what's going on here? Solve 5(3y-4) - 10y <= 4(2y-1) - 16

[allegansveritatem]

Here is the problem:

13. Solve \(\displaystyle 5\, (3y\, -\, 4)\, -\, 10y\, \leq\, 4\, (2y\, -\, 1)\, -\, 16\)

Here is what I did with it:

. . . . .\(\displaystyle 5\, (3y\, -\, 4)\, -\, 10y\, \leq\, 4\, (2y\, -\, 1)\, -\, 16\)

. . . . .\(\displaystyle 15y\, -\, 20\, -\, 10y\, \leq\, 8y\, -\, 4\, -\, 16\)

. . . . .\(\displaystyle 5y\, -\, 20\, \leq\, 8y\, -\, 20\)

. . . . .\(\displaystyle 5y\, \leq\, 8y\)

. . . . .\(\displaystyle y\, \leq\, \dfrac{8}{5}\, y\)

My question is: Is my work correct? And if so, did I carry it far enough? If not, how to carry it further. Or, is my work correct but the book answer simply states my solution in another form?

 

[Steven G]

Here is the problem:

13. Solve \(\displaystyle 5\, (3y\, -\, 4)\, -\, 10y\, \leq\, 4\, (2y\, -\, 1)\, -\, 16\)

Here is what I did with it:

. . . . .\(\displaystyle 5\, (3y\, -\, 4)\, -\, 10y\, \leq\, 4\, (2y\, -\, 1)\, -\, 16\)

. . . . .\(\displaystyle 15y\, -\, 20\, -\, 10y\, \leq\, 8y\, -\, 4\, -\, 16\)

. . . . .\(\displaystyle 5y\, -\, 20\, \leq\, 8y\, -\, 20\)

. . . . .\(\displaystyle 5y\, \leq\, 8y\)

. . . . .\(\displaystyle y\, \leq\, \dfrac{8}{5}\, y\)

My question is: Is my work correct? And if so, did I carry it far enough? If not, how to carry it further. Or, is my work correct but the book answer simply states my solution in another form?

When you solve for y, in the end there should only be a y on one side of the equal or inequality sign. So continue...

Alternatively you can think about what your result is saying: y<=(8/5)y. Now 8/5 >1. So you inequality is saying that y is less than or equal to (8/5)y. If y<0, that inequality is wrong. If y=0, then the inequality is correct. If y>0, then the inequality is correct. Hence the answer is y>=0

 

[Dr.Peterson]

Here is the problem:

13. Solve \(\displaystyle 5\, (3y\, -\, 4)\, -\, 10y\, \leq\, 4\, (2y\, -\, 1)\, -\, 16\)

Here is what I did with it:

. . . . .\(\displaystyle 5\, (3y\, -\, 4)\, -\, 10y\, \leq\, 4\, (2y\, -\, 1)\, -\, 16\)

. . . . .\(\displaystyle 15y\, -\, 20\, -\, 10y\, \leq\, 8y\, -\, 4\, -\, 16\)

. . . . .\(\displaystyle 5y\, -\, 20\, \leq\, 8y\, -\, 20\)

. . . . .\(\displaystyle 5y\, \leq\, 8y\)

. . . . .\(\displaystyle y\, \leq\, \dfrac{8}{5}\, y\)

My question is: Is my work correct? And if so, did I carry it far enough? If not, how to carry it further. Or, is my work correct but the book answer simply states my solution in another form?

You must go further. Collect terms with y on one side: y <= (8/5)y becomes 0 <= (8/5)y - y, which is 0 <= (3/5)y. What do you do next?

 

[Harry_the_cat]

Going back to line \(\displaystyle 5y\leq 8y\) , you should continue with:

\(\displaystyle 0\leq3y\), subtracting 5y from both sides

\(\displaystyle y\geq0\), dividing both sides by 3 and rewriting in conventional order​

​

 

[allegansveritatem]

When you solve for y, in the end there should only be a y on one side of the equal or inequality sign. So continue...

Alternatively you can think about what your result is saying: y<=(8/5)y. Now 8/5 >1. So you inequality is saying that y is less than or equal to (8/5)y. If y<0, that inequality is wrong. If y=0, then the inequality is correct. If y>0, then the inequality is correct. Hence the answer is y>=0

I see it now. I don't know why I stopped short of isolating the variable. Thanks for the tip

 

[allegansveritatem]

You must go further. Collect terms with y on one side: y <= (8/5)y becomes 0 <= (8/5)y - y, which is 0 <= (3/5)y. What do you do next?

I would divide both sides by 3/5. Yes, I see it now. Somehow I was blind to this while doing the problem. It could have something to do with not wanting to fool with that zero. Thanks for pointing the way here.

 

[allegansveritatem]

Going back to line \(\displaystyle 5y\leq 8y\) , you should continue with:

\(\displaystyle 0\leq3y\), subtracting 5y from both sides

\(\displaystyle y\geq0\), dividing both sides by 3 and rewriting in conventional order​

​

Yes, I see that now. Thanks.