← GED Mathematics

GED Mathematics · Chapter 6 · pp. 318–411

Equations, Inequalities, and Functions

This is the algebra heart of the GED® Mathematical Reasoning test. You will solve equations and inequalities, turn word problems into equations, solve quadratics, draw lines on the coordinate plane, find their slope and equation, and use functions to model real situations.

Learning objectives

By the end of this chapter you can:

  1. Solve one-step and multi-step equations, and write an equation from a word problem.
  2. Solve inequalities, including compound ones, and show the answer on a number line.
  3. Multiply and factor simple quadratic expressions, and solve quadratic equations.
  4. Use guess-and-check to answer harder multiple-choice algebra questions.
  5. Plot points on the coordinate plane and graph a line from its equation.
  6. Find the slope of a line and write its equation in slope-intercept and point-slope form.
  7. Solve a system of two linear equations by graphing and by substitution.
  8. Continue patterns, use functions and function notation, and find a function's domain and range.

In this chapter: Lesson 1 Equations · 2 Equation word problems · 3 Inequalities · 4 Quadratic equations · 5 Algebra problem solving · 6 The coordinate plane · 7 Graphing a line · 8 Slope of a line · 9 Slope and equations · 10 Systems of linear equations · 11 Patterns and functions · 12 Function applications · 13 Function notation

🎯 About the test

The GED® formula sheet gives you several of this chapter's formulas: the slope formula, the slope-intercept and point-slope forms of a line, the standard form of a quadratic equation and the quadratic formula. You still need to know what each letter means and when to use which one.

✏️ How to use this page

The tabs group the 13 lessons into topics. Each has worked examples and a short Try it. The Slope tab has a tool that finds the equation of a line from two points and draws it. The book's practice sets are in this chapter's printable assignment, and there is a quiz at the end.

Next: Lessons 1–2, equations and word problems →

Lesson 1 · Equations

🔑 Key ideas

  • An equation says that two expressions are equal.
  • To solve it, get the variable on its own by using inverse operations on both sides.
  • Check by putting your answer back into the original equation.

Solving an equation means finding the value of the variable that makes it true. Think of the equation as a balance: whatever you do to one side, you must do to the other. To undo an operation, use its inverse.

If the variable is…undo it by…
having a number addedsubtracting that number from both sides
having a number subtractedadding that number to both sides
multiplied by a numberdividing both sides by it
divided by a numbermultiplying both sides by it

📘 Example 1 · One-step equations

x − 23 = 41
   + 23  + 23
x      = 64
  1. 23 is subtracted from x, so add 23 to both sides: x = 64. Check: 64 − 23 = 41 ✔
  2. = 7. x is divided by 2, so multiply both sides by 2: x = 14.
  3. 5x = 75. x is multiplied by 5, so divide both sides by 5: x = 15.

💡 Not sure on a multiple-choice algebra question? Put each answer choice into the equation. The one that makes it true is correct.

Multi-step equations

  1. Simplify each side: remove brackets with the distributive property, then combine like terms.
  2. Add or subtract to get the variable terms on one side and the numbers on the other.
  3. Multiply or divide to get the variable on its own.

📘 Example 2 · Solve 3x + x + 5 = 25

  1. Combine like terms: 4x + 5 = 25.
  2. Subtract 5 from both sides: 4x = 20.
  3. Divide both sides by 4: x = 5.
  4. Check: 3(5) + 5 + 5 = 15 + 10 = 25 ✔

📘 Example 3 · Solve 2(x + 6) = 30 − 4x

  1. Distribute the 2: 2x + 12 = 30 − 4x.
  2. Add 4x to both sides: 6x + 12 = 30.
  3. Subtract 12: 6x = 18. Divide by 6: x = 3.
  4. Check: 2(3 + 6) = 18 and 30 − 12 = 18 ✔

Writing an equation from words

Write each part of the sentence in symbols, and put an equals sign where the sentence says is. Keep subtraction and division in the order the words give: “the difference between x and y” is x − y, and “the quotient of x and y” is , not the other way round.

📘 Example 4 · The product of a number and 6 is 44 more than twice the number

  1. Write it: 6x = 2x + 44.
  2. Subtract 2x from both sides: 4x = 44. Divide by 4: x = 11.
  3. Check: 6(11) = 66 and 2(11) + 44 = 66 ✔

Lesson 2 · Equation word problems

🔑 Key ideas

  • Decide which quantity x stands for. Write the other quantities in terms of x.
  • Write and solve an equation, then answer the question asked. Finding x is often not the last step.

📘 Example 1 · Twice as many women as men, 24 students in all. How many are women?

  1. Let x = the number of men. Then 2x = the number of women.
  2. x + 2x = 24, so 3x = 24 and x = 8.
  3. The question asks for women: 2 × 8 = 16.

Consecutive numbers follow each other in counting order, such as 7, 8, 9. Write them as x, x + 1, x + 2. Consecutive even or odd numbers go up by 2: x, x + 2, x + 4.

📘 Example 2 · Three consecutive numbers add up to 105. What is the greatest?

  1. x + (x + 1) + (x + 2) = 105, so 3x + 3 = 105.
  2. 3x = 102 and x = 34. The numbers are 34, 35 and 36.
  3. The greatest is 36.

📘 Example 3 · Tickets: $12 adults, $8 children. 200 sold for $2240. How many adult tickets?

  1. Let x = adult tickets. The rest are children's tickets: 200 − x.
  2. Money from each kind adds up to the total: 12x + 8(200 − x) = 2240.
  3. 12x + 1600 − 8x = 2240, so 4x = 640 and x = 160 adult tickets.
  4. Check: 160 × $12 = $1920, and 40 × $8 = $320. $1920 + $320 = $2240 ✔

✏️ Try it

  1. Solve 4(3 + 2x) + 8 = 92.
  2. Solve 5p − 2 = 6p − 9.
  3. Julia has 24 coins, all dimes or quarters, worth $4.50 in total. How many are dimes?
  4. The Bulldogs won twice as many games as they lost. They played 36 games with no ties. How many did they win?
Answers

1. 12 + 8x + 8 = 92, so 8x = 72 and x = 9 · 2. Subtract 5p: −2 = p − 9, so p = 7 · 3. 0.10x + 0.25(24 − x) = 4.50, so −0.15x = −1.50 and x = 10 dimes · 4. x + 2x = 36, so they lost 12 and won 24

Next: Lesson 3, inequalities →

Lesson 3 · Inequalities

🔑 Key ideas

  • An inequality uses >, ≥, < or ≤ to compare two expressions that are not equal.
  • The symbol always points to the smaller side.
  • Solve it like an equation, but reverse the symbol if you multiply or divide by a negative number.
SymbolMeansOn a number line
>greater thanopen dot, arrow to the right
≥greater than or equal toclosed dot, arrow to the right
<less thanopen dot, arrow to the left
≤less than or equal toclosed dot, arrow to the left

📘 Example 1 · Solve 3x + 2 < 8

  1. Subtract 2 from both sides: 3x < 6.
  2. Divide both sides by 3: x < 2. Any number less than 2 works.
  3. Check with a number that fits, such as 1: 3(1) + 2 = 5, and 5 < 8 ✔

📘 Example 2 · Solve −2n − 5 ≥ 3

  1. Add 5 to both sides: −2n ≥ 8.
  2. Divide both sides by −2 and reverse the symbol: n ≤ −4.
  3. Check with −5: −2(−5) − 5 = 10 − 5 = 5, and 5 ≥ 3 ✔

Why reverse? Start with a true statement, 2 < 3. Multiply both sides by −1 and you get −2 and −3. But −2 is greater than −3, so the symbol has to turn round: −2 > −3.

💡 The most common mistake is forgetting to reverse, or reversing when you didn't divide by a negative. Always check with a number that fits your answer.

Graphing the solution

An inequality usually has many solutions, so we often draw them on a number line. A closed dot means the number is included (≤ or ≥). An open dot means it is not (< or >).

−6−5−4−3−2−10123456
x < 2 (open dot: 2 is not included)
−6−5−4−3−2−10123456
n ≤ −4 (closed dot: −4 is included)
−6−5−4−3−2−10123456
x ≥ 1
−6−5−4−3−2−10123456
2 < x < 4 (a compound inequality)

Compound inequalities

A compound inequality joins two inequalities. Split it into two, solve each, then put the answers back together.

📘 Example 3 · Solve 3x + 4 < 5x < 16 + x

3x + 4 < 5x      5x < 16 + x
     4 < 2x      4x < 16
     2 < x        x < 4
  1. Left part: subtract 3x, then divide by 2. So 2 < x.
  2. Right part: subtract x, then divide by 4. So x < 4.
  3. Together: 2 < x < 4. Any number between 2 and 4 works, such as 3: 13 < 15 < 19 ✔

✏️ Try it

  1. 3x − 7 > 5
  2. −4(x + 2) < 24
  3. 3 < 5x − 27 < 53
  4. A square's perimeter is P = 4s. The perimeter is at most 64 inches. What can the side s be?
Answers

1. 3x > 12, so x > 4 · 2. −4x − 8 < 24, so −4x < 32; divide by −4 and reverse: x > −8 · 3. Add 27: 30 < 5x < 80, so 6 < x < 16 · 4. 4s ≤ 64, so s ≤ 16 inches (and more than 0)

Next: Lessons 4–5, quadratic equations and problem solving →

Lesson 4 · Quadratic equations

🔑 Key ideas

  • A quadratic equation has a squared variable, such as x2. It can have two solutions.
  • To solve one, make one side 0 and factor the other side.
  • Then find the values of x that make each factor equal 0.

Factoring out a common term

To factor an expression is to write it as a product. Start by looking for a term that divides into every part.

📘 Example 1 · Factor 15x2 + 9x

  1. Both terms can be divided by 3x: 15x2 ÷ 3x = 5x, and 9x ÷ 3x = 3.
  2. So 15x2 + 9x = 3x(5x + 3). Multiply back out to check.

Multiplying two brackets: FOIL

Multiply every term in the first bracket by every term in the second. FOIL helps you not miss one: First, Outer, Inner, Last.

📘 Example 2 · Multiply (x + 3)(x − 5)

FirstOuterInnerLast
x · x = x2x · (−5) = −5x3 · x = 3x3 · (−5) = −15

Combine the middle terms: x2 − 5x + 3x − 15 = x2 − 2x − 15.

Solving by factoring

Factoring runs FOIL backwards. For x2 + bx + c, look for two numbers that multiply to c and add to b. Then use this fact: if two factors multiply to 0, at least one of them is 0.

📘 Example 3 · Solve x2 − 3x = 4

  1. Make one side 0. Subtract 4: x2 − 3x − 4 = 0.
  2. Find two numbers that multiply to −4 and add to −3: −4 and 1. So (x − 4)(x + 1) = 0.
  3. Set each factor to 0: x − 4 = 0 gives x = 4; x + 1 = 0 gives x = −1.
  4. The solutions are 4 and −1. Check: 16 − 12 = 4 ✔ and 1 + 3 = 4 ✔

💡 Factoring is often trial and error. List the factor pairs of the last number and test which pair adds to the middle number.

The quadratic formula

When an equation won't factor neatly, use the quadratic formula. It is on the GED® formula sheet, along with the standard form of a quadratic equation, ax2 + bx + c = 0.

x =

📘 Example 4 · Example 3 again, with the formula

  1. In x2 − 3x − 4 = 0, a = 1, b = −3 and c = −4.
  2. b2 − 4ac = 9 − 4(1)(−4) = 9 + 16 = 25, and √25 = 5.
  3. x = , so x = = 4 or x = = −1. The same answers.

✏️ Try it

  1. Multiply (2x + 1)(x − 2).
  2. Factor x2 + 5x − 14.
  3. Solve x2 − x = 20.
  4. What is the only positive solution of 2x2 − 7x − 30 = 0?
Answers

1. 2x2 − 4x + x − 2 = 2x2 − 3x − 2 · 2. (x + 7)(x − 2) · 3. x2 − x − 20 = (x − 5)(x + 4) = 0, so x = 5 or −4 · 4. (2x + 5)(x − 6) = 0, so x = 6 (the other solution, −2.5, is negative)

Lesson 5 · Algebra problem solving

🔑 Key ideas

  • Guess-and-check is a way to answer a hard multiple-choice question.
  • Try an answer choice in the problem. If it works, you have the answer. If not, try another.

On a multiple-choice question the right answer is already on the screen. Instead of solving from scratch, you can test the choices. This works especially well for quadratics and for word problems where the equation is hard to write.

📘 Example 1 · Which is a solution of x2 − x = 6? A. 4   B. 3   C. 0   D. −1

  1. A: 42 − 4 = 12. Not 6.
  2. B: 32 − 3 = 6 ✔ The answer is B, 3. You can stop here.

📘 Example 2 · Terry is 10 years older than Tomas. Twenty years ago, Terry was twice as old as Tomas. How old is Terry now?

A. 25   B. 30   C. 40   D. 45

  1. A: Terry 25, Tomas 15. Twenty years ago Tomas wasn't born. ✘
  2. B: Terry 30, Tomas 20. Twenty years ago they were 10 and 0. ✘
  3. C: Terry 40, Tomas 30. Twenty years ago they were 20 and 10, and 20 is twice 10 ✔ The answer is C, 40.

💡 Use your judgment. If you can see how to solve a problem quickly, just solve it; testing four choices can take longer.

✏️ Try it (guess-and-check)

  1. A number divided by 2 is 12 less than the number. What is the number? (12, 20, 24 or 28)
  2. Nelson is twice as old as Maria. Six years ago he was five times as old as she was. How old was Nelson six years ago? (5, 10, 15 or 20)
  3. A group spent $440 on 20 passes: adults $25, children $15. How many children's passes? (5, 6, 9 or 14)
Answers

1. 24: 24 ÷ 2 = 12, and 24 − 12 = 12 ✔ · 2. 10: Maria is 8 and Nelson 16 now; six years ago they were 2 and 10, and 10 = 5 × 2 ✔ · 3. 6: 14 adults × $25 = $350, 6 children × $15 = $90, total $440 ✔

Next: Lessons 6–7, the coordinate plane and graphing a line →

Lesson 6 · The coordinate plane

🔑 Key ideas

  • The coordinate plane is made by two number lines that cross: the horizontal x-axis and the vertical y-axis.
  • They cross at the origin, (0, 0).
  • Every point has an ordered pair (x, y). The x value always comes first.

To find a point, start at the origin. The first number tells you how far to move right (positive) or left (negative). The second tells you how far to move up (positive) or down (negative). The axes split the plane into four quadrants, numbered I to IV anticlockwise from the top right.

−6−4−2246−6−4−2246xy I II III IV M (2, 3) P (−1, −3) A (−2, 1) B (3, −2)
Four points and the four quadrants

📘 Example 1 · Reading and plotting points

  1. M is 2 spaces right of the origin and 3 up: (2, 3).
  2. P is 1 space left and 3 down: (−1, −3).
  3. To plot A (−2, 1): from the origin count 2 left, then 1 up.
  4. To plot B (3, −2): count 3 right, then 2 down.

💡 If a grid on the test isn't labelled, the origin is still (0, 0). Count right or left first, then up or down.

✏️ Try it

−6−4−2246−6−4−2246xy C D E F
  1. Write the ordered pair for C, D, E and F.
  2. Which quadrant is each of C, D and E in?
Answers

1. C (4, 1) · D (−3, 2) · E (−2, −4) · F (0, −3) · 2. C is in I, D in II, E in III. F is on the y-axis, so it is in no quadrant.

Lesson 7 · Graphing a line

🔑 Key ideas

  • A linear equation has two variables, x and y, with no exponents. Its graph is a straight line.
  • To find a point on the line, choose a value for x and solve for y.
  • You need at least two points to draw the line. A third point is a good check.

📘 Example 1 · Graph y = 2x − 1

xy = 2x − 1point
−12(−1) − 1 = −3(−1, −3)
02(0) − 1 = −1(0, −1)
12(1) − 1 = 1(1, 1)
22(2) − 1 = 3(2, 3)
−6−4−2246−6−4−2246xy (2, 3) (0, −1) y = 2x − 1

Make a table of values, plot the points and draw a straight line through them. If one point is off the line, re-check its working.

💡 If y isn't on its own, isolate it first. For 2x + y = 5, subtract 2x from both sides: y = −2x + 5.

You don't always need to draw. To test whether a line passes through a point, put the point's x into the equation and see if you get its y.

📘 Example 2 · Which line passes through A (5, 3)?

  1. y = 2x − 3: 2(5) − 3 = 7. Not 3.
  2. y = −x + 4: −5 + 4 = −1. Not 3.
  3. y = x − 2: 5 − 2 = 3 ✔ This line passes through A.

✏️ Try it

  1. For y = −x + 3, find y when x = −1, 0 and 2.
  2. Which line passes through (2, −1): y = x + 1 or y = −2x + 3?
  3. Rewrite 3x + y = 7 with y on its own.
Answers

1. 4, 3 and 1, giving (−1, 4), (0, 3), (2, 1) · 2. y = −2x + 3, because −2(2) + 3 = −1 (the other gives 3) · 3. y = −3x + 7

Next: Lessons 8–9, slope and line equations →

Lesson 8 · Slope of a line

🔑 Key ideas

  • Slope measures how steep a line is: the ratio of rise to run.
  • Reading left to right, a line going up has a positive slope; a line going down has a negative slope.
  • Find slope by counting squares on a graph, or with the slope formula.

Picture a road up a hill. The rise is how far it climbs, and the run is how far it goes forward. A big rise over a short run is a steep road. Slope = rise ÷ run.

−6−4−2246−6−4−2246xy run 2 rise 1 P
Line P: up 1 for every 2 across. Slope =
−6−4−2246−6−4−2246xy run 2 rise −6 s
Line s: down 6 for every 2 across. Slope = = −3
−6−4−2246−6−4−2246xy y = 3: slope 0 x = −2: undefined
Horizontal: slope 0. Vertical: slope undefined.

The slope formula

The slope formula is on the GED® formula sheet. For two points (x1, y1) and (x2, y2) on a line:

m =

It is the difference of the y's over the difference of the x's. Either point can be point 1; just subtract in the same order on top and bottom.

📘 Example 1 · Slope of the line through (1, 4) and (7, 6)

  1. Let (x1, y1) = (1, 4) and (x2, y2) = (7, 6).
  2. m = = = .
  3. The slope is positive, so the line rises: up 1 for every 3 to the right.

📌 Two special cases to remember

  • A horizontal line, like flat road, has slope 0 (no rise).
  • A vertical line has an undefined slope (no run, and you can't divide by 0).

✏️ Try it

  1. Find the slope through (0, 2) and (4, 0).
  2. Find the slope through (4, 2) and (2, 2).
  3. Find the slope through (6, 1) and (0, 3).
  4. Line L passes through (1, 0) with slope 3. Which point is also on it: (0, 3), (1, 3), (2, 3) or (2, 5)?
Answers

1. = − · 2. = 0 (horizontal) · 3. = − · 4. (2, 3): one step right from (1, 0) means 3 steps up

Lesson 9 · Slope and equations

🔑 Key ideas

  • Knowing the slope and one point on a line, you can write its equation two ways.
  • Slope-intercept form: y = mx + b.
  • Point-slope form: y − y1 = m(x − x1).

📄 On the GED® formula sheet

FormulaWhat the letters mean
slope m = (x1, y1) and (x2, y2) are two points on the line
y = mx + bm = slope, b = y-intercept (where the line crosses the y-axis)
y − y1 = m(x − x1)m = slope, (x1, y1) = a point on the line

On the computer test, open the sheet from the formula button on the toolbar.

📘 Example 1 · Slope-intercept form: m = 4, through (2, 3)

  1. Put m, x and y into y = mx + b: 3 = 4(2) + b.
  2. Isolate b: 3 = 8 + b, so b = −5.
  3. Write the equation with x and y left as letters: y = 4x − 5.
−6−4−2246−6−4−2246xy (2, 3) b = −5

📘 Example 2 · Point-slope form: through (4, 3) with slope 2

  1. The point is (x1, y1) = (4, 3) and m = 2.
  2. Put them into the form: y − 3 = 2(x − 4).
  3. Solve for y to get slope-intercept form: y − 3 = 2x − 8, so y = 2x − 5.

📘 Example 3 · From two points: (−1, 0) and (2, −3)

  1. Find the slope first: m = = = −1.
  2. Use one point, say (−1, 0): 0 = −1(−1) + b, so 0 = 1 + b and b = −1.
  3. The line is y = −x − 1. Check with the other point: −2 − 1 = −3 ✔

🧪 Try it: two-point line tool

Type two points. The tool finds the slope, writes the equation of the line through them and draws it.

Point 1
Point 2

    Slope as a unit rate

    When a graph shows a real situation, the slope is a unit rate: how many units of y there are for one unit of x, such as dollars per gallon or miles per hour. Write it as a whole number or a decimal, not a fraction.

    📘 Example 4 · The price of gas

    Two receipts: 3 gallons cost $7.50, and 5 gallons cost $12.50.

    1. m = = = 2.5.
    2. The gas costs $2.50 per gallon.

    The line also passes through (0, 0), since 0 gallons cost $0. That is why 7.50 ÷ 3 gives the same answer.

    012345651015gallonscost ($) (3, 7.50) (5, 12.50)

    📘 Example 5 · A falling price

    A phone costs $680 when it comes out and $570 two years later, falling at a steady rate. The points are (0, 680) and (2, 570).

    m = = = −55. The price drops by $55 a year.

    ✏️ Try it

    1. Write the equation, in slope-intercept form, of the line through (1, −2) with slope −4.
    2. Write the point-slope equation of the line through (2, 1) with slope 3. Then solve it for y.
    3. Find the equation of the line through (−3, 0) and (−2, 4).
    4. A caterer orders 150 lb of clams to serve 100 people, and 30 lb for 20 people. How many pounds is that per serving?
    Answers

    1. −2 = −4(1) + b, so b = 2: y = −4x + 2 · 2. y − 1 = 3(x − 2), so y = 3x − 5 · 3. m = 4 ÷ 1 = 4; 0 = 4(−3) + b, so b = 12: y = 4x + 12 · 4. (150 − 30) ÷ (100 − 20) = 120 ÷ 80 = 1.5 lb per serving

    Next: Lesson 10, systems of linear equations →

    Lesson 10 · Systems of linear equations

    🔑 Key ideas

    • Two linear equations together make a system. Its solution is the point where the two lines cross, written (x, y).
    • You can solve a system by graphing both lines and reading where they meet.
    • Or by substitution: solve one equation for a variable and put that into the other equation.

    Solve by graphing

    1. Make an x–y table (a T-chart) for each equation. Put x = 0 and find y; then put y = 0 and find x.
    2. Plot the points and draw both lines.
    3. Read off the point where they cross.

    📘 Example 1 · Graph 2x + y = 4 and x − y = −1

    2x + y = 4x − y = −1
    xyxy
    0401
    20−10

    The lines cross at (1, 2). Check in both: 2(1) + 2 = 4 ✔ and 1 − 2 = −1 ✔

    −6−4−2246−6−4−2246xy (1, 2) 2x + y = 4 x − y = −1

    Solve by substitution

    📘 Example 2 · The same system, by substitution

    1. Solve the first equation for y: 2x + y = 4 becomes y = 4 − 2x.
    2. Put that into the second equation: x − (4 − 2x) = −1, so 3x − 4 = −1, 3x = 3 and x = 1.
    3. Put x = 1 back into y = 4 − 2x: y = 2.
    4. The solution is (1, 2), the same point the graph showed.

    💡 Asked whether a point solves a system? Put it into both equations. It must make both true.

    💡 Watch the brackets when you substitute an expression: −(4 − 2x) is −4 + 2x.

    ✏️ Try it: solve by substitution

    1. 7x − y = 22 and 4x + 2y = 10
    2. x + y = 9 and 2x − 3y = 8
    3. y = 3x + 15 and 5x − 2y = −26
    Answers

    1. y = 7x − 22, so 4x + 14x − 44 = 10, x = 3, y = −1: (3, −1) · 2. y = 9 − x, so 2x − 27 + 3x = 8, x = 7, y = 2: (7, 2) · 3. 5x − 6x − 30 = −26, so x = −4, y = 3: (−4, 3)

    Next: Lessons 11–13, patterns and functions →

    Lesson 11 · Patterns and functions

    🔑 Key ideas

    • A pattern (or sequence) is a list of numbers made by a rule.
    • A function is a rule that links each input number to exactly one output number.
    • To use a function, substitute the value you know and solve.

    To find a pattern's rule, ask: what do I do to each term to get the next one? Start by finding the difference between neighbouring terms.

    📘 Example 1 · Find the 7th term of 1, 2, 4, 8, 16, …

    1. The rule: each term is 2 times the one before.
    2. 6th term: 16 × 2 = 32. 7th term: 32 × 2 = 64.

    A street vendor sells pizza at $1.50 a slice. The cost depends on (is a function of) the number of slices.

    Slices, n123456
    Cost, c$1.50$3.00$4.50$6.00$7.50$9.00

    As a function: c = $1.50n. Give it any number of slices and it gives one cost.

    📘 Example 2 · Using a function

    1. y = 3x + 5. Find y when x = 9: y = 3(9) + 5 = 32.
    2. n = 2m − 1. Find n when m = 12: n = 2(12) − 1 = 23.

    Lesson 12 · Function applications

    🔑 Key ideas

    • Functions are used at work to find costs, profit, wages and taxes, and to compare plans.
    • To use one, you must know what each variable stands for. The question tells you.

    📘 Example 1 · A project's finishing cost

    Finishing a project costs $2000 a day for the first 7 days and $3500 a day after that. For more than 7 days the cost is C = $2000d + $1500(d − 7), where d is the number of days. What does a 12-day project cost?

    C = 2000(12) + 1500(12 − 7) = 24,000 + 7,500 = $31,500.

    Check another way: 7 days × $2000 = $14,000, and 5 more days × $3500 = $17,500. $14,000 + $17,500 = $31,500 ✔

    📘 Example 2 · Comparing two health clubs

    Healthstars charges $250 to join plus $8 a month: P = 250 + 8m. Freedom charges $25 a month: P = 25m. Nita will go for 18 months.

    1. Healthstars: 250 + 8(18) = 250 + 144 = $394.
    2. Freedom: 25(18) = $450.
    3. Healthstars is cheaper over 18 months, by $56, even with the joining fee.

    The graph shows the lines cross between 14 and 15 months. For a shorter stay, Freedom would be cheaper.

    04812162024100200300400500600monthsprice ($) $394 $450 Healthstars Freedom

    💡 There is often more than one way to solve a function problem. Solve with the function, then check another way.

    Lesson 13 · Function notation

    🔑 Key ideas

    • In function notation the output is written f(x), said “f of x”, as in f(x) = x + 4.
    • The domain is the set of input values the function is defined for, or that make sense.
    • The range is the set of outputs the function can give from its domain.

    f(x) does not mean f times x. It is a name for the output. f(4) means “the output when the input is 4”: if f(x) = 3x − 5, then f(4) = 3(4) − 5 = 7.

    📘 Example 1 · What is the domain of f(x) = ?

    1. You can't divide by 0, so find the inputs that make the bottom 0: x2 − 1 = 0 when x = 1 or x = −1.
    2. The domain is all real numbers except 1 and −1.

    📘 Example 2 · What is the range of f(x) = x2?

    1. Any number can be squared, so the domain is all real numbers.
    2. A square is never negative. The smallest output is f(0) = 0.
    3. As x moves away from 0 in either direction, x2 grows without limit.
    4. The range is all numbers greater than or equal to 0.
    −6−4−2246−6−4−2246xy f(x) = x² lowest: 0

    📘 Example 3 · Working backwards from the output

    Each cup of oats makes 24 cookies: f(c) = 24c. Rafael's family made 180 cookies. How many cups did they use?

    Put the output in for f(c): 180 = 24c, so c = 180 ÷ 24 = 7.5 cups.

    📘 Example 4 · Domain and range of a real situation

    Take the project cost from Lesson 12: f(d) = $2000d + $1500(d − 7). The formula is only meant for jobs of 7 days or more, so the domain is d ≥ 7. The smallest cost is f(7) = $14,000, and there is no upper limit, so the range is f(d) ≥ $14,000.

    ⚠️ Book correction

    The book gives this domain as “d ≤ 7”. It should be d ≥ 7: the formula is for jobs of 7 days or more, which is why the smallest cost is at d = 7.

    ✏️ Try it

    1. What is the next term? 21, 26, 31, 36, …
    2. What is the next term? 1000, 500, 250, 125, 62.5, …
    3. The Chimney Sweep charges C = $25 + (s − $15), where s is the cost of extra services. Jan buys a $89 smoke guard. What does she pay?
    4. Job 1 pays P = $9.75h and Job 2 pays P = $70 + $8.40h. For 30 hours, how much more does Job 2 pay?
    5. If g(x) = , which input is not in the domain?
    Answers

    1. Add 5 each time: 41 · 2. Halve each time: 31.25 · 3. 25 + (89 − 15) = $99 · 4. Job 1: $292.50. Job 2: 70 + 252 = $322. Difference $29.50 · 5. x = 3, because it makes the bottom 0

    Next: summary and quiz →

    Chapter summary

    • Equations: do the same to both sides. Simplify, then add or subtract, then multiply or divide. Check by substituting.
    • Word problems: let x be one quantity and write the others in terms of it. After finding x, answer the question that was asked.
    • Inequalities: solve like equations, but reverse the symbol when you multiply or divide by a negative. Open dot for < and >, closed dot for ≤ and ≥.
    • Quadratics: multiply brackets with FOIL. To solve, make one side 0, factor, and set each factor to 0. The quadratic formula is on the formula sheet.
    • Guess-and-check: on a hard multiple-choice question, try the answer choices.
    • Coordinate plane: points are (x, y), x first. To graph a line, make a table of at least two points.
    • Slope: m = rise ÷ run = . Horizontal lines have slope 0; vertical lines are undefined. In a real situation the slope is a unit rate.
    • Line equations: y = mx + b and y − y1 = m(x − x1), both on the formula sheet.
    • Systems: the solution is where the lines cross. Find it by graphing or by substitution, and check it in both equations.
    • Functions: one output for each input. f(x) is the output. The domain is the allowed inputs; the range is the possible outputs.

    📄 Practice questions

    A selection of the book's practice questions for Lessons 1–13 and its chapter practice questions are in the printable assignment for this chapter.

    Mini-quiz

    1. Solve x − 23 = 41.
    2. Solve 5y + 3(y + 2) = 54.
    3. Three consecutive numbers add up to 105. What is the greatest of them?
    4. Solve −2n − 5 ≥ 3.
    5. When must you reverse the inequality symbol?
    6. What are the solutions of x² − 3x − 4 = 0?
    7. Which point is 3 units left of the origin and 2 units up?
    8. What is the slope of the line through (1, 4) and (7, 6)?
    9. What is the slope of a vertical line?
    10. A line has slope 2 and passes through (3, 1). What is its equation?
    11. Where do the lines y = x + 1 and 2x + y = 4 meet?
    12. If f(x) = 3x − 5, what is f(4)?

    Source: GED® Test Prep Mathematics, Chapter 6 (Equations, Inequalities, and Functions). Explanations, the graphs, the Try it answers, the line tool and the quiz are Guru Jazzy's own. GED® is a registered trademark of the American Council on Education; this page is not made or endorsed by GED Testing Service.