To isolate \(\displaystyle x\) on one side, subtract seven from both sides

\(\displaystyle X+7\left- ( 7\right )=45\left- ( 7\right )\)

\(\displaystyle X=45-7=38\)

To solve the equation for \(\displaystyle Y\), you must isolate \(\displaystyle Y\) on one side.  Here you can divide both sides by 3.

\(\displaystyle 3Y=6\)

\(\displaystyle \frac{3Y}{3}=\frac{69}{3}\)

\(\displaystyle Y=23\)

To solve equations, you must perform the same operations on both sides.

\(\displaystyle n+5=11\)

Subtract 5 from both sides.

\(\displaystyle n+5-5=11-5\)

\(\displaystyle n=6\)

\(\displaystyle -5x = 25\)

In order to solve a one step equation, we want to isolate the variable. To do this, we pay special attention to what is being done to the variable. In this case we are multiplying the variable by \(\displaystyle -5\). To "undo" this, we use the inverse of multiplication, which is division; thus we divide by \(\displaystyle -5\) on both sides. Note that we are dividing by \(\displaystyle -5\) since that is what is being multiplied; don't be tricked into dividing by \(\displaystyle +5\).

\(\displaystyle \frac{-5x}{-5}=\frac{25}{-5}\)

\(\displaystyle x=-5\)

Dividing both sides by \(\displaystyle -5\), we get \(\displaystyle x=-5\).

To solve for  we must get all of the numbers on the other side of the equation as .

To do this in a problem where  is being multiplied by a number, we must divide both sides of the equation by the number.

In this case the number is \(\displaystyle 3\) so we divide each side of the equation by \(\displaystyle 3\) to make it look like this \(\displaystyle \frac{3x}{3}=\frac{45}{3}\).

The \(\displaystyle 3\)s cancel to leave  by itself.

Then we perform the necessary division to get the answer of \(\displaystyle x=15\).

To solve for  we must get all of the numbers on the other side of the equation as .

To do this in a problem where  is being subtracted by a number, we must add the number to both sides of the equation.

In this case the number is \(\displaystyle 18\) so we add \(\displaystyle 18\) to each side of the equation to make it look like this \(\displaystyle x-18+18=24+18\)

The \(\displaystyle 18\)s cancel on the left side and leave  by itself.

Then we perform the necessary addition to get the answer of \(\displaystyle x=42\).

Remembering the final goal is to get the \(\displaystyle x\) alone on one side of the equation.

To get rid of the \(\displaystyle -5\), we perform the opposite operation. We add \(\displaystyle 5\) to each side of the equation, cancelling out the \(\displaystyle 5\) on the left side.

This brings the equation to \(\displaystyle x =25\). 

If \(\displaystyle x=7\) you put \(\displaystyle 7\) in for \(\displaystyle x\). Therefore it becomes \(\displaystyle 7^{2}\)\(\displaystyle +5\).

 \(\displaystyle 7 \times 7= 49\)

Finally,\(\displaystyle 49 +5 =54\)

First you multiply the coefficients \(\displaystyle (2\cdot 2)\).

Then you multiply \(\displaystyle x^{3} \cdot x^4\) According to the laws of exponents, when you multiply two numbers with similar bases (both \(\displaystyle x\) in this case) then you simply add the exponents \(\displaystyle (3+4 =7)\).

Therefore, the final answer is \(\displaystyle 4x^7\). 

To solve for \(\displaystyle x\) we must get all of the numbers on the other side of the equation as \(\displaystyle x\).

To do this in a problem where a number is being added to \(\displaystyle x\), we must subtract the number from both sides of the equation.

In this case the number is \(\displaystyle 5\) so we subtract \(\displaystyle 5\) from each side of the equation to make it look like this \(\displaystyle x+5-5=15-5\)

The \(\displaystyle 5\)'s on the left side cancel to get \(\displaystyle x=15-5\)

Then we perform the necessary subtraction to get the answer of  \(\displaystyle x=10\).