Consider the following situation:
A company sells products that cost $5.00. Excusing any of their production costs, generate a linear equation modeling the company’s revenue if they sell 13 products.
In this case, there is a linear relationship between the number of products sold and the total revenue generated. If 1 product is sold, the revenue is $5.00; if 3 products are sold, the revenue is 3 * 5.00 = 15.00. We can model this relationship as a function of the number of products sold, with each product netting $5.00:
Revenue = $5.00 * number of products sold
Let the variable p represent the unknown number of products sold, and R(p) represent the revenue generated:
R(p) = 5.00 * p is the linear equation modeling this situation.
Always remember to consider what the independent variable is (in this case number of products sold), and what the dependent variable is (in this case the revenue generated), and create an equation that relates them appropriately.
Consider the following single-variable equation:
2 − 3(x + 2) − 4x = 3 + 4x
To solve this equation, manipulate the equation such that a single x is on one side of the equation, and a single value is on the other side. In this case, we will begin by distributing the −3 through the parentheses:
2 − 3x − 6 − 4x = 3 + 4x
Now, combine like terms on the left side of the equation:
−4 − 7x = 3 + 4x
Now, either add 7x to both sides of the equation, subtract 4x from both sides of the equation, add 4 to both sides of the equation, or subtract 3 from both sides of the equation. To avoid a negative sign in the variable’s coefficient, we will add 7x to both sides:
−4 = 3 + 11x
Because we want all of the numbers on one side and only 1x on the other side, now subtract 3 from both sides. Remember that any operation performed on one side of the equation must also be performed on the other side of the equation to maintain equality:
−7 = 11x
Notice that the variable is currently multiplied by 11. To undo this operation, and to isolate just 1x, it is necessary to divide the right side by 11; and, to keep the equation equal, we must also divide the left side by 11:
−7/11 = 11x/11
−7/11 = x
Our answer can be verified by substituting −7/11 into our original x and checking to see if the equation is true:
2 − 3(−7/11 + 2) − 4 * −7/11 = 3 + 4 * −7/11
2 − 3(15/11) + 28/11 = 3 − 28/11
2 − 45/11 + 28/11 = 3 − 28/11
22/11 − 45/11 + 28/11 = 33/11 − 28/11
5/11 = 5/11
The solution is correct.
On exams, presuming enough time is available, always verify your answer using the method above.
Consider the following linear inequality in one variable:
3 + 2(4 − x) ≤ 4x − 1
As with linear equations, linear inequalities are solved by manipulating the inequality such that the variable is on one side of the inequality and a value is on the other side. One important thing to remember when working with inequalities is that any time the inequality is multiplied or divided by a negative value, the direction of the inequality switches. For example, −4x ≥ 8 becomes x ≤ −2.
Let’s solve the inequality from above:
Distribute the 2: 3 + 8 − 2x ≤ 4x − 1
Combine like terms: 11 − 2x ≤ 4x − 1
Rearrange: 12 ≤ 6x
Divide both sides by 6: 2 ≤ x
As before, this solution can be verified by selecting a value for x that is larger than or equal to 2 and testing if it yields a true statement. We will leave the verification to you.
You will also be tested on your understanding of graphs of linear inequalities. The solution determined above would graph as a closed circle at x = 2 with an arrow pointing to the right to indicate that all x values larger than 2 are also part of the solution set. The closed circle indicates that 2 is also part of the solution set. An open circle, contrastingly, indicates that the value where the circle is located is not part of the solution set.
Consider the following situation:
In a chemical reaction, the concentration of reactant A decreases as the concentration of reactant B increases. It is determined that there is a linear relationship between the concentrations of the reactants as the reaction progresses. At one point, there are found to be 10 parts of reactant A for every 1 part of reactant B. At another point, there are found to be 4 parts of A for every 4 parts of B. Express the relationship between the two reactants using a linear function.
From the information provided, it can be seen that the concentrations of the reactants exhibit an inverse relationship. We can rewrite the relative concentrations as points on an AB-coordinate plane, in which the A-axis represents the concentration of A, and the B-axis represents the concentration of B.
We are given two points then: (10,1) and (4,4). From our knowledge of linear equations, we can determine the slope of the line that relates these two variables:
Slope =
(y2 − y1)/(x2 − x1) = (4 − 1)/(4 − 10) = 12
Now that we have the slope, we can use the point-slope formula to determine the linear function describing the relationship:
(y − 4) = 1/2(x − 4)
y − 4 = 12x − 2
y = 12x + 2
where y is the concentration of B, and x is the concentration of A.
Consider the following problem:
3xy + 4x − 2y + 3(x − y) = 4y + 2x
Solve for y in terms of x.
Because we have only one equation, but two variables, we can only express one variable in terms of the other. To solve for y in terms of x means to end up with an equation with y on one side and everything else on the other.
The first step, in this case, entails distributing the 3 through the parentheses so that every term is clarified:
3xy + 4x − 2y + 3x − 3y = 4y + 2x
Now, move every term containing a y to one side of the equation and everything else to the other side of the equation:
3xy − 2y − 3y − 4y = 2x − 4x − 3x
Combine like terms:
3xy − 9y = −5x
Notice that y can be factored out of the expression on the left:
y(3x − 9) = −5x
We can now divide both sides by the parentheses to solve for y in terms of x:
y = −5x/(3x − 9)
Consider the following linear inequality:
−3x − 4(y + 3) > 2y + 2(3 − x)
As with a linear equation with two variables, we will begin by manipulating the inequality to solve for y in terms of x:
Distribute the 4:
−3x − 4y − 12 > 2y + 6 − 2x
Combine like terms through rearrangement:
−4y − 2y > −2x + 3x + 6 + 12
−6y > x +18
Recall, when multiplying or dividing both sides of an inequality by a negative value, the inequality sign switches directions:
−6y/−6 < x + 18/−6
y < (−1/6)x − 3
To graph this solution, we first find two points that lie along the line. Because this is already in slope-intercept form, we know that (0,−3) is one point on the line, and using the slope of −16, we can subtract 1 from the y value of the y-intercept and add 6 to the x value of the y intercept:
(0 + 6, −3 + −1) = (6, −4)
Plot the points (0, −3) and (6, −4), and because our inequality has a less than symbol, draw a dotted line extending through both points. Because y is less than the graphed line, the solution set is every point below the dotted line. The solution set is indicated by a shaded region.
Consider the following system of equations:
4x + y = 6
−3x − 2y = 8
Three methods of solving this system should be available to you: elimination, substitution, and graphing.
The goal in solving a system by elimination is to multiply one of the equations by a value such that combining both equations will result in the elimination of one of the variables. In this case, by multiplying the top equation by 2, the coefficients of the y variable will cancel and the x variable can be found:
2(4x + y = 6) becomes 8x + 2y = 12
Adding the first equation with the second yields:
5x = 20
x = 4
Now that x has been solved, its value can be substituted into either equation to solve for y:
4(4) + y = 6
16 + y = 6
y = −10
The solution set is (4, −10).
Solving by elimination is not always the most efficient method. Sometimes it is best to solve a system using substitution. The goal of solving by substitution is to express one of the variables in terms of the other. The expression can then be substituted into the other equation, and the value for the variable can be found:
4x + y = 6
−3x − 2y = 8
Solve the first equation for y:
y = −4x + 6
Now substitute this expression for y into the other equation:
−3x − 2(−4x + 6) =8
−3x + 8x − 12 = 8
5x = 20
x = 4
Substituting this value of x into either equation will yield the value of y, −10.
Solving a system of equations using the method of graphing entails graphing both lines and finding their point of intersection. There are three cases of solutions:
The two lines do not intersect. This happens when the two lines are parallel but have different y or x intercepts. In this case, there is no solution.
The two lines intersect at one point. This happens when the two lines are distinct but do not share the same slope. In this case, there is one solution.
The two lines intersect at every point. This happens when the two lines are the same. In this case, there are infinite solutions.
Determining a solution set is equivalent to finding all values that will satisfy an equation or inequality. It is worthwhile to recognize the relationship between the number and type of variable or variables present in an equation or inequality. It is also valuable to understand the solution sets possible for systems of equations.
Consider the following equations:
x + 1 = 2 and x + 1 = x
The solution set for the first is {1} because only when x = 1 is the equation true. The solution set for the second is the null set, meaning that there is no value that will make the equation true—there is no solution.
In the case of an equation containing two first-degree variables, such as (y=mx), the solution set will be multiple points that lie in a straight line. Contrastingly, the solution set of a linear inequality will either be the collection of points above or below a line, with the points along the line also being part of the solution depending on whether the inequality contains an equal sign.
A solution set can always be verified by substituting values back into the original expression to determine whether the equation produces a true statement.
Problems that present two distinct pieces of information that can be represented with variables are prime candidates for solving by way of a system of equations. Consider the following problem:
Pens cost 50 cents and pencils cost 25 cents. Amanda purchased a total of 9 pens and pencils for a cost of $4.25. What is the number of pens she bought and what is the number of pencils she bought?
This problem presents two pieces of information—the total number of pens and pencils and the prices of each. Both the associated costs and the total amount are fixed values, but the number of pens and pencils are both unknown. Let’s use x to represent the number of pens and y to represent the number of pencils. We can generate a system of equations as follows:
50x + 25y = 425
x + y = 9
Solving the second equation for y and substituting the expression into the first equation:
y = 9 − x
50x + 25(9 − x) = 425
50x + 225 − 25x = 425
25x = 200
x = 8
Substituting this value into the second equation:
8 + y = 9
y = 1
Amanda bought 8 pens and 1 pencil.
In other words, what do they stand for in terms of the context given in the problem or question? Let’s analyze the previous problem statement and its corresponding system of equations.
Pens cost 50 cents, and pencils cost 25 cents. Amanda purchased a total of 9 pens and pencils for a cost of $4.25. What is the number of pens she bought, and what is the number of pencils she bought?
The system of equations are:
50x + 25y = 425
x + y = 9
The first equation corresponds to the associated cost of the pens and pencils. Notice that the original problem statement presents the information in terms of dollars and cents. However, by multiplying each dollar amount by 100, we can eliminate the decimals and make the numbers easier to work with. Notice that the coefficient of the x, which we assumed to be the unknown number of pens is 50 — this is because we were given that the cost of every pen was 50 cents (and multiplying by 100 gives 50). Likewise, the coefficient of the y, which represents the unknown number of pencils, is 25.
The product of the unknown number of pens with the cost per pen gives the total cost of the pens, and the product of the unknown number of pencils with the cost per pencil gives the total cost of the pencils. This combined cost represents the total cost of pens and pencils, which is given in the problem statement.
The problem statement also gives us the total number of pens and pencils purchased. Because x represents the unknown number of pens and y represents the unknown number of pencils, x + y is equivalent to the total number of pens and pencils.
By understanding which elements of a problem represent constants and which elements represent variables, it is manageable to generate an equation or system of equations that models the situation.
Linear equations provide information about the slope and intercept(s) of a particular line.
The slope is the ratio of the change in y values to the change in x values, often written as m = (y2 − y1)/(x2 − x1), where (x1, y1) and (x2, y2) are points on the line. The slope is commonly remembered by the expression rise over run.
The intercepts of a line are the points at which a line crosses the x or y axis. These values can be determined by setting the x or y variable equal to 0 and solving for the remaining variable.
To match a given linear equation with its corresponding graph, begin by finding the slope and intercept(s) of the equation, and then find the graph that exhibits the same slope and intercept(s).
From a provided graph, you can determine both the slope and any axes intercepts present. In order to match a given graph with its corresponding equation, begin by selecting two points along the line and then determine the ratio of the change in the y values to the change in the x values.
Next, find the y-intercept of the line (in the case where no y-intercept is present, then the line is vertical and contains only one variable, x).
From these two values, the slope-intercept form of the line can be written: y = mx + b, where m is the slope, and b is the y-intercept. If the equation generated does not match an available answer choice, it may be necessary to algebraically manipulate the equation so that it does match an answer choice.
Occasionally you will be asked to match the description of a graph with its corresponding equation. For situations like this, it is important to understand common terminology, such as, “The following data exhibits a linear/exponential/inverse, etc. relationship…”
As a result, familiarity with these terms is important.
The x or y intercept of a function is the point at which the function crosses the x or y axis, respectively. These points are located by evaluating the function with x or y equal to 0. When x is equal to 0, you are solving for the y-intercept. When y is equal to 0, you are solving for the x-intercept.
Linear equations produce straight-line graphs. Lines can be positive and slope upward as the x value increases, or they can be negative and slope downward as the x value increases.
Quadratic equations produce parabolas. Parabolas can be positive, have a vertex as an absolute minimum, and extend upwards to infinity; or, they can be negative, have a vertex as an absolute maximum, and extend downwards to infinity. Quadratic equations will always have a y-intercept, but there are different cases regarding their x-intercepts. A quadratic that has only imaginary solutions does not cross the x-axis. A quadratic that has only one real solution has its vertex along the x-axis. A quadratic that has two solutions crosses the x-axis at two points. The x intercepts of a quadratic are also called the zeroes of the quadratic.
Cubic equations can be positive or negative. As the x value increases, positive cubic functions extend from negative infinity toward a point of inflection before curving upward to positive infinity. Negative cubic functions exhibit the opposite—as the x value increases, they extend from positive infinity to a point of inflection before curving downward to negative infinity.
Familiarity with parent functions and the general rules dictating horizontal and vertical translations, as well as dilations and scaling factors, enables a quick assessment of how a function will graph.
Consider the parent function of a linear equation: y = x
This plots as a line with a slope of 1. But, if we add a constant to either side of the equation, the line will vertically shift either up or down, depending on the sign of the constant.
Multiplying the x variable by a constant will produce another type of change in the line’s graph. If we multiply it by a negative 1, the slope of the line becomes negative, meaning as the x values increase, the y values decrease. Multiplying the x variable by a large positive constant increases the steepness of the line, whereas multiplying it by a very small positive constant decreases the steepness of the line.
Similar translations take place when other functions are manipulated in the same manner. You should become familiar with whether a function will shift horizontally or vertically, as well as whether a function will be stretched or dilated.
Some general trends can be deduced from a specific example. Consider the following:
f(x) = −(x − 4)2 + 3
This should be recognized as the vertex form of a parabola. Usually, a parabola has its vertex at the origin, (0, 0), but in this case, because of the negative sign in front of the parentheses, the subtraction of 4 inside the parentheses, and the addition of 3, the vertex of the parabola will be shifted.
The addition or subtraction of a constant from the x variable produces a horizontal shift. In this case, the x value of the vertex will be shifted 4 places to the right. If instead, the parentheses contained (x + 4), the x value of the vertex would be shifted 4 places to the left. The shift is in the opposite direction of the sign. The addition of 3 on the outside results in an upward vertical translation. If, instead, there was a −3, the vertical translation would be downward. The negative sign in front of the parentheses produces an inversion — rather than the parabola extending upward and possessing an absolute minimum, the parabola extends downward and possesses an absolute maximum.