Distribution

[chromechris]

[MATH]\frac{y-6}{4}-\frac{2y+5}{2}=-7[/MATH]
[MATH](4)\frac{y-6}{4}-(4)\frac{2y+5}{2}=(4)-7[/MATH]
Where I have an issue: ()I don't know whether to distribute [MATH(2)][/MATH] or [MATH](-2)[/MATH] to the [MATH](2y+5)[/MATH] (How do I decide what to distribute?)
[MATH]y-6-(2)2y+5=-28[/MATH]
What I want to do it, but I get a wrong answer:
[MATH]y-6-4y+10=-28[/MATH][MATH]y=\frac{32}{3}[/MATH]
A different way I tried it and got the correct answer:
[MATH]y-6-4y-10=-28[/MATH][MATH]y=4[/MATH]

 

[pka]

[MATH]\frac{y-6}{4}-\frac{2y+5}{2}=-7[/MATH][MATH](4)\frac{y-6}{4}-(4)\frac{2y+5}{2}=(4)-7[/MATH]

If you would write as \(4\left(\dfrac{y-6}{4}\right)-4\left(\dfrac{2y+5}{2}\right)=4(-7)\), I think it is clearer as to what to do.
\((y-6)-2(2y+5)=-28\) is that not clear now?

 

[chromechris]

It is, thank you!

 

[HallsofIvy]

I would not think "distribute" at all but rather "multiply on both sides by the least common denominator". Here the least common denominator is "4" so you do what pka suggested.

 

[chromechris]

I would not think "distribute" at all but rather "multiply on both sides by the least common denominator". Here the least common denominator is "4" so you do what pka suggested.

Right. I was more confused with y−6−(2)(2y+5)=−28 in regards to multiplying either [MATH]2[/MATH] vs [MATH]-2[/MATH] to the [MATH]2y+5[/MATH]...................................(edited - those parentheses are important)