Mathematics Study Guide for the ACT

Algebra: Part 1

Algebra is the part of mathematics that deals with the study of symbols and the rules associated with manipulating them. In algebra, you use letters in place of numbers when solving problems.

One main benefit of algebra is that it allows us to solve equations and inequalities that contain unknowns. For example, with algebra, you can solve the equation \(2x + 13 = 5x - 9\) for the unknown variable \(x\). You can also solve more complex problems such as a system of equations or inequalities.

Algebra is also extensively used in the field of geometry to calculate dimensions of shapes and to solve problems relating to angles, volumes, and areas. You have to use algebra in calculus as well. Without algebra, you can’t delve very deeply into mathematics.

The Basic Premise of Algebra

Algebra uses formulas and equations to represent real-world scenarios and problems and solve for unknown variables.

At the heart of algebra, we have variables and constants. A variable is a symbol that represents an unknown quantity. A variable is usually denoted by a lowercase letter of the alphabet, such as \(x\), \(y\), \(a\), or \(b\). The value of a variable can change depending on the problem or equation at hand.

A constant is a value that remains fixed and does not change. For example, the numbers \(10\), \(42\), \(\pi\), and \(-1\) are examples of constants.

Equations are mathematical statements that establish the equality of two expressions. An equation has an \(=\) sign and expressions on both sides of it. For example, \(2x + 10 = 20\) is an example of an equation. Here, \(2\), \(10\), and \(20\) are the constants and \(x\) is the variable. We can solve for the unknown using algebraic techniques.

Solving Equations

Solving an equation involves finding the values of the variables that satisfy the equation, making the expressions on either side of the equation equal. Here are the steps for solving a linear equation (an equation where the variable is raised to the power of \(1\) and there are no other variable exponents). Follow the steps as we solve this equation:

\[3x - 6 = 12\]

1. Simplify both sides. You can do this by combining like terms and using the distributive property to expand.

There is nothing to simplify in this equation. Let’s move to the next step.

1. Isolate the variable term. This is done by performing reverse operations on it. For example, if the variable is being added to a number, subtract that number from both sides of the equation, and if it is being subtracted from a number, add it. If the variable is being multiplied or divided by a number, divide or multiply, respectively, both sides of the equation by that number:

\[3x - 6 + 6 = 12 + 6\] \[3x = 18\]

1. Simplify both sides of the equation again to find the value of the variable. Basically, solve it:

\[\frac{3x}{3} = \frac{18}{3}\] \[x = 6\]

1. Check the answer. Plug the value that you found into the variable of the original equation and check if both sides are equal or not:

\[3x - 6 = 12\] \[3(6) - 6 \stackrel{?}{=} 12\] \[18 - 6 \stackrel{?}{=} 12\] \[12 = 12\]

Note: Not all linear equations will have a solution, and some may have infinitely many solutions.

Solving Word Problems

Solving word problems involves translating a word problem into an equation, solving it, and then interpreting the solution. Here are the steps to follow:

1. Read the problem carefully and identify keywords for operations before attempting to solve it.

2. Assign variables to unknown quantities.

3. Translate the words into equations. Skim the problem for clues that are going to help you achieve this.

4. Use algebraic techniques to solve the equation for the unknown variable.

5. Make sure your answer makes sense and it is reasonable in the context of the problem.

Translating Words to Symbols

The key to solving word problems is to be able to translate words into mathematical symbols or expressions. Here are some common keywords and phrases to look for when translating words to symbols:

Build Equations with One Variable

In beginning algebra you will see word problems that require you to build equations with one variable. This means that there is only one value that is unknown. Building equations with one variable can help us solve problems involving rates, distances, proportions, and other mathematical concepts. To build equations with a single variable, you can follow these three steps:

1. Identify the unknown. This is the variable we want to solve for.

2. Translate the word problem into an equation using the variable.

3. Solve the equation.

Let us build an equation for the following problem:

What is a car’s average speed if it travels \(200\) miles in four hours?

The unknown is the car’s average speed. Let’s assign it the variable \(x\). The basic distance equation is \(d = rt\), where \(d\) is the distance, \(r\) is the rate (speed), and \(t\) is the time. From the problem, we see that \(d = 200\) and \(t = 4\). So, we can write the distance equation as:

\[d = rt\] \[200 = x(4)\] \[200 = 4x\]

This is our equation. Solving it, we find that the car’s average speed is \(50\) miles per hour.

With practice and by adhering to the outlined steps, you can develop a skill for constructing equations with a single variable, which will help you resolve problems more effectively.

Analyze and Draw Conclusions

Using properties of algebra, we can analyze and draw conclusions in math. In algebra, we use various properties to simplify expressions, solve equations, and make sense of problems. Algebraic properties refer to the rules that govern the manipulation of algebraic expressions. Here, we list the six algebraic properties:

• commutative property一This property states that the order of addition or multiplication does not affect the result:

\[a + b = b + a\] \[a \times b = b \times a\]

• associative property一This property tells us that the way in which terms are grouped does not affect the result of multiplication or addition:

\[(a+b) + c = a + (b+c)\] \[(a \times b) \times c = a \times (b \times c)\]

• distributive property一This property states that multiplying a number by a sum (or difference) is the same as multiplying the number by each term of the sum (or difference) and then adding (or subtracting) the results:

\[a \times (b+c) = ab + ac\] \[a \times (b-c) = ab - ac\]

• identity property一This property tells us that the sum of any number and \(0\) is the number itself, whereas the product of any number and \(1\) is the number itself:

\[a + 0 = a\] \[a \times 1 = a\]

• inverse property一This property states that for any number, \(a\), there is an additive inverse, \(-a\), such that \((a) + (-a) = 0\), and a multiplicative inverse, \(\frac{1}{a}\), such that \(a \times \frac{1}{a} = 1\).

• zero property一This property says that the product of any number and \(0\) is always \(0\):

\[a \times 0 = 0\]

Evaluating Expressions

Evaluating expressions is the process of finding the value of an algebraic expression for a certain variable or number. It involves simplifying the algebraic expression by performing arithmetic operations and then replacing the variable(s) with specific values.

Combining Like Terms

If there are multiple terms in an algebraic expression with the same variables and exponents, we can merge them into a single term. This is known as “combining like terms.” By combining like terms, we simplify an expression and make it less complex.

For example, consider the expression \(-8x + 6y + 2x + y\). Here, \(-8x\) and \(2x\) are like terms, as are \(6y\) and \(y\). We can combine the like terms and get the simplified version of the expression:

\[-8x + 6y + 2x + y\] \[= -8x + 2x + 6y + y\] \[= -6x + 7y\]

Substitution

Substitution in evaluating expressions means to replace the variable in an expression with a given value. For example, if you want to evaluate the expression \(13x + 4y\) for \(x = 1\) and \(y = 2\), you will substitute \(1\) for \(x\) and \(2\) for \(y\) and simplify the expression:

\[13x + 4y\] \[= 13(1) + 4(2)\] \[= 13 + 8\] \[= 21\]

Operations with Binomials and Polynomials

Binomials and polynomials are algebraic expressions that are composed of constants, variables, and exponents. A binomial is an algebraic expression with two terms, such as \(x^2 + 3x\). A polynomial is any expression that has two or more terms, such as \(y^2 + 2y + 3\) or \(-7x + 2m +m^2 + 3\). We can perform addition, subtraction, and multiplication on binomials and polynomials.

Note: A binomial is a type of polynomial. Similarly, a trinomial is a specific type of polynomial with three terms.

Adding

Adding binomials or polynomials requires combining like terms. To do this, we look at all the terms and identify which terms have the same variable and exponent. These are like terms, a name that refers to terms that have identical variables and exponents. For example, the terms \(6y^2\) and \(-y^2\) are like terms. When adding like terms, we simply add their coefficients and keep the variable and exponent the same:

\[6y^2 - y^2\] \[= (6-1)y^2\] \[= 5y^2\]

Subtracting

Subtracting binomials and polynomials is only slightly different from adding. To subtract one expression from another, we change the sign of the second expression and then combine like terms. For example, if we want to subtract \(4x^2 + 2x\) from \(-3x^2 + 3\), we write the problem in the following way and simplify:

\[(-3x^2+3) - (4x^2 + 2x)\] \[= -3x^2 + 3 -4x^2 - 2x\] \[= (-3-4)x^2 + 3 - 2x\] \[= -7x^2 - 2x + 3\]

Multiplying

Multiplying binomials or polynomials requires the use of the distributive property. To multiply expressions, you must multiply each term in the first expression by each term in the second expression. Then, you combine like terms and simplify. For example, consider the expression \((2x+1)(x^2-x+2)\). Here, we are multiplying a binomial, \(2x+1\), by a polynomial, \(x^2 -x + 2\). In order to find the product, first, we will multiply each term of the second expression by the first term of the first expression, \(2x\). Then, we multiply each term of the second expression by the second term of the first expression, \(1\). Last, we combine like terms and simplify the answer:

\[(2x+1)(x^2-x+2)\] \[= (2x)(x^2) + (2x)(-x) + (2x)(2) + (1)(x^2) + (1)(-x) + (1)(2)\] \[= 2x^{3} - 2x^2 + 4x + x^2 - x + 2\] \[= 2x^3 -2x^2 + x^2 + 4x - x + 2\] \[= 2x^{3} + (-2+1)x^{2} + (4-1)x + 2\] \[= 2x^{3} - x^{2} + 3x + 2\]