From the second equation and by solving for \(\displaystyle x\) in terms of \(\displaystyle y\) we get

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

Replacing \(\displaystyle x\) in the first equation with

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

we get \(\displaystyle 2\left ( -5 + 2y \right ) - y = 2\)

Solving for \(\displaystyle y\) one gets \(\displaystyle y = 4\)

Replacing \(\displaystyle y\) with 4 in the above equation:

\(\displaystyle -5 + 2\left ( 4 \right )=3\)

Hence the solution to the above system of linear equations is \(\displaystyle \left ( 3, 4 \right )\).

We would like to eliminate \(\displaystyle x\).

Hence we multiply the first equation by \(\displaystyle -3\) which gives us the following equations:

\(\displaystyle - 3x +9y = -9\)

\(\displaystyle 3x + 2y = -2\)

Adding the above two equations eliminates the variable \(\displaystyle x\). We are left with:

\(\displaystyle 11y = -11\)

and so \(\displaystyle y=1\)

Replacing \(\displaystyle y\) with \(\displaystyle y=-1\) in the original equation gives us

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

solving for \(\displaystyle x\) gives us

\(\displaystyle x = 0\)

Hence the solution is \(\displaystyle \left ( 0, -1 \right )\).

These two equations represent the same line and they "cross' at infinitely many points.  Therefore these systems are dependent.

 \(\displaystyle Simplify \; x^{3} + 3x^{2} - 6x + 10 -(x^{2} +x + 13)\)
\(\displaystyle \fn_cm Distribute \; -1 \; over \; x^2+x+13.\)\(\displaystyle -(x^2+x+13) \rightarrow \; -x^2-x-13\)\(\displaystyle \fn_cm \fn_cm -(x^2+x+13) \rightarrow \; =-x^2-x-13\)
\(\displaystyle Expression \; after \; distribution: \;-x^2-x-13+x^3+3 x^2-6 x+10\)
\(\displaystyle Group \; like \; terms \; in \; x^3+3 x^2-x^2-x-6 x-13+10.\)
\(\displaystyle \fn_cm \rightarrow x^3+(3 x^2-x^2)+(-6 x-x)+(10-13)\)\(\displaystyle \fn_cm \fn_cm Combine \; like \; terms \; 3 x^2-x^2\rightarrow2x^2\)

\(\displaystyle Combine \; like \; terms \; -6 x-x\rightarrow-7x\)

 \(\displaystyle Evaluate \; 10-13\rightarrow-3\)
\(\displaystyle Answer: \; x^3+2 x^2-7 x-3\)

 \(\displaystyle \fn_cm Re-arrange \; terms: \; -3-7x+2x^2+x^3\)

The expression on the left shows eight being multiplied by an expression in parentheses; that expression is the difference of an unknown number and three. The whole right expression is therefore worded as "Eight multiplied by the difference of a number and three"; the equality symbol and the forty round out the answer.

\(\displaystyle x-9\) can be written as "the difference of a number and nine"; the expression on the left shows that difference being divided by five, or, alternatively stated, "five divided into the difference of a number and nine". "Is equal to eighty " rounds out the sentence.

First, calculate the total number of men.

\(\displaystyle \small 60_{total}-32_{women}=28_{men}\)

Next, calculate the number of blonde men.

\(\displaystyle \small 28_{men}-16_{not\ blonde}=12_{blonde\ men}\)

Now we can calculate the number of blonde women.

\(\displaystyle \small 20_{blonde}-12_{blonde\ men}=8_{blonde\ women}\)

Use the distance equation, where \(\displaystyle \small d=rt\). \(\displaystyle \small d\) is distance (mi), \(\displaystyle \small r\) is rate (mph), and \(\displaystyle \small t\) is time (h). In this instance, the train is traveling in two segments, so we will add the distances traveled during each segment.

\(\displaystyle \small d=r_1t_1+r_2t_2\)

The first rate is 60mph, and the first time is 2.5 hours. The second rate is 40mph, and the second time is 1 hour.

\(\displaystyle \small d=(60)(2.5)+(40)(1)\)

\(\displaystyle \small d=150+40=190\)

Suppose one number is \(\displaystyle \small x\).

The other number is "three less than twice the other number." Twice our other number is \(\displaystyle \small 2x\), and three less than twice this number is \(\displaystyle \small 2x-3\).

We are told that the sum of these two numbers is forty.

\(\displaystyle x+(2x-3)=40\)

Now we can simplify.

\(\displaystyle 3x-3=40\)

Write this number with the decimal point after it.

\(\displaystyle 130,000,000,000,000,000,000,000.\)

Move the decimal point to the position after the first nonzero digit, which here is the 1. Note that the resulting number, after truncating the trailing zeroes, is 

\(\displaystyle 1.3\).

Also note that the number of places the decimal point moved is 23. Since the point was moved left the exponent will be 23, and the correct choice is 

\(\displaystyle 1.3 \times 10 ^{23}\).