\(\displaystyle 8x-5-4x=-x+10\) 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:

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

\(\displaystyle 9=3x\)

\(\displaystyle 3=x\)