=0.60(20x)$$

I would begin by first multiplying the $$\displaystyle 20x$$ by the $$\displaystyle 0.60$$.
As such we would have,

$$\displaystyle 15+\frac{1}{2}=12x$$

Then we would need to condense the left side of the equation.

$$\displaystyle \frac{31}{2}=12x$$

Then I would divide the 12 over.

$$\displaystyle \frac{1}{12}\times(\frac{31}{2})=(12x)\times \frac{1}{12}$$

Then we would get the final answer that everyone else has arrived at!

$$\displaystyle x=\frac{31}{24}$$

#### gjay822

##### New member
Solve for x and then your need to distribute the numbers to get the value of the variable.