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Section 1.3 Variables, Expressions, and Equations

Subsection 1.3.1 Overview

In algebra, a variable represents a number whose value is not specified. Sometimes, variables are introduced to describe the process for a calculation that depends on another value. Other times, variables are used to find a quantity of interest that makes an equation true. In this section, we review the concepts of variables, expressions, equations, and basic strategies for solving equations.

Subsection 1.3.2 Expressions and Properties of Algebra

In algebra, we often use variables as placeholders for numerical values. The variable is given a symbol, usually a letter like \(x\text{,}\) that takes the place of the value. Some times, this is because we want to describe a calculation without referencing a specific value. Other times, we want a specific but unknown value and use a variable as a name.

Example 1.3.1

Maybe you have seen a calculation described similarly to the following.

Think of a number. Add five. Double the result. Subtract four. Divide the answer in half. Subtract your original number.

Use the variable \(x\) to represent the number chosen. Write out the formula that describes this calculation.

Solution

We use parentheses to emphasize the order of calculations. Start with \(x\) and add five to get \(x+5\text{.}\) Doubling this means to multiply by two to get \(2(x+5)\text{.}\) Subtracting four gives \(2(x+5)-4\text{.}\) Dividing in half means divide this by two, resulting in a fraction, \(\displaystyle \frac{2(x+5)-4}{2}\text{.}\) We end by subtracting the original value \(x\text{.}\) The formula that matches this calculation would be

\begin{equation*} \frac{2(x+5)-4}{2}-x\text{.} \end{equation*}

An expression is any formula involving numbers and variables, such as the formula in the previous example. An expression itself represents a numerical value. When an expression involves variables, the value of the expression itself is unknown until the values of all variables are known. We say that the expression is a dependent variable and often represent its value by a symbol.

Example 1.3.2

For the previous example, we could introduce a variable \(y\) to represent the final number. That gives a dependent variable defined by the expression

\begin{equation*} y=\frac{2(x+5)-4}{2}-x\text{.} \end{equation*}

Dependent variables often depend on more than one variable.

Example 1.3.3

For a business that earns money by selling a number of items each of which is sold for the same price, the revenue (money brought in through sales) is computed by multiplying the number of items sold by the price of each item. We can summarize this statement defining revenue using algebra if we represent the state variables by symbols. Let \(R\) represent the revenue, let \(n\) represent the number of items sold, and let \(p\) represent the price of each item. The product \(n \cdot p\) is the expression representing the product of the number of items and the price per item. We use the equation

\begin{equation*} R = n \cdot p \end{equation*}

to describe that the revenue is computed by this expression.

We see that the revenue \(R\) is defined here as a dependent variable based on the values of \(n\) and \(p\text{.}\) If we know the value of both \(n\) and \(p\text{,}\) then we will know the value of \(R\text{.}\) For example, if the company sells \(n=1000\) items and sets a price of \(p=1.25\) dollars, then the revenue is \(R = 1000 \cdot 1.25 = 1250\) dollars.

Different expressions can represent the same value. For example, \(x+x\) and \(2x\) are different expressions—they describe different calculations—but they always have the same value. We say that two expressions are equivalent if they result in the same value for all possible values of the involved variables. The properties of algebra describe the rules for how to create equivalent expressions.

Elementary Properties of Algebra

For any expressions \(x\text{,}\) \(y\text{,}\) and \(z\text{,}\) the following expressions are equivalent.

  • Additive identity (zero): \(x+0=x\text{.}\)
  • Multiplicative identity (one): \(1 \cdot x = x\text{.}\)
  • For every value \(x\text{,}\) there is an additive inverse value written \(-x\) so that \(x+-x=0\text{.}\) It is always the case that \(-x = -1 \cdot x\text{.}\)
  • For every non-zero value \(x\) (\(x \ne 0\)), there is a multiplicative inverse value written \(\div x\) so that \(x \cdot \div x=1\text{.}\) When \(x \ne 0\text{,}\) we have \(\div x = \frac{1}{x}\text{,}\) which is why the inverse of \(x\) is also called the reciprocal.
  • Commutative properties: \(x+y = y+x\) and \(x \cdot y = y \cdot x\text{.}\)
  • Associative properties: \((x+y)+z = x+(y+z)\) and \((x \cdot y) \cdot z = (x \cdot y) \cdot z\text{.}\)
  • Distributive property of multiplication over addition: \(x \cdot (y+z) = x \cdot y + x \cdot z\text{.}\)
Example 1.3.4

Our pick-a-number example with the dependent variable

\begin{equation*} y=\frac{2(x+5)-4}{2}-x \end{equation*}

could be used as a mind-reader trick. If a volunteer does the math correctly, you can always guess the final number \(y\text{.}\) Use algebra properties to determine what you should predict.

Solution

We start with the distributive property to rewrite \(2(x+5) = 2x+10\text{.}\) Subtracting 4 is equivalent to adding \(-4\text{.}\) We can then use the associative property to rewrite \((2x+10)+-4 = 2x+(10+-4) = 2x+6\text{.}\) Dividing by two is equivalent to multiplying by \(\frac{1}{2}\text{.}\) This allows us to use the distributive property again,

\begin{align*} \frac{2(x+5)-4}{2} = \frac{2x+6}{2} &= \frac{1}{2}(2x+6) = \frac{1}{2}(2x) + \frac{1}{2}(6)\\ &= (\frac{1}{2} \cdot 2) x + \frac{6}{2} = x + 3 \end{align*}

The second line used the associative property and the multiplicative identity. Can you see where? The final step in the calculation is to subtract \(x\text{,}\)

\begin{equation*} y = \frac{2(x+5)-4}{2}-x = x+3 - x = (x+-x)+3 = 3\text{.} \end{equation*}

In conclusion, we have discovered \(y=3\text{.}\) No matter what number is originally chosen, the final result of the described calculation will always be three.

Subsection 1.3.3 Equations

An equation is a logical statement that two expressions are equal. As a logical statement, an equation can be true or false.

Example 1.3.5

\(1+1=3\) is an equation. The two expressions are \(1+1\) and \(3\text{.}\) This equation is a false statement because the values of the expressions are different.

When an equation involves variables, the truth of the statement depends on the values of the variables. If we specify particular values for each variable, then we calculate the exact numerical value of each expression and then test if the values are equal. For some values of the variables, the equation may be false; for other values, the equation will be true. If the equation is true for all possible values of the variables, the equation is called an identity. A solution to the equation is a set of values for the variables in the equation that makes the statement true. The solution set of an equation is the set of all possible solutions.

Example 1.3.6

\(x+1=3\) is an equation with a variable \(x\text{.}\) When \(x=1\text{,}\) or when \(x\) represents the value 1, the equation is the same as our earlier example. In that case, the equation is false. However, when \(x=2\text{,}\) the equation corresponds to \(2+1=3\text{,}\) which is true. The value \(2\) is in the solution set.

Example 1.3.7

The statement \(2x+5=13\) is an equation involving a single variable, \(x\text{.}\) We can test the equation using different values for \(x\text{.}\) For \(x=1\text{,}\) the expression \(2x+5\) has a value \(2(1)+5=7\text{.}\) Since \(7 \ne 13\text{,}\) the equation is false and \(x=1\) is not a solution. For \(x=4\text{,}\) the expression \(2x+5\) has a value \(2(4)+5=13\text{.}\) We see that \(x=1\) is a solution because the expressions \(2x+5\) and \(13\) have the same value. The value \(1\) is in the solution set.

Testing values to find solutions is impractical because there are infinitely many different values possible for each variable. Finding a solution by guessing would be a stroke of luck. In addition, finding one solution would not tell you if there were more solutions not yet found.

Instead, we use algebra to find solutions by solving the equation. You would have learned many strategies for solving equations in an earlier algebra class. Rather than attempt to address every strategy, we will focus on the overarching principles.

Most of these strategies rely on a principle of finding equivalent equations. Equations are equivalent when they true or false for exactly the same values of variables. The symbol \(\Leftrightarrow\) can be used to say that statements are equivalent.

You may have learned that an equation is like a balance or scale. The two expressions are like two masses being balanced against one another. The equation is true if the masses are in balance. We create an equivalent equation if we apply the same operation to both sides of the equation, so long as the operation is invertible.

Balanced Operations Result in Equivalent Equations

The following operations can be used to create equivalent equations, where each variable represents arbitrary expressions.

  • Balanced Addition: \(a=b\) is equivalent to \(a+c=b+c\text{.}\)
  • Balanced Subtraction: \(a=b\) is equivalent to \(a-c=b-c\text{.}\)
  • Balanced Multiplication: \(a=b\) is equivalent to \(a \cdot c=b \cdot c\text{,}\) so long as \(c \ne 0\text{.}\)
  • Balanced Division: \(a=b\) is equivalent to \(\displaystyle \frac{a}{c}=\frac{b}{c}\text{,}\) so long as \(c \ne 0\text{.}\)

Because multiplication and division include a condition \(c \ne 0\text{,}\) the new equation might have extra solutions corresponding to values where \(c=0\) that are not solutions to the original equation. These extraneous should not be confused with actual solutions.

In addition to the balanced arithmetic operations, we will later learn about invertible or one-to-one functions. An invertible function can be applied to both sides of an equation to create an equivalent equation, so long as the expressions have values in the function domain.

The primary strategy for solving an equation is to create an equivalent equation where the variable is isolated. If a variable appears only once in an equation, then our strategy would be to apply balanced operations until one side of the equation only has that variable. Generally, we can use the inverse operation for the last operation in the expression based on the order of operations.

Example 1.3.8

Consider the earlier equation \(2x+5=13\text{.}\) Use balanced operations to solve the equation.

Solution

The variable \(x\) only appears in the expression \(2x+5\text{.}\) Because order of operations applies multiplication before addition, the operation of addition \(+5\) would be the last operation. The inverse operation is to add \(-5\text{,}\) which we do in a balanced way.

\begin{equation*} 2x+5 = 13 \quad \Leftrightarrow \quad 2x+5+-5=13+-5 \quad \Leftrightarrow \quad 2x=8 \end{equation*}

The last operation in the expression \(2x\) is now multiplication by 2. The next balanced operation is to multiply by the inverse \(\frac{1}{2}\text{.}\)

\begin{equation*} 2x+5 = 13 \quad \Leftrightarrow \quad 2x=8 \quad \Leftrightarrow \quad \frac{1}{2} \cdot 2x = \frac{1}{2} \cdot 8 \quad \Leftrightarrow \quad x = 4 \end{equation*}

The equation \(x=4\) has isolated the variable, so the only solution is \(x=4\text{.}\) The solution set \(\{x : 2x+5=13\}\)—the set of values \(x\) that make \(2x+5=13\) true—has a single value \(\{x : 2x+5=13\} = \{4\}\text{.}\)

If an equation has the variable appearing in multiple locations, we generally have two strategies to consider. One strategy—isolating a variable—is to find an equivalent equation where the variable only appears once. To do this, we use balanced operations to put terms with the variable on the same side of the equation. We then use algebra properties, if possible, to solve for that variable.

Example 1.3.9

Solve the equation \(\displaystyle \frac{3x}{x+2} = 2\text{.}\)

Solution

The equation has the variable \(x\) appear twice. For the expression on the left to be defined, we know \(x+2 \ne 0\text{.}\) We can use balanced multiplication and multiply both sides of the equation by \(x+2\) to find an equivalent equation

\begin{equation*} 3x = 2(x+2)\text{.} \end{equation*}

(This is also called cross-multiplication.) The right expression can be rewritten to obtain

\begin{equation*} 3x=2x+4\text{,} \end{equation*}

which can be solved as

\begin{equation*} x=4\text{.} \end{equation*}

We check our answer by testing the truth of the original equation. With \(x=4\text{,}\) our equation is

\begin{equation*} \frac{3(4)}{4+2} = 2\text{.} \end{equation*}

Because \(3(4)=12\) and \(4+2=6\) and \(12 \div 6 = 2\text{,}\) the equation is true. The solution set is

\begin{equation*} \{x : \frac{3x}{x+2}=2 \} = \{4\}\text{.} \end{equation*}

The other common strategy—factoring—is to find an equivalent equation with one expression exactly zero and the other expression is factored. The factoring strategy is based on the properties of zero in relation to multiplication. When non-zero numbers are multiplied, the product is also non-zero. The only way a product can equal zero is if one of the factors is zero.

Consequently, when an equation is written as a product equalling zero, we can identify all solutions for each factor individually equal to zero.

Example 1.3.11

Solve the equation \(x^3=4x\text{.}\)

Solution

Because the variable \(x\) appears as a cube \(x^3\) and alone as \(x\text{,}\) isolating the variable will not be a successful strategy. We use factoring instead, which requires moving all terms to one side. The balanced operation would be to add \(-4x\) to both sides,

\begin{equation*} x^3 - 4x = 0\text{.} \end{equation*}

Now that we have an equivalent equation written as an expression equal to zero, we need to factor our expression. The expression has a common factor \(x\) in all terms, so we write

\begin{equation*} x(x^2-4) = 0\text{.} \end{equation*}

The factoring principle tells us that solutions satisfy either \(x=0\) or \(x^2-4=0\text{.}\) We can continue to factor:

\begin{equation*} x(x+2)(x-2)=0\text{.} \end{equation*}

Now, each factor might equal zero leading to a different solution: \(x=0\text{,}\) \(x=-2\text{,}\) or \(x=2\text{.}\) Because these are the only values that make a factor equal zero, they are the only solutions. The solution set is

\begin{equation*} \{x : x^3-4x \} = \{-2, 0, 2\}\text{.} \end{equation*}

In addition to the general strategies of isolating a variable and factoring an expression equal to zero, we have a special strategy for solving a quadratic polynomial equal to zero, even when it does not easily factor. This strategy is to use the quadratic formula.

I like to remember the quadratic formula using the tune to “Pop Goes the Weasel”: “\(x\) equals negative \(b\) plus or minus square-root \(b\)-squared minus \(4ac\) all over \(2a\text{.}\)”

Example 1.3.13

Solve \(2x^2 = 3x + 5\text{.}\)

Solution

First, we need to write an equivalent equation that is in the standard quadratic form. This requires all terms on the left side of the equation,

\begin{equation*} 2x^2 - 3x - 5 =0\text{.} \end{equation*}

Once in this form, we identify \(a=2\text{,}\) \(b=-3\) and \(c=-5\text{.}\) Knowing the coefficients \(a\text{,}\) \(b\text{,}\) and \(c\text{,}\) we can apply the quadratic formula:

\begin{equation*} x = \frac{+3 \pm \sqrt{(-3)^2-4(2)(-5)}}{2(2)}\text{.} \end{equation*}

We then simplify the terms

\begin{equation*} x = \frac{3 \pm \sqrt{9+40}}{4} = \frac{3 \pm \sqrt{49}}{4}\text{.} \end{equation*}

The notation \(\pm\) indicates that we find two solutions be either adding or subtracting the square-root. So one solution will be

\begin{equation*} x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}\text{,} \end{equation*}

and the other solution will be

\begin{equation*} x = \frac{3 - 7}{4} = \frac{-4}{4} = -1\text{.} \end{equation*}

Simple quadratic problems can often be solved directly by factoring. In this case, we would have the following equivalent equations

\begin{gather*} 2x^2-3x-5 = 0\\ 2x^2-5x+2x-5 = 0\\ x(2x-5) + (2x-5) = 0\\ (x+1)(2x-5) = 0 \end{gather*}

The solutions then solve \(x+1=0\) or \(2x-5=0\text{,}\) giving \(x=-1\) and \(x=\frac{5}{2}\text{.}\)

This example illustrates a strategy of grouping to factor quadratics. We multiplied the coefficients \(a=2\) and \(c=-5\) to get \(ac=-10\text{.}\) We then looked for factors of \(-10\) that added to \(-3\text{,}\) in this case \(-5+2=-3\text{.}\) We then rewrite \(-3x\) as \(-3x = -5x + 2x\) and group what are now four terms into two groups. Once we pull out common factors from the groups, we can identify the factors which are used in simpler equations.

The next example reminds you to be careful about what you think will be equivalent equations.

Example 1.3.14

Solve the equation

\begin{equation*} \frac{x}{x-3} = \frac{3x-4}{x-3}\text{.} \end{equation*}
Solution

A common strategy for this equation that two fractions are equal is to cross-multiply. That is, multiply the \(x\) in the numerator on the left by the \(x-3\) in the denominator on the right, and then multiply the \(3x-4\) in the numerator on the right by the \(x-3\) in the denominator on the left. Then we can use the factoring method.

\begin{gather*} x(x-3) = (3x-4)(x-3)\\ x^2-3x = 3x^2-9x-4x+12\\ x^2-3x = 3x^2-13x+12\\ 0 = 2x^2-10x+12\\ 2(x^2-5x+6) = 0\\ 2(x-2)(x-3) = 0 \end{gather*}

This final equation is factored. The equation \(2=0\) has no solution. The equations \(x-2=0\) and \(x-3=0\) have solutions \(x=2\) and \(x=3\text{,}\) respectively. However, because the denominators were the same, \(x-3\text{,}\) our solution \(x=3\) was actually an extraneous solution.

Multiplying an equation involving fractions by an expression involving \(x\) always risks introducing extraneous solutions, particularly if it changes the domain of the expressions. Factoring is always preferable and only slightly more challenging. To use factoring, we find an equivalent equation by adding expressions to get zero on one side,

\begin{equation*} \frac{x}{x-3} - \frac{3x-4}{x-3} = 0\text{.} \end{equation*}

The common denominator is a common inverse factor, allowing us to combine the fractions,

\begin{equation*} \frac{x}{x-3} - \frac{3x-4}{x-3} = \frac{x-(3x-4)}{x-3} = \frac{x-3x+4}{x-3} = \frac{-2x+4}{x-3}\text{.} \end{equation*}

Consequently, our equation is equivalent to

\begin{equation*} \frac{-2x+4}{x-3} = 0\text{.} \end{equation*}

The factors are \(-2x+4\) and the multiplicative inverse of \(x-3\text{,}\) which can never equal zero. The only solution is the solution to \(-2x+4=0\) or \(x=2\text{.}\) (A quotient equals zero only if the numerator equals zero and the denominator is non-zero.)

Finally, if an equation is equivalent to an equation that is always false, then the equation has no solutions. The solution set is the empty set, \(\varnothing = \{\}\text{.}\)

Example 1.3.15

Find the solution set for the equation \(\displaystyle \frac{c}{c+3} = 1\text{.}\)

Solution

When we cross-multiply the equation by the expression \(c+3\) (assuming \(c \ne -3\)), we get an equation

\begin{equation*} c = c+3 \end{equation*}

which is equivalent to

\begin{equation*} 0 = 3\text{.} \end{equation*}

Both of these equivalent equations are never true. There are no solutions to the original equation. The solution set is the empty set \(\varnothing\text{.}\)

Subsection 1.3.4 Systems of Equations and Constraints

An equation involving multiple variables establishes a relation between the variables. We often have multiple equations, each of which constrains the possible values for the variables. A useful strategy for solving equations is to isolate a dependent variable in one equation and then substitute the resulting value or formula into the other equation. That equation then has one variable which can be solved using the usual methods.

Example 1.3.16

The equation

\begin{equation*} u^2+v^2=16+6u \end{equation*}

forms a relation between variables \(u\) and \(v\text{.}\)

  • Find the possible values for \(v\) when \(u=-1\text{.}\)
  • Find the possible values for \(u\) when \(v=2\text{.}\)
  • Find the possible values for \(u\) and \(v\) when \(u=v\text{.}\)
Solution

First, to solve the equation when \(u=-1\text{,}\) we substitute the value of \(u=-1\) and then use algebra to isolate \(v\text{.}\)

\begin{gather*} u^2+v^2=16+6u\\ (-1)^2+v^2 = 16 + 6(-1)\\ 1+v^2=10\\ v^2=9\\ v=\pm 3 \end{gather*}

There are two values for \(v\) when \(u=-1\text{.}\) The solutions are the states \((u,v)=(-1,3)\) and \((u,v)=(-1,-3)\text{.}\)

Next, to solve the equation when \(v=2\text{,}\) we substitute the value \(v=2\text{.}\) However, because \(u\) appears in the equation with terms \(u^2\) and \(6u\text{,}\) we can not combine terms to isolate \(u\text{.}\) Instead, we need to use the quadratic formula.

\begin{gather*} u^2+v^2=16+6u\\ u^2+(2)^2 = 16 + 6u\\ u^2+4=16+6u\\ u^2-6u-12=0\\ u=\frac{6 \pm \sqrt{(-6)^2-4(-12)}}{2}\\ u=\frac{6 \pm \sqrt{84}}{2} = \frac{6 \pm 2\sqrt{21}}{2}\\ u=3 \pm \sqrt{21} \end{gather*}

Again, two states are solutions, \((u,v)=(3+\sqrt{21},2)\) and \((u,v)=(3-\sqrt{21},2)\text{.}\)

Finally, the equation \(u=v\) is a constraint involving both variables. Because \(v\) is shown as a dependent variable in the constraint \(v=u\text{,}\) we substitute \(u\) in place of \(v\) in the original equation.

\begin{gather*} u^2+v^2=16+6u\\ u^2+(u)^2 = 16 + 6u\\ 2u^2-6u-16=0\\ u^2-3u-8=0\\ u=\frac{3 \pm \sqrt{(-3)^2-4(1)(-8)}}{2}\\ u=\frac{3 \pm \sqrt{9+32}}{2} = \frac{3 \pm \sqrt{41}}{2} \end{gather*}

For each value of \(u\text{,}\) we have \(v=u\text{.}\) One solution would be \((u,v) = (\frac{3 + \sqrt{41}}{2},\frac{3 + \sqrt{41}}{2})\) while the other solution would be \((u,v) = (\frac{3 - \sqrt{41}}{2},\frac{3 - \sqrt{41}}{2})\text{.}\)

Equations allow us to answer questions involving variables whose relation is known.

Example 1.3.17

Is it possible to enclose an area of 25 m2 in a rectangle with perimeter of 18 m? If so, how?

Solution

In this problem, we need to identify the relevant variables for the system and the equations that constrain the state. We are working with a rectangle, which is characterized by a length and a width. Let us draw a figure and use variables \(L\) for the length and \(W\) for the width.

The perimeter \(P\) and the area \(A\) can be considered to be dependent variables, defined by the equations

\begin{gather*} P=2L+2W,\\ A=L \cdot W. \end{gather*}

The problem gives us two additional pieces of information, \(P=18\) and \(A=25\text{.}\) When we substitute those values of the state into the equations, we have two equations for two variables:

\begin{equation*} 2L+2W=18, \qquad L \cdot W = 25. \end{equation*}

In order to solve these equations, we use one equation to isolate one of the variables, say \(L\text{,}\) and then substitute the resulting expression into the other equation.

\begin{gather*} 2L+2W=18 \qquad \Rightarrow \qquad L=9-W\\ L\cdot W = 25 \qquad \Rightarrow \qquad (9-W)W = 25 \end{gather*}

Then we solve the equation that only involves \(W\text{.}\)

\begin{gather*} (9-W)W = 25\\ 9W-W^2=25\\ W^2-9W+25=0\\ W=\frac{9 \pm \sqrt{(-9)^2-4(25)}}{2}\\ W=\frac{9 \pm \sqrt{81-100}}{2}=\frac{9\pm \sqrt{-19}}{2} \end{gather*}

When solving this quadratic formula, we have the square-root of a negative number giving complex numbers.

In conclusion, we found that there are no real solutions. This means that it is not possible to create a rectangle with a perimeter of 18 m and an area of 25 m2.

Subsection 1.3.5 Summary

  • Variables in algebra are symbols that represent a placeholder for a number. Variable names can be letters, symbols, or words. Uppercase and lowercase letters are different symbols and should not be interchanged.

  • An expression is any value or a formula that represents a value. An equation is a logical statement that two expressions are equal.

  • A solution to an equation is a state (values specified for all variables) that makes the equation true. The solution set is the set of all possible solutions.

  • Solving an equation for the value of one variable when the value of the other variable is given allows us to predict the state of a system.

Subsection 1.3.6 Exercises

1

Show using the elementary properties of addition and multiplication why \(2(x+3)-1=2x+5\text{.}\)

2

Show using the elementary properties of addition and multiplication why \((x+3)(x-1)=x^2+2x-3\text{.}\)

3

A student made a mistake writing \(\displaystyle \frac{3x+1}{x} = \frac{3+1}{1} = 4\text{.}\) What did the student do? Why was it incorrect?

4

A student made a mistake writing \(\displaystyle x \cdot \frac{2x+1}{x+3} = \frac{2x^2+x}{x^2+3x}\text{.}\) What did the student do? Why was it incorrect?

Rewrite each of the following expressions as an equivalent sum instead of as a product.

5

\(4(x+3)\)

6

\(3(x-2)(x+4)\)

7

\((x-1)(x-2)(x-3)\)

Rewrite each of the following expressions as an equivalent factored expression.

8

\(3x-15\)

9

\(4x^2-6x\)

10

\(x^2-5x-14\)

11

\(5x^2-8x+3\)

12

\(5x^2-8x+3\)

Without solving the equation, which of the following values are in the solution set?

13

\(x^2-2x = x+4\)

  1. \(x=-2\)
  2. \(x=-1\)
  3. \(x=0\)
  4. \(x=1\)
  5. \(x=2\)
14

\(z^3-5z = 2z^2-6\)

  1. \(z=-2\)
  2. \(z=-1\)
  3. \(z=0\)
  4. \(z=1\)
  5. \(z=2\)

Find the solution set for each equation.

15

\(3t+5 = t-2\)

16

\(\displaystyle \frac{4u}{u+5} = 3\)

17

\(x^2-2x = x+4\)

18

\(y^2-y = y+5\)

19

\(p^2-1 = 3(p-2)\)

20

\(z^3-5z = 2z^2-6\)

21

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

22

\(\displaystyle \frac{4x^2-5x}{x-2} = 4x+3\)

Find the solutions to the system of equations.

23

\(3x+2y=15\) and \(y=3\text{.}\)

24

\(2x-5y=7\) and \(x+2y=9\text{.}\)

25

\(x^2-y=4x\) and \(2x+y=3\text{.}\)

26

Is it possible to enclose an area of 25 m2 using a rectangle with perimeter of 25 m? If so, how?

27

Is it possible to enclose an area of 50 m2 in two congruent rectangles that share an edge such that the total length of edges is 40 m (counting the shared edge only once)? If so, how?