Need help with algebra exponent problem

[150mph]

I don't know what this kind of problem is called, looked through all my notes and can't find where or if I learned it previously, and struggling with the solution process. Although I gave it a try (2nd img.), I didnt get very far...

 

[Deleted member 4993]

I don't know what this kind of problem is called, looked through all my notes and can't find where or if I learned it previously, and struggling with the solution process. Although I gave it a try (2nd img.), I didnt get very far...View attachment 15341View attachment 15342

Can you simplify:

(y - 1)² = 2y² - y - 29

2y² - y - 29 - (y - 1)² = 0

expand (y - 1)² and simplify.

What kind of equation (name) you are getting?

 

[150mph]

Thanks. Sorry, I'm a beginner and I dont follow how in your first equation you can change the problem to this? The equation name...IDK...

 

[Deleted member 4993]

Thanks. Sorry, I'm a beginner and I dont follow how in your first equation you can change the problem to this? The equation name...IDK...

(y - 1)² = 2y² - y - 29 ..................................................................This was the equation you were given

0 = 2y² - y - 29 - (y - 1)² ............................................................The terms on the left-hand-side [(y-1)²] was brought to the right-hand-side

Now expand (y - 1)² and simplify.

 

[Dr.Peterson]

I don't know what this kind of problem is called, looked through all my notes and can't find where or if I learned it previously, and struggling with the solution process. Although I gave it a try (2nd img.), I didnt get very far...

Please note that [MATH]2y^2[/MATH] does not mean [MATH](2y)^2 = (2y)(2y)[/MATH]. You changed it to [MATH]4y^2[/MATH], which clearly is not the same as [MATH]2y^2[/MATH].

This is a common mistake for beginners, so it's important to learn early not to do it.

 

[150mph]

I don't remember learning a rule that allows moving part of the equation to the other side of the equal sign, changing it to subtraction(?) and replacing it with zero. (What is this process called?) No wonder I'm confused. If I knew what this type of problem was called, I could research though... Im a beginner, and didnt get far with "expand, simplify"

0 = 2y² - y - 29 - (y - 1)²
0 = (2y)(2y) - y - 29 - (y - 1) (y - 1)
0 = (2y)(2y) - 1y - 29- (1y - 1)(1y -1)

btw, is 2y² not the same as (2y)(2y) ? 2y² = 4y then?

 

[Deleted member 4993]

I don't remember learning a rule that allows moving part of the equation to the other side of the equal sign, changing it to subtraction(?) and replacing it with zero. (What is this process called?) No wonder I'm confused. If I knew what this type of problem was called, I could research though... Im a beginner, and didnt get far with "expand, simplify"

0 = 2y² - y - 29 - (y - 1)²
0 = (2y)(2y) - y - 29 - (y - 1) (y - 1)
0 = (2y)(2y) - 1y - 29- (1y - 1)(1y -1)

btw, is 2y² not the same as (2y)(2y) ? 2y² = 4y then?

Can you solve for x from the following equation:

2 * x + 6 = 11

If you can, please post the solution step-by-step.

 

[150mph]

IDK what the "*" represents... but here's what I guessed if its just a multiply symbol:

2x+6 - 6 = 11- 6
2x= 5
2x divided by 2 = 5 divided by 2
x = 2.5

 

[lev888]

I don't remember learning a rule that allows moving part of the equation to the other side of the equal sign, changing it to subtraction(?) and replacing it with zero.

Are you sure? Then what did you do here?
2x + 6 = 11
2x+6 - 6 = 11- 6
You moved 6 from the left hand side to the right, changed the sign and replaced it with zero (6-6).

 

[Deleted member 4993]

IDK what the "*" represents... but here's what I guessed if its just a multiply symbol:

2x+6 - 6 = 11- 6
2x= 5
2x divided by 2 = 5 divided by 2
x = 2.5

"*" represents multiplication symbol - to avoid confusion with 'x' or '\(\displaystyle \cdot\)'

You solved the problem correctly.

Now tell me - why did you move '6' from one side to the other side with a "-" sign (your question in response #6 that said " moving part of the equation to the other side of the equal sign, changing it to subtraction...."). I did exactly the same operation in response #4 (subtracted (y-1)² from both sides). The left-hand-side becomes 0 after the subtraction.

Now about the expansion:

do you know the formula:

(a + b)² = (a + b) * (a + b) = a * (a + b) + b * (a + b) = a * a + a * b + a * b + b * b = a² + 2*a*b + b²

Do you see that? Please work with these problems with pencil/paper - don't just stare at the screen!

 

[Harry_the_cat]

Are you sure? Then what did you do here?
2x + 6 = 11
2x+6 - 6 = 11- 6
You moved 6 from the left hand side to the right, changed the sign and replaced it with zero (6-6).

What you actually did was subtract 6 from both sides.

 

[150mph]

Ok that makes response #4 clearer to see a step was omitted, because the rule I’ve used so far allows getting rid of a single number on one side of an equals sign by subtracting it from one side and adding it to the other, or vice versa. I don’t think I’d learned yet it also allows moving whole or partial equations like you did

I had some simplistic exposure to the expansion formula - one more thing to try to commit to memory…

Ps - I struggle trying to understand problems on paper for some time before I give up and take the long route to type out my issues and ask for help online

0 = 2y² -y -29 -(y - 1)²
0 = (2y)(2y) -y -29 -(y-1)(y-1)

IDK how to proceed, but here’s a try:

0 = 4y -y - 29 -(y-1)(y-1) ------- don’t understand what expanding (y - 1)² does for me here
0 = 3y -29 -(y-1)(y-1)
0 +29 = 3y -29 +29 -(y-1)(y-1)
29 = 3y -(y-1)(y-1)

 

[lev888]

0 = 4y -y - 29 -(y-1)(y-1) ------- don’t understand what expanding (y - 1)2 does for me here

You don't know what it does or how to do it? Please look up the formula. You'll be able to further simplify the right side.

 

[150mph]

You don't know what it does or how to do it?

I expanded (y-1)² into (y-1)(y-1) as suggested, I just don't understand why it was suggested because I don't know how to use the expanded version to solve this problem. If the goal after my last equation is to simplify, isn't (y - 1)² the simplest?

 

[lev888]

I expanded (y-1)² into (y-1)(y-1) as suggested, I just don't understand why it was suggested because I don't know how to use the expanded version to solve this problem. If the goal after my last equation is to simplify, isn't (y - 1)² the simplest?

That's not quite what we had in mind. Please read this:

Simplifying (a+b)^2 and (a-b)^2

In one step, you should be able to go from (a+b)^2 to a^2 + 2ab + b^2 , and from (a-b)^2 to a^2 - 2ab + b^2 . Free, unlimited, online practice. Worksheet generator. Time yourself for mastery (Algebra Pinball)!

www.onemathematicalcat.org

 

[150mph]

I'll try again from where I left off, replacing my simplification of (y-1)² to -(y-1)(y-1) with that webpage's explanation, best as I can figure it:

29 = 3y - y² + y-1 + y-1 -1² or
29 = 3y - y² + 2y-1 -1²

then what

 

[lev888]

I'll try again from where I left off, replacing my simplification of (y-1)² to -(y-1)(y-1) with that webpage's explanation, best as I can figure it:

29 = 3y - y² + y-1 + y-1 -1² or
29 = 3y - y² + 2y-1 -1²

then what

Continue - group like terms, etc. You'll end up with a quadratic equation. Solve it and you are done.

 

[Harry_the_cat]

You don't have to "take \(\displaystyle (y-1)^2\) to the other side" before you expand. Personally, I would expand first.
\(\displaystyle (y-1)^2 = 2y^2-y-29\)
\(\displaystyle y^2 -2y+1 =2y^2 -y-29\)
\(\displaystyle 0=2y^2-y-29-y^2+2y-1\)
\(\displaystyle 0=y^2+y -30\)
Now it's in the usual form of a quadratic equation. You should be able to factorise and solve, or use the quadratic formula to solve

 

[Harry_the_cat]

Ok that makes response #4 clearer to see a step was omitted, because the rule I’ve used so far allows getting rid of a single number on one side of an equals sign by subtracting it from one side and adding it to the other, or vice versa. I don’t think I’d learned yet it also allows moving whole or partial equations like you did

I had some simplistic exposure to the expansion formula - one more thing to try to commit to memory…

Ps - I struggle trying to understand problems on paper for some time before I give up and take the long route to type out my issues and ask for help online

0 = 2y² -y -29 -(y - 1)²
0 = (2y)(2y) -y -29 -(y-1)(y-1) ......................\(\displaystyle 2y^2 \neq (2y)(2y)\neq4y\)

IDK how to proceed, but here’s a try:

0 = 4y -y - 29 -(y-1)(y-1) ------- don’t understand what expanding (y - 1)² does for me here
0 = 3y -29 -(y-1)(y-1)
0 +29 = 3y -29 +29 -(y-1)(y-1)
29 = 3y -(y-1)(y-1)

see red comment above