Subtract \(\displaystyle 3y\) from both sides.

\(\displaystyle 3y-7-3y = 4y-3y\)

Simplify both sides.

\(\displaystyle -7 =y\)

The answer is:  \(\displaystyle -7\)

Multiply both sides by the least common denominator to cancel out the fractions.

\(\displaystyle 6(\frac{1}{2}x +\frac{7}{3}) = \frac{8}{3}\cdot6\)

\(\displaystyle 3x+14 = 16\)

Subtract 14 on both sides.

\(\displaystyle 3x+14 -14= 16-14\)

\(\displaystyle 3x=2\)

Divide by 3 on both sides.

\(\displaystyle \frac{3x}{3}=\frac{2}{3}\)

The answer is:  \(\displaystyle \frac{2}{3}\)

Given the following equation, find the value of b when \(\displaystyle w=-3\)

\(\displaystyle b-2=\frac{5w+12}{3}+4\)

This may look a little messy up front, but don't worry, it's not overly complicated.

All we need to do is plug in -3 for w and solve for b.

Let's begin:

\(\displaystyle b-2=\frac{5(-3)+12}{3}+4\)

\(\displaystyle b-2=\frac{-15+12}{3}+4\)

Now add our numbers:

\(\displaystyle b-2=\frac{-3}{3}+4\)

Now we can simplify our fraction to get:

\(\displaystyle b-2=-1+4\)

From her on out it is just subtraction and addition:

\(\displaystyle b-2=3\)

\(\displaystyle b=3+2=5\)

So our answer is 5

\(\displaystyle \sqrt{3x+4}-2=6\)

Start by adding \(\displaystyle 2\) to both sides.

\(\displaystyle \sqrt{3x+4}=8\)

Next, square both sides to get rid of the square root.

\(\displaystyle 3x+4=64\)

Subtract \(\displaystyle 4\) from both sides.

\(\displaystyle 3x=60\)

Finally, divide both sides by \(\displaystyle 3\).

\(\displaystyle x=20\)

\(\displaystyle \sqrt{3x-1}+5=9\)

Start by subtracting \(\displaystyle 5\) from both sides.

\(\displaystyle \sqrt{3x-1}=4\)

Now, square both sides of the equation to get rid of the square root.

\(\displaystyle 3x-1=16\)

Add \(\displaystyle 1\) to both sides.

\(\displaystyle 3x=17\)

Divide both sides by \(\displaystyle 3\).

\(\displaystyle x=\frac{17}{3}\)

\(\displaystyle \sqrt{x+15}-2=4\)

Start by adding \(\displaystyle 2\) to both sides.

\(\displaystyle \sqrt{x+15}=6\)

Next, square both sides to get rid of the square root.

\(\displaystyle x+15=36\)

Finally, subtract \(\displaystyle 15\) from both sides.

\(\displaystyle x=21\)

To solve for \(\displaystyle \small x\), all you have to do is simplify your equation until you get a number on one side and your \(\displaystyle \small x\) on the other.

So for starters we already have all of our numbers on one side: \(\displaystyle \small x=7+5-2\). Now all we need to do is add and subtract when called for in order to get our answer.

Start by adding \(\displaystyle \small 7\) to \(\displaystyle \small 5\)

\(\displaystyle \small x=(7+5)-2\)

\(\displaystyle \small x=12-2\)

Now subtract \(\displaystyle \small 2\) from our newly formed \(\displaystyle \small 12\)

\(\displaystyle \small x=10\)

Your answer is \(\displaystyle \small x=10\)

In order to solve for \(\displaystyle \small x\), all you have to do is simplify your equation so that your \(\displaystyle \small x\) is on one side and your number is on the other.

For this equation, we have one number accompanying our \(\displaystyle \small x\). That is \(\displaystyle \small -6\): \(\displaystyle \small x-6=5\cdot3\).

Let's move that \(\displaystyle \small -6\) over to the other side by adding it on both sides.

\(\displaystyle \small x=5\cdot3-6\)

Now that all of our numbers are on one side, it's time to simplify. Multiplication comes before addition, so we need to multiply the \(\displaystyle \small 5\) and \(\displaystyle \small 3\) together before we touch the \(\displaystyle \small 6\).

\(\displaystyle \small x=(5\cdot 3)-6\)

\(\displaystyle \small x=15-6\)

Now we can add the \(\displaystyle \small 6\) to \(\displaystyle \small 15\).

\(\displaystyle \small x=21\)

Our answer is \(\displaystyle \small x=21\)

In order to solve for \(\displaystyle \small x\), we must simplify our equation so that we have \(\displaystyle \small x\) on one side and our final number on the other side. 

Before we can do any simplifying, we must get \(\displaystyle \small x\) by itself. Currently our \(\displaystyle \small x\) is being added by \(\displaystyle \small 6\), but it's also being divided by \(\displaystyle \small 2\). We can't touch the \(\displaystyle \small +6\) yet, so let's move the \(\displaystyle \small 2\) by multiplying both sides.

\(\displaystyle \small \frac{x+6}{2}=6\)

\(\displaystyle \small x+6=6\cdot2\)

Now let's move that \(\displaystyle \small 6\) on the \(\displaystyle \small x\) side by subtracting it on both sides.

\(\displaystyle \small x+6=6\cdot2\)

\(\displaystyle \small x=6\cdot2-6\)

Now that we have all our numbers to one side, it's time to start simplifying. Multiplication comes before subtraction, so we're going to multiply our \(\displaystyle \small 6\) and \(\displaystyle \small 2\) together.

\(\displaystyle \small x=(6\cdot 2)-6\)

\(\displaystyle \small x=12-6\)

Now we can subtract our \(\displaystyle \small 6\) from \(\displaystyle \small 12\).

\(\displaystyle \small x=12-6\)

\(\displaystyle \small x=6\)

Our answer is \(\displaystyle \small x=6\)

Solve this equation for \(\displaystyle x\) by isolating the \(\displaystyle x\) on the left side of the equation. This can be done by first, subtracting \(\displaystyle 3x\) from both sides:

\(\displaystyle 6x- 17 = 3x\)

\(\displaystyle 6x- 17- 3x = 3x - 3x\)

\(\displaystyle 6x- 3x - 17 =0\)

Collect like terms by subtracting coefficients of \(\displaystyle x\):

\(\displaystyle (6-3)x-17=0\)

\(\displaystyle 3x - 17 =0\)

Add 17 to both sides:

\(\displaystyle 3x - 17+ 17 =0+ 17\)

\(\displaystyle 3x= 17\)

Divide both sides by 3:

\(\displaystyle 3x \div 3= 17 \div 3\)

\(\displaystyle x = \frac{17}{3}= 5 \frac{2}{3}\),

the correct choice.