# Thread: solve each equation algebraically

1. ## solve each equation algebraically

1/3x^2-5/6=1

3x√ x-1 +2√ (x-1)^3=0 ......... square root symbol is over the x-1 and x-1^3

x^3-26x^3/2=27

2x^2(2x-1)^1/3=4x(2x-1)^4/3

15/2^x-1=7

3^x-1 + 3^x+1 =30

Any help on any of thse problems would be appreciated!

2. ## Re: solve each equation algebraically

I can't read the first, third, fourth, fifth and sixth questions. Some brackets would be nice... LaTeX would be even nicer...

\displaystyle \displaystyle \begin{align*} 3x\sqrt{x-1} + 2\sqrt{(x - 1)^3} &= 0 \\ 3x\sqrt{x - 1} + 2\sqrt{x-1}\sqrt{(x - 1)^2} &= 0 \\ 3x\sqrt{x - 1} + 2\sqrt{x - 1}|x - 1| &= 0 \\ \sqrt{x - 1}\left(3x + 2|x - 1|\right) &= 0 \\ \sqrt{x - 1} = 0 \textrm{ or } 3x + 2|x - 1| &= 0 \end{align*}

Go from here...

3. ## Re: solve each equation algebraically

Hello, etran3358!

I'll guess what the last one means . . .

3^x-1+3^x+1=30

3^x/3+3.3^x=30

10/3.3^x=30

3^x=9

3^x=3^2

x=2

If my solution is hard to read and understand,
it's because I typed it the way you did ...
no parentheses, no spaces.

This is what I meant . . .

We are given: .3^{x-1} + 3^{x+1} = 30

which equals: .(3^x)/3 + 3(3^x) = 30

add: .(10/3)(3^x) = 30

Multiply by 3/10: .3^x = 9

Hence: .3^x = 3^2

Therefore: .x = 2

4. ## Re: solve each equation algebraically

$\displaystyle 1/3 x^2-5/6x=1$ is #1

5. ## Re: solve each equation algebraically

#3 is $\displaystyle x^3 -26x ^ 3/^ 2 = 27$ it's 26 x raished to the 3/2 power

6. ## Re: solve each equation algebraically

$\displaystyle 2x^2 (2x-1) ^1 / ^3 = 4x(2x-1) ^ 4 / ^3$

7. ## Re: solve each equation algebraically

$\displaystyle 15 / 2 ^x -1 =7$

8. ## Re: solve each equation algebraically

again i apologize if it's hard to read/understand ... i'm new to this site

also the last problem reads as 15 ALL OVER 2^x -1 which is equal to 7

any help would be well appreciated!

10. ## Re: solve each equation algebraically

Hello, etran3358!

You still need parentheses,
or learn to use LaTeX more accurately.

$\displaystyle 1/3 x^2-5/6x=1$ is #1.

Can you see that this looks like: .$\displaystyle \frac{1}{3x^2} -= \frac{5}{6x} \:=\:1\,?$

But it might be: .$\displaystyle \frac{1}{3}x^2 - \frac{5}{6}x \:=\:1$

11. ## Re: solve each equation algebraically

Hello, etran3358!

$\displaystyle 2x^2 (2x-1) ^{\frac{1}{3}} \:=\: 4x(2x-1) ^{\frac{4}{3}}$

We have: .$\displaystyle 2x^2(2x-1)^{\frac{1}{3}} - 4x(2x-1)^{\frac{4}{3}} \:=\:0$

Factor: . .$\displaystyle 2x\cdot(2x-1)^{\frac{1}{3}}\big[x - 2(2x-1)\big] \:=\:0$

. . . . . . . .$\displaystyle 2x\cdot(2x-1)^{\frac{1}{3}}\cdot(2-3x) \:=\:0$

Therefore:

. . $\displaystyle 2x \:=\:0 \quad\Rightarrow\quad x \:=\:0$

. . $\displaystyle (2x-1)^{\frac{1}{3}} \:=\:0 \quad\Rightarrow\quad 2x-1 \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{1}{2}$

. . $\displaystyle 2 - 3x \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{2}{3}$