Help solving a linear equation in one variable

[kmooney]

15+1/2=0.60(20x)

 

[MarkFL]

Re: pre-algebra help

Hello and welcome to MHB!

Do you have any thoughts on how you should begin?

 

[kmooney]

Re: pre-algebra help

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

 

[Jameson]

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]

Re: pre-algebra help

the first one

 

[MarkFL]

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]

Re: pre-algebra help

1/2

 

[MarkFL]

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]

Re: pre-algebra help

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

 

[MarkFL]

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]

o sorry hit it by mistake

 

[MarkFL]

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]

put the 12 over the same denominator

 

[kmooney]

15/1+1/2=16/3

 

[MarkFL]

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]

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]

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]

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]

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]

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