To solve for the variable we must move all constants to the other side of the equation. Since we have absolute value bars in the particular equation we will first move all constants that are outside of the absolute value bars. To do this apply the oppisite operation. In this particular case, subtract twelve from both sides.

\(\displaystyle | 3y - 17| + 12 = 49\)

\(\displaystyle | 3y - 17| + 12 - 12 = 49 - 12\)

\(\displaystyle | 3y - 17| = 37\)

Now we will deal with the absolute value bars which state that the quantity inside them can be positive of negative.

Either

\(\displaystyle 3y - 17= -37\) or \(\displaystyle 3y - 17= 37\).

Solve each separately:

\(\displaystyle 3y - 17= -37\)

\(\displaystyle 3y - 17+ 17 = -37 + 17\)

\(\displaystyle 3y = -20\)

\(\displaystyle \frac{3y}{3} = \frac{-20}{3}\)

\(\displaystyle y = -6 \frac{2}{3}\)

\(\displaystyle 3y - 17= 37\)

\(\displaystyle 3y - 17+17 = 37+17\)

\(\displaystyle 3y = 54\)

\(\displaystyle \frac{3y}{3} = \frac{54}{3}\)

\(\displaystyle y = 18\)

 Therefore, \(\displaystyle y = -6 \frac{2}{3}\) or \(\displaystyle y = 18\).

The correct choice is  \(\displaystyle y = -6 \frac{2}{3}\).

To solve for the variable we must move all constants to the other side of the equation. Since we have absolute value bars in the particular equation we will first move all constants that are outside of the absolute value bars. To do this apply the oppisite operation. In this particular case, subtract 125 from both sides.

\(\displaystyle 125 - |4t-98| = 113\)

\(\displaystyle 125 - |4t-98| - 125 = 113 - 125\)

\(\displaystyle - |4t-98| =-12\)

\(\displaystyle |4t-98| =12\)

\(\displaystyle 4t-98 = -12\) or \(\displaystyle 4t-98 = 12\)

Solve separately:

\(\displaystyle 4t-98 = -12\)

\(\displaystyle 4t-98 + 98 = -12 + 98\)

\(\displaystyle 4t = 86\)

\(\displaystyle \frac{4t}{4} = \frac{86 }{4}\)

\(\displaystyle t= 21\frac{1}{2}\)

\(\displaystyle 4t-98 = 12\)

\(\displaystyle 4t-98 + 98 = 12 + 98\)

\(\displaystyle 4t = 110\)

\(\displaystyle \frac{4t}{4} = \frac{110 }{4}\)

\(\displaystyle t= 27\frac{1}{2}\)

To figure out the total portraits done by James, you must multiply rate \(\displaystyle p\) by the number of months \(\displaystyle m\).  This equation would be \(\displaystyle pm=t\).  To solve for the number of months \(\displaystyle m\) you must divide both sides by \(\displaystyle p\). This gives us \(\displaystyle m=\frac{t}{p}\).

The first step is to simplify your equation.  You distribute \(\displaystyle 8\) to get \(\displaystyle y-2=8x-24\).  Then to get \(\displaystyle y\) by itself you add \(\displaystyle 2\) to both sides ending with \(\displaystyle y=8x-22\).  Finally insert the \(\displaystyle x\)-coordinate to get \(\displaystyle 8(5)-22\) which equals \(\displaystyle 18\).

To get \(\displaystyle a\) by itself, divide each side by \(\displaystyle 3\) to get \(\displaystyle a-6=27\).  Then add \(\displaystyle 6\) to both sides to get \(\displaystyle a=33\).

The absolute value of a nonnegative number is the number itself; the absolute value of a negative number is its positive opposite.

By substitution, 10.5 can be seen to be a solution of each of the equations in the five choices. For example:

\(\displaystyle |2x + 6| = 27\)

\(\displaystyle |2 \cdot 10.5 + 6| = 27\)

\(\displaystyle |21 + 6| = 27\)

\(\displaystyle |27 | = 27\) - a true statement.

That 10.5 is a solution of the other four equations can be proved similarly. Therefore, the question is essentially to choose the equation with \(\displaystyle -18.5\) as its other solution. Again, we can do this using substitution in each equations. We see that \(\displaystyle |2x + 8 | = 29\) is correct as follows:

\(\displaystyle |2x + 8 | = 29\)

\(\displaystyle |2 (-18.5) + 8 | = 29\)

\(\displaystyle |-37 + 8 | = 29\)

\(\displaystyle |-29| = 29\) - a true statement.

Similar substitution in the other four statements shows that \(\displaystyle -18.5\) is not a solution of any of them; for example, in \(\displaystyle |2x + 9| = 30\):

\(\displaystyle |2x + 9| = 30\)

\(\displaystyle |2 (-18.5) + 9| = 30\)

\(\displaystyle |-37+ 9| = 30\)

\(\displaystyle |-28| = 30\)

\(\displaystyle 28 = 30\) - a false statement.

Convert \(\displaystyle 75 ^{\circ } F\) to the Celsius scale by setting \(\displaystyle F = 75\) in the conversion formula and solving for \(\displaystyle C\):

\(\displaystyle 1.8 C + 32 = F\)

\(\displaystyle 1.8 C + 32 = 75\)

\(\displaystyle 1.8 C + 32 - 32 = 75 - 32\)

\(\displaystyle 1.8 C = 43\)

\(\displaystyle 1.8 C \div 1.8 = 43 \div 1.8\)

\(\displaystyle C \approx 23.9\)

The question is therefore asking for the number of days that the temperature topped \(\displaystyle 23.9^{\circ } C\). Examine the graph below:

The high temperature was greater than \(\displaystyle 23.9^{\circ } C\) on Tuesday, Friday, and Saturday - three different days.

Tommy's current age is represented by t, and Sara's is represented by s. In five years, both Tommy's and Sara's ages will be increased by five. Thus, in five years, we can represent Tommy's age as \(\displaystyle t + 5\)and Sara's as \(\displaystyle s + 5\).

The problem tells us that Tommy's age in five years will be twice as great as Sara's in five years. Thus, we can write an algebraic expression to represent the problem as follows:

\(\displaystyle t + 5 = 2(s + 5)\)

In order to solve for t, first simplify the right side by distributing the 2.

\(\displaystyle t + 5 = 2s + 10\)

Then subtract 5 from both sides.

\(\displaystyle t = 2s + 5\)

The answer is \(\displaystyle t = 2s + 5\).

In order to solve this, we need to set up an equation. \(\displaystyle 549\:\text{m}=\frac{10\:\text{m}}{\text{sec}}\cdot x\), where \(\displaystyle x\) is time. All we need to do is divide by \(\displaystyle 10\: \frac{\text{m}}{\text{sec}}\) on each side.

\(\displaystyle 549\:\text{m}=\frac{10\:\text{m}}{\text{s}}\cdot x\)

\(\displaystyle \frac{549\:\text{m}}{\frac{10\:\text{m}}{\text{sec}}}=x\)

\(\displaystyle \frac{549\:\text{m\:sec}}{10\:\text{m}}=x\)

\(\displaystyle x=\frac{549\:\text{sec}}{10}=54.9\:\text{sec}\)

First we need to create equations that represent the path's of the Rocket's.

For Rocket \(\displaystyle A\), the equation is

\(\displaystyle y=25x\)

For Rocket \(\displaystyle B\), the equation is

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

In order to solve for the time the Rocket's cross, we need to set the equations equal to each other.

\(\displaystyle 25x=16x+3\)

Now solve for \(\displaystyle x\)

\(\displaystyle 9x=3\)

\(\displaystyle x=\frac{1}{3}\)