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

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.

Solve this equation for \(\displaystyle x\) by isolating the \(\displaystyle x\) on the left side of the equation. This can be done by first,multiplying 5 by each expression within the parentheses:

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

\(\displaystyle 5\cdot x+5 \cdot 3= 2x\)

\(\displaystyle 5x+15= 2x\)

Subtract \(\displaystyle 2x\) from both sides:

\(\displaystyle 5x+15-2x= 2x - 2x\)

\(\displaystyle 5x-2x+15= 0\)

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

\(\displaystyle (5-2)x+15= 0\)

\(\displaystyle 3x+15 = 0\)

Subtract 15 from both sides:

\(\displaystyle 3x+15-15 = 0 - 15\)

\(\displaystyle 3x= -15\)

Divide both sides by 3:

\(\displaystyle 3x \div 3 = -15 \div 3\)

\(\displaystyle x= -5\),

the correct response.

\(\displaystyle \sqrt{3x-3}+5=4\)

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

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

Square both sides of the equation to get rid of the square root.

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

Add \(\displaystyle 3\) to both sides.

\(\displaystyle 3x=4\)

Divide both sides of the equation by \(\displaystyle 3\).

\(\displaystyle x=\frac{4}{3}\)

\(\displaystyle \sqrt{2x+5}-12=10\)

Start by adding \(\displaystyle 12\) to both sides.

\(\displaystyle \sqrt{2x+5}=22\)

Square both sides of the equation to get rid of the square root.

\(\displaystyle 2x+5=484\)

Subtract \(\displaystyle 5\) from both sides.

\(\displaystyle 2x=479\)

Divide both sides by \(\displaystyle 2\).

\(\displaystyle x=239.5\)

Rearrange the following equation so that it is solved for "b"

\(\displaystyle \frac{3b}{4x}=12y+6t\)

This problem may look intimidating, but don't be overwhelmed! Read the problem carefully, all we need to do is get the b all by itself.

To do this, let's first multiply both sides by 4x.

\(\displaystyle (4x)\frac{3b}{4x}=12y+6t(4x)\rightarrow 3b=(4x)(12y+6t)\)

Next, we simply need to divide both sides by 3 to get the b all by itself.

\(\displaystyle (\div 3)3b=(4x)(12y+6t)(\div3)\rightarrow b=\frac{4x(12y+6t)}{3}\)

\(\displaystyle b=\frac{4x(12y+6t)}{3}\)

One last thing, we can simplify the denominator and get rid of our three by dividing a three out of the 12 and the 6.

This yields:

\(\displaystyle b=4x(4y+2t)\)

Starting with

\(\displaystyle \frac{x+7}{2x-1}=3\)

We must get rid of the fraction by multiplying both sides by (2x-1)

\(\displaystyle \frac{x+7}{2x-1}(2x-1)=3(2x-1)\)

The terms cancel on the right, and we must distribute on the left

\(\displaystyle x+7=3\cdot2x-3\cdot1\)

Simplify 

\(\displaystyle x+7=6x-3\)

Subtract x from both sides to get the variables all on one side

\(\displaystyle x+7-x=6x-3-x\)

combine like terms

\(\displaystyle 7=5x-3\)

Now add 3 to both sides to get the term with the variable by itself

\(\displaystyle 7+3=5x-3+3\)

Simplify

\(\displaystyle 10=5x\)

And finally, divide both side by 5 to get the variable all by itself

\(\displaystyle \frac{10}{5}=\frac{5x}{5}\)

\(\displaystyle 2=x\)