$\displaystyle 8x-5-4x=-x+10$ can be simplified to become

$\displaystyle 4x-5=-x+10$

Then, you can further simplify by adding 5 and $\displaystyle x$ to both sides to get $\displaystyle 5x=15$.

Then, you can divide both sides by 5 to get $\displaystyle x=3$.

Turn the word problem into math.  Start at the end--what is the greatest negative integer?  -1!

$\displaystyle (-1)^2\cdot4\cdot\frac{1}{2}+8-6=1\cdot4\cdot\frac{1}{2}+8-6=2+8-6=4$

$\displaystyle 18\div\frac{1}{6}\cdot\frac{9}{72}\cdot\frac{4}{27}\cdot(-8+7)^9=18\cdot6\cdot\frac{1}{8}\cdot\frac{4}{27}\cdot-1$

Remember to cancel as you go--the 18 will cancel with the 27, the 4 will cancel with the 8, and so on:

$\displaystyle 18\cdot6\cdot\frac{1}{8}\cdot\frac{4}{27}\cdot-1= 2\cdot6\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot-1$

Continue to reduce:

$\displaystyle 2\cdot6\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot-1= 2\cdot1\cdot-1=-2$

To solve for $\displaystyle x$, you must first combine the $\displaystyle x$'s on the right side of the equation. This will give you $\displaystyle \ 6x-1=9x+8$.

Then, subtract $\displaystyle 8$ and $\displaystyle 6x$ from both sides of the equation to get $\displaystyle -9=3x$.

Finally, divide both sides by $\displaystyle 3$ to get the solution $\displaystyle x=-3$.

First, use the distributive property to simplify the right side of the equation: $\displaystyle 3x+5=6x-4$. Then, subtract $\displaystyle 3x$ and add 4 to both sides to separate the $\displaystyle x$'s and the integers to get $\displaystyle 9=3x$. Divide both sides by 3 to get $\displaystyle x=3$.

To solve for $\displaystyle x$, first divide both sides by $\displaystyle a$: $\displaystyle x+b=\frac{c}{a}$. Then, subtract both sides by $\displaystyle b$ to get $\displaystyle x=\frac{c}{a}-b$.

$\displaystyle (4^2)^x= 4^{2x}=64$

$\displaystyle 64= 4\cdot 4\cdot 4=4^3$

$\displaystyle 4^{2x}=4^3$

$\displaystyle 2x = 3$

$\displaystyle x = \frac{3}2{}$

1.  First simplify the first expression:

$\displaystyle \frac{4^2 +6\cdot 9}{4+3(2)}=\frac{4\cdot 4+54}{4+6}=\frac{16+54}{10}=\frac{70}{10}=7$

2.  Then, simplify the next two expressions:

$\displaystyle \frac{7}{14}\cdot\frac{50}{25}=\frac{1}{2}\cdot\frac{2}{1}=1$

3.  Finally, add and subtract:

$\displaystyle 7+1-1=7$

1.  First solve for the numerator by plugging in -2 for x:

$\displaystyle (-2)^2-(-2)^3-(-(-2)-4)=4+8+2= 14$

2.  Then, solve the denominator by combining the fractions:

$\displaystyle \frac{1}{x}+\frac{x}{1}= \frac{-1}{2}+\frac{-2}{1}=\frac{-1}{2}+\frac{-4}{2}=\frac{-5}{2}$

3.  Finally, "rationalize" the complex fraction by multiplying top and bottom by -2/5:

$\displaystyle \frac{14}{\frac{-5}{2}}\cdot\frac{\frac{-2}{5}}{\frac{-2}{5}}=\frac{\frac{-28}{5}}{1}=\frac{-28}{5}$

$\displaystyle \frac{x*7 +5*4+1}{y*2+4}= \frac{7x +21}{2y +4}=3$

Multiply both sides by the denominator (2y +4) to cancel it:

$\displaystyle \frac{7x +21}{2y +4}*(2y +4)=3*(2y+4)$

$\displaystyle 7x + 21=6y +12$

Now, use substitution to solve for x:

$\displaystyle \frac{x}{y}=\frac{12}{20}-->20x=12y-->10x=6y$

Substitute 10x for 6y in the first equation:

$\displaystyle 7x+21=(10x)+12$

$\displaystyle 9=3x$

$\displaystyle 3=x$