ORCCA Linear Equations and Inequalities Chapter Review

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Section 2.6 Linear Equations and Inequalities Chapter Review

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Solving Multistep Linear Equations.

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There is a regular process to use for solving a linear equation.

Simplify: 🔗
Simplify the expressions on each side of the equation by distributing and combining like terms.
Separate: 🔗
Use addition or subtraction to separate the terms so that the variable terms are on one side of the equation and the constant terms are on the other side of the equation.
Clear the Coefficient: 🔗
Use multiplication or division to eliminate the variable term’s coefficient.
Check: 🔗
Check the solution in the original equation. Substitute values into the original equation and use the order of operations to simplify both sides. It’s important to use the order of operations alone rather than properties like the distributive law. Otherwise you might repeat the same arithmetic errors you (might have) made while solving, and fail to catch an incorrect solution.
Summarize: 🔗
State the solution set. Or in the case of an application problem, summarize the result in a complete sentence using appropriate units.

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Simplifying expressions, evaluating expressions, and solving equations are distinct algebra tasks.

An expression like $10 - 3 (x + 2)$ can be simplified to $- 3 x + 4$ (as in Example 2.1.14). However we cannot “solve” an expression like this. It is incorrect to say you will “solve $10 - 3 (x + 2)$ ”.
An expression like $10 - 3 (x + 2)$ can be evaluated, but only once you have a number to use in place of the variable. This is what happens in Example 2.1.15, where $x = 2,$ and the expression evaluates to $- 2 .$
An equation connects two expressions with an equal sign. In Example 2.1.16, $10 - 3 (x + 2) = x - 16$ has one expression on either side of equal sign. You can solve this equation, because it is an equation. You can also solve inequalities. You just cannot solve an expression like $10 - 3 (x + 2) .$
When we solve the equation $10 - 3 (x + 2) = x - 16,$ we are looking for a number which makes those two expressions evaluate to the same value. In Example 2.1.16, we found the solution was $5 .$ That number $5$ makes both $10 - 3 (x + 2)$ and $x - 16$ evaluate to the same number (and that number is $- 11$ if you were curious.)

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Checkpoint 2.6.1.

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Quotes are given from a student following a quiz. Fill in the blanks with the appropriate vocabulary terms.

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(a)

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The algebra quiz was fun because you had to

evaluate
simplify
solve

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all these equations.

Explanation.

Typically if you have an equation, your task is to solve that equation.

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(b)

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The last exercise seemed hard at first, but it became more clear once you

evaluated
simplified
solved

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each side.

Explanation.

You can simplify the two sides of an equation.

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(c)

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At first your answer to #4 of the quiz might be pretty messy looking, but it looks more reasonable after you

evaluate
simplify
solve

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it.

Explanation.

An expression can be simplified.

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(d)

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One of the word problems was asking you to take a given expression for the height of a ball and

evaluate
simplify
solve

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it for

t = 7 .

Explanation.

When you have a specific number, you can use that to evaluate an expression with one variable by substituting that number in for the variable.

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Checkpoint 2.6.2.

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Solve the equation.

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- 8 A - 8 = - 56

Explanation.

\begin{aligned} - 8 A - 8 & = - 56 \\ - 8 A - 8 + 8 & = - 56 + 8 \\ - 8 A & = - 48 \\ \frac{- 8 A}{- 8} & = \frac{- 48}{- 8} \\ A & = 6 \end{aligned}

The solution set is

{6} .

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Checkpoint 2.6.3.

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One of the tires on your car looks a little flat. You measure its air pressure and are alarmed to see it so low at

22 p s i .

You have a portable device that can pump air into the tire increasing the pressure at a rate of

2.3 \frac{p s i}{m i n} .

How long will it take to fill the tire to the manual’s recommended pressure of

32 p s i ?

Explanation.

This is a rate scenario. The air pressure starts out at

22,

and increases with a rate of

2.3 .

Eventually the pressure reaches

32 .

So the equation we set up is:

22 + 2.3 t = 32

where

t

is the number of minutes the pump runs. Now we follow the standard steps to solve this equation.

\begin{aligned} 22 + 2.3 t & = 32 \\ 22 + 2.3 t - 22 & = 32 - 22 \\ 2.3 t & = 10 \\ \frac{2.3 t}{2.3} & = \frac{10}{2.3} \\ t & = 4.34782608695652 \end{aligned}

So it will take

4.34783 m i n

for the pump to fill the tire.

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Solving Multistep Linear Inequalities.

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Solving a linear inequality is much like solving a linear equation. Two noteworthy differences are that multiplication/division by a negative number requires reversing the direction of the inequality symbol, and checking a solution takes more effort.

Simplify: 🔗
Simplify the expressions on each side of the inequality by distributing and combining like terms.
Separate: 🔗
Use addition or subtraction to separate the terms so that the variable terms are on one side of the inequality and the constant terms are on the other side of the inequality.
Clear the Coefficient: 🔗
Use multiplication or division to eliminate the variable term’s coefficient. If you multiply or divide each side by a negative number, switch the direction of the inequality symbol.
Check: 🔗
A solution to a linear inequality has a “boundary number”. Using the original inequality, check (1) a number less than the boundary number, (2) the boundary number itself, and (3) a number greater than the boundary number to confirm what should and shouldn’t be solutions are all working as expected. (This can take time, so use your judgment about when you might get away with skipping this checking.)
Summarize: 🔗
State the solution set. Or in the case of an application problem, summarize the result in a complete sentence using appropriate units.

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Checkpoint 2.6.4.

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Solve the inequality. Graph the solution set, and write the solution set using both interval notation and set-builder notation.

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9 L + 1 > - 44

Explanation.

\begin{aligned} 9 L + 1 - 1 & > - 44 - 1 \\ 9 L & > - 45 \\ \frac{9 L}{9} & > - 5 \\ L & > - 5 \end{aligned}

So in set-builder notation, the solution set is

{L ∣ L > - 5} .

We can graph this by shading all numbers greater than

- 5 .

Since the sign in the inequality does not allow for

L

to equal

- 5,

we use a parenthesis at

- 5 .

We read the graph from left to right to tell us the interval notation for the solution set:

(- 5, \infty) .

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Checkpoint 2.6.5.

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Solve the inequality. Graph the solution set, and write the solution set using both interval notation and set-builder notation.

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- 4 - 7 R > - 25

Explanation.

\begin{matrix} - 4 - 7 R + 4 > - 25 + 4 \\ - 7 R > - 21 \\ \frac{- 7 R}{- 7} < \frac{- 21}{- 7} \\ R < 3 \end{matrix}

So in set-builder notation, the solution set is

{R ∣ R < 3} .

We can graph this by shading all numbers less than

3 .

Since the sign in the inequality does not allow for

R

to equal

3,

we use a parenthesis at

3 .

We read the graph from left to right to tell us the interval notation for the solution set:

(- \infty, 3) .

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Checkpoint 2.6.6.

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Tyler is driving on the highway, and presently has 15 gal of gasoline in their tank. Their car, under ideal conditions, uses gas at a rate of 0.048 gal/mi. When the tank reaches only one gallon of gas, the low gas light will turn on and Tyler will start looking for a gas station. How far will they drive before this happens?

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(a)

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Write an inequality to represent this situation, using

x

to represent how many miles Tyler might drive before the low gas light turns on.

Explanation.

This is a rate exercise, and we can use

(initial value) \pm (rate) \cdot variable ? (final value)

The question mark in the above is not an equal sign, because we are trying to set up an inequality. The initial value for the gas in their tank is

15,

Tyler is losing gas (so we will subtract, not add), the rate is

0.048,

and the gas amount we are interested in reaching is

1 .

We do not know far Tyler can drive until the gas gets that low, and we decide to use

x

to represent that distance. The rate of gas consumption we are using is only valid under ideal conditions. So the quantity

15 - 0.048 x

will be greater than or equal to the actual amount of gas left in the tank after driving

x

miles. And we want this to reach

1 .

So the inequality we set up is:

15 - 0.048 x \geq 1

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(b)

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Solve this inequality. At most how far will Tyler drive before the low gas light turns on?

Explanation.

\begin{matrix} 15 - 0.048 x - 15 \geq 1 - 15 \\ - 0.048 x \geq - 14 \\ \frac{- 0.048 x}{- 0.048} \leq \frac{- 14}{- 0.048} \\ x \leq 291.667 \end{matrix}

So Tyler can drive at most

291.667

miles.

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(c)

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Use interval notation to express the number of miles Tyler might drive before the low gas light turns on.

Explanation.

Tyler can drive at most

291.667

miles, so the right end of the interval notation for this interval will be

291.667,

with a bracket. We know Tyler has more than 1 gallon of gas when this all started, so they can drive at least some positive number of miles. So the interval notation, accounting for the context of this exercise, is

(0, 291.667] .

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Linear Equations and Inequalities with Fractions.

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For both equations and inequalities, it is often helpful to “clear denominators” if there are any fractions present. This is done by identifying the “least common denominator” for the fractions that are present, and multiplying on each side by that number. This lets you avoid some fraction arithmetic that could lead to human error.

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A proportional equation uses two ratios that should be equal to each other, for situations where two quantities change together. One example might use a ratio of how much of a substance is dissolved in how much of a certain liquid. Typically one ratio has two known numbers, and the other ratio has one known number and one unknown variable.

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Checkpoint 2.6.7.

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Solve the given equation. Practice clearing the denominator(s) as a first step.

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\frac{2}{3} x + \frac{3}{2} = 4 x + 6

Explanation.

This equation has terms with denominators

2

and

3 .

The least common multiple of these is

6 .

So the first step in solving this equation can be to multiply on each side by

6 .

\begin{aligned} 6 (\frac{2}{3} x + \frac{3}{2}) & = 6 (4 x + 6) \\ 4 x + 9 & = 24 x + 36 \\ 4 x + 9 - 9 & = 24 x + 36 - 9 \\ 4 x & = 24 x + 27 \\ 4 x - 24 x & = 24 x + 27 - 24 x \\ - 20 x & = 27 \\ x & = - \frac{27}{20} \end{aligned}

The solution set is

{\frac{- 27}{20}} .

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Checkpoint 2.6.8.

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Solve the given inequality. Practice clearing the denominator(s) as a first step.

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\frac{7}{20} x - \frac{4}{15} \leq \frac{5}{12}

Explanation.

This inequality has terms with denominators

20,

15,

and

12 .

The least common multiple of these is

60 .

So the first step in solving this inequality can be to multiply on each side by

60 .

\begin{aligned} 60 (\frac{7}{20} x - \frac{4}{15}) & \leq 60 (\frac{5}{12}) \\ 21 x - 16 & \leq 25 \\ 21 x - 16 + 16 & \leq 25 + 16 \\ 21 x & \leq 41 \\ x & \leq \frac{41}{21} \end{aligned}

The solution set is

{x ∣ x \leq \frac{41}{21}}

using set-builder notation. In interval notation, we have

(- \infty, \frac{41}{21}] .

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Checkpoint 2.6.9.

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An old cookbook has a recipe for soup that uses

10

g of flour for thickener. The recipe makes

1.5

gal of soup, but you are adjusting it to make only

2.75

gal. How much flour should you use?

Explanation.

Let

x

be how many grams of flour we should use. The amount of flour is proportional to the amount of soup, so we can compare the ratios of flour to soup for our batch with the original recipe:

\frac{x}{2.75} = \frac{10}{1.5}

These denominators are not whole numbers, and we decide not to try using the least common multiple. Instead, we just multiply on each side by both denominators:

\begin{aligned} 1.5 \cdot 2.75 \cdot \frac{x}{2.75} & = 1.5 \cdot 2.75 \cdot \frac{10}{1.5} \\ 1.5 x & = 27.5 \\ x & = \frac{27.5}{1.5} \\ x & = 18.3333333333333 \end{aligned}

So we should use

18.3333 g

of flour.

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Special Solution Sets.

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An equation or inequality might reduce to an unambiguously true statement like

2 = 2

4 < 9 .

If it does, then the solution set is all real numbers. This can be written as

(\infty, \infty)

R .

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Another special thing that can happen is that an equation or inequality can reduce to an unambiguously false statement like

2 = 5

4 \geq 9 .

When this happens, then there is no solution at all. We can say that the solution set is “empty”. The solution set can be written as

{}

\emptyset .

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Checkpoint 2.6.10.

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Solve the equation for its variable.

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- 2 t + 7 - 9 t = - 1 - 11 t + 8

Explanation.

We try to use the standard steps for solving an equation:

\begin{aligned} - 2 t + 7 - 9 t & = - 1 - 11 t + 8 \\ - 11 t + 7 & = - 11 t + 7 \\ 7 & = 7 \end{aligned}

This is an outright true equation, so all real numbers are solutions. The solution set is

(- \infty, \infty),

which we can write as

R .

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Checkpoint 2.6.11.

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Solve the equation for its variable.

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\frac{5}{4} Q = \frac{5}{4} Q + 3

Explanation.

We try to use the standard steps for solving an equation:

\begin{aligned} \frac{5}{4} Q & = \frac{5}{4} Q + 3 \\ 0 & = 3 \end{aligned}

This is an outright false equation, so there are no solutions. We could write the solution set as

{}

\emptyset .

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Checkpoint 2.6.12.

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Solve the inequality for its variable.

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- 13 + 3 (3 - 3 b) \leq - 13 b - (4 - 4 b)

Explanation.

We try to use the standard steps for solving an inequality:

\begin{aligned} - 13 + 3 (3 - 3 b) & \leq - 13 b - (4 - 4 b) \\ 9 - 13 - 9 b & \leq 4 b - (13 b + 4) \\ - (4 + 9 b) & \leq - (9 b + 4) \\ - 4 & \leq - 4 \end{aligned}

This is an outright true inequality. So all real numbers are solutions.

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Isolating a Linear Variable.

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Equations might relate two or more variables to each other. When you have an equation like that (for example

2 x + 3 y = 4 z

) you can solve for any one of the variables, isolating it on one side of an equal sign. The example could be solved for

y,

which gives

y = \frac{4 z - 2 x}{3} .

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The process for doing this is not different than solving a one-variable equation. You should just keep clear which variable you are solving for.

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Checkpoint 2.6.13.

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Isolate the indicated variable.

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L

Y L + F = c

Explanation.

We need to keep our focus on

L

and take algebra steps that bring us closer to isolating

L .

\begin{aligned} Y L + F & = c \\ Y L + F - F & = c - F \\ Y L & = c - F \\ L & = \frac{c - F}{Y} \end{aligned}

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Checkpoint 2.6.14.

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Solve the equation

A = \frac{1}{2} b h

for

b .

Explanation.

We need to keep our focus on

b

and take algebra steps that bring us closer to isolating

b .

\begin{aligned} A & = \frac{1}{2} b h \\ 2 A & = b h \\ \frac{2 A}{h} & = b \\ b & = \frac{2 A}{h} \end{aligned}

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Checkpoint 2.6.15.

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Isolate

y

in the standard form linear equation (a topic covered later in this book).

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5 x - 9 y = - 7

Explanation.

\begin{aligned} 5 x - 9 y & = - 7 \\ 5 x - 9 y - 5 x & = - 7 - 5 x \\ - 9 y & = - 7 - 5 x \\ \frac{- 9 y}{- 9} & = \frac{- 7 - 5 x}{- 9} \\ y & = \frac{5}{9} x + \frac{7}{9} \end{aligned}

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This exercise walks you through an economics problem. Fill in the blanks with the appropriate vocabulary terms.

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(a)

🔗

If the revenue from selling

x

items is

2 x

dollars but the cost of production is

0.5 x,

then the profit is

2 x - 0.5 x .

But that can be

evaluated
simplified
solved

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(b)

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We need to know how much profit there was after

1000

items, so we can

evaluate
simplify
solve

🔗

the profit expression.

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(c)

🔗

If we want to work out how many sales would give us a profit of

$ 3000,

we can

evaluate
simplify
solve

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the equation

1.5 x = 3000 .

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(d)

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At first we find that it would take

\frac{3000}{1.5}

sales, but we can

evaluate
simplify
solve

🔗

that result.

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Exercise Group.

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Solve the equation.

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2.

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3 i - 4 = 9

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3.

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- 4.3 p + 3.3 = 3.6

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Exercise Group.

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Solve the equation.

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4.

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- 7 u - 5 = - 2 u - 40

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5.

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3 (A - 5) = - 8 A - 4

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6.

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- 6 F + 7 = 9 F - 6

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7.

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5 (L - 1) + 4 L = - 4 (3 L + 6) + 1

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8.

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- 2 (R - 4) - (7 R - 6) = - 7

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9.

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You planted a young tree in front of your house, and it was

4

feet tall. Ever since, it has been growing by

\frac{11}{12} f t

each year. How many years will it take for the tree to grow to be 10 feet tall?

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10.

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Alyssa puts a pot of water from the tap onto the stove and turns the burner all the way up. The water temperature starts at

61 ℉

and climbs steadily up to the boiling point of

212 ℉,

raising at a rate of

27 \frac{℉}{min} .

How long will it take for the pot to boil?

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11.

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On a cold snowy day, the temperature in your home is a cozy

70 ℉,

but then you lose power and heating. Your home temperature begins to drop at a rate of

11 \frac{℉}{hour} .

How long will it take before your home is

39 ℉ ?

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12.

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For El’s 8th birthday party, their parents rented a venue that charges a flat fee of

$ 80

plus

$ 15

per guest. Ultimately it cost El’s parents

$ 290 .

How many guests were there?

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13.

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A tool shed is for sale in a state where sales tax applies. The sales tax rate is

5.2 %

and the total was

$ 482 .

What was the price before sales tax?

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14.

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One year, the median rent for a one-bedroom apartment in a city was reported to be

$ 1,340 .

This was reported to be an increase of

2 %

over the previous year. Based on this reporting, what was the median rent for of a one-bedroom apartment the previous year?

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Section 2: Solving Multistep Linear Inequalities

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Exercise Group.

🔗

Solve the inequality. Graph the solution set, and write the solution set using both interval notation and set-builder notation.

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15.

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- 3 + 7 F < - 45

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16.

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- 8 - 8 K < - 32

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17.

🔗

6 + 6 R - 8 \leq - 14

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18.

🔗

- 2 X - 8 + 9 X - 7 \geq - 57

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19.

🔗

- 6 d - 8 \leq - 3 d - 14

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20.

🔗

7 i - 8 + 9 i - 9 \geq - 9 i - 4 - 7 i - 77

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21.

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You hail a taxi and can only pay with cash. This cab service charges a flat fee of

$ 15.00

and then charges

$ 4.00

per mile. No tip is expected. You are carrying a total of

$ 95.00

in cash with you. You want to know how many miles you can afford.

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(a)

🔗

Write an inequality to represent this situation, using

x

to represent how many miles you can afford.

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(b)

🔗

Solve this inequality. At most how many miles can you afford?

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(c)

🔗

Use interval notation to express the number of miles you might ride that day.

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22.

🔗

The population of a certain country grew by

7 %

over the course of the past decade. One city in this country grew in population too, but at an even faster rate than the country grew overall. The city’s current population is

92438 .

What might its population have been ten years ago? (Note: the population ten years ago is known to have been at least

10,000 .

)

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(a)

🔗

Write an inequality to represent this situation, using

x

to represent the city’s population from ten years ago.

🔗

(b)

🔗

Solve this inequality. What’s the most the population could have been ten years ago?

🔗

(c)

🔗

Use interval notation to express what the city’s population could have been ten years ago.

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Section 3: Equations and Inequalities with Fractions

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Exercise Group.

🔗

Solve the given equation. Practice clearing the denominator(s) as a first step.

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23.

🔗

7 x + 4 = \frac{1}{7}

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24.

🔗

\frac{1}{30} x + 8 = \frac{7}{12}

🔗

25.

🔗

\frac{4}{35} x + \frac{5}{42} = \frac{7}{30}

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26.

🔗

\frac{1}{12} x - \frac{3}{20} = - \frac{4}{15} x - 1

🔗

Exercise Group.

🔗

Solve the given inequality. Practice clearing the denominator(s) as a first step.

🔗

27.

🔗

12 x + 10 \geq \frac{9}{7}

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28.

🔗

\frac{7}{2} x - 9 \leq \frac{1}{2}

🔗

29.

🔗

\frac{7}{3} x + 9 < \frac{5}{3} x + 11

🔗

30.

🔗

x + \frac{7}{12} > - \frac{1}{20} x - \frac{8}{15}

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31.

🔗

You planted a young tree in front of your house, and it was

5

feet tall. Ever since, it has been growing by

\frac{5}{12} f t

each year. How many years will it take for the tree to grow to be 11 feet tall?

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32.

🔗

In one college math course, the final grade (out of

100

points) is calculated by adding together one-half of the homework average, one-sixth of the midterm score, and one-third of the final exam score.

🔗

Jenifer has 83 for her homework average and 87 for her midterm score. Her goal is to finish the course with a final grade of 80. What must she score on the final exam?

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33.

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Lachlan and Precious went to the weekend market to buy apples in bulk for a school event, canning lots of applesauce. Lachlan bought 55 lb of apples and paid $62.70. Precious has bagged her apples and has 83 lb of apples ready. How much will it Precious’s bag of apples cost?

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34.

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To estimate the health of the black-tailed deer population in the Jewell Meadow Wildlife Area, the Oregon Department of Fish and Wildlife caught, tagged, and released