Equations and inequalities · applications

Applications: Equations and Inequalities

16 question types · Secondary 2 · each worked step by step with a figure that follows the steps

O-Level · SAT · GCSE Higher

01

The Mark Needed on the Next Test to Reach a Mean

methodWrite the Mean as a Fraction, Then Multiply Both Sides by the Denominator

Priya scored 68, 77 and 62 marks on her first three science tests. Each test is marked out of 100. (a) After the fourth test her mean mark for the four tests is 72. Find her mark on the fourth test. (b) After a fifth test her mean mark for the five tests is 75. Find her mark on the fifth test.

Four tests687762xa mean of 72 over 4 tests(207 + x)/4=72
The three marks add up to 207, so the mean of the four marks is 207 + x4 = 72.
Let the fourth mark be x. The first three marks add up to 68 + 77 + 62 = 207, so the mean of the four marks is 207 + x4. The mean is 72, so 207 + x4 = 72.
step 1 of 6

The mean is the sum of the marks divided by the number of tests. With the unknown mark written as a letter, the mean is an equation with a fraction in it. Multiplying both sides by the denominator removes the fraction.

  1. Let the fourth mark be x. The first three marks add up to 68 + 77 + 62 = 207, so the mean of the four marks is 207 + x4. The mean is 72, so 207 + x4 = 72.
  2. Multiply both sides by 4 to remove the fraction: 207 + x = 288.
  3. Subtract 207 from both sides: x = 81. (a) Her fourth mark is 81. Check: 68 + 77 + 62 + 81 = 288 and 288 ÷ 4 = 72.
  4. The four marks add up to 288. Let the fifth mark be y. The mean of the five marks is 75, so 288 + y5 = 75.
  5. Multiply both sides by 5: 288 + y = 375. Subtract 288 from both sides: y = 87.
  6. (b) Her fifth mark is 87. Check: 375 ÷ 5 = 75, and 87 is not more than 100, so it is a possible mark.

answer(a) 81 marks; (b) 87 marks

techniqueEquations with Fractions · Forming Equations

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Writing 207 + x4 = 72. All four marks are added before the sum is divided by 4, so the whole sum 207 + x is the numerator.
  • Dividing by 4 again in part (b). There are now five tests, so the sum 288 + y is divided by 5.
02

The Missing Parallel Side of a Trapezium-Shaped Wall

methodChange the Subject of the Formula First, Then Substitute

The end wall of a shed is a trapezium. The area of a trapezium is given by the formula A = 12(a + b)h, where a and b are the lengths of the two parallel sides and h is the distance between them. (a) Make b the subject of the formula. (b) The wall has an area of 42 m2. One of its parallel sides is 5 m long and the distance between the parallel sides is 6 m. Find the length of the other parallel side.

abhAA=1/2 (a + b)h2A=(a + b)hmultiply both sides by 2
Multiply both sides of A = 12(a + b)h by 2: 2A = (a + b)h.
Start from A = 12(a + b)h. Multiply both sides by 2 to remove the fraction: 2A = (a + b)h.
step 1 of 5

To make b the subject, undo what the formula does to b, one operation at a time, doing the same to both sides. The bracket (a + b) is kept as a single quantity until it stands alone.

  1. Start from A = 12(a + b)h. Multiply both sides by 2 to remove the fraction: 2A = (a + b)h.
  2. Divide both sides by h, so that the bracket stands alone: 2Ah = a + b.
  3. Subtract a from both sides: 2Ah − a = b. (a) b = 2Ah − a.
  4. Substitute A = 42, h = 6 and a = 5: b = 2 × 426 − 5 = 846 − 5 = 14 − 5 = 9.
  5. (b) The other parallel side is 9 m long. Check with the original formula: 12 × (5 + 9) × 6 = 12 × 14 × 6 = 42 m2.

answer(a) b = 2Ah − a; (b) 9 m

techniqueChanging the Subject of a Formula · Treating a Bracket as a Single Quantity

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Subtracting a first and writing A − a = 12bh. The side a is inside the bracket, so it is multiplied by 12h as well. The bracket must stand alone before a is subtracted.
  • Working out 2 × 426 − 5. Only 2A is divided by h, and a is subtracted afterwards: 846 − 5 = 9.
03

A Mobile Top-Up Rule Written as a Formula and Inverted

methodTurn the Words into a Formula in Two Letters, Then Make the Other Letter the Subject

A mobile company has this rule for a top-up: a handling fee of $2 is taken off the amount paid, and every dollar that remains buys 5 minutes of calls. (a) Write a formula for the number of minutes, m, bought by a top-up of p dollars, and find the minutes bought by a top-up of $20. (b) Make p the subject of the formula, and find the top-up that buys 150 minutes.

p dollars$2p − 2 dollars, 5 minutes eachm=5(p − 2)
The fee leaves p − 2 dollars, and each of them buys 5 minutes: m = 5(p − 2).
A top-up of p dollars leaves p − 2 dollars after the fee. Each of those dollars buys 5 minutes, so m = 5(p − 2).
step 1 of 5

Follow the rule in the order the words give it: the fee is taken off first, and what remains is multiplied by 5. A bracket keeps that order. To find the top-up from the minutes, the formula is rearranged so that p is the subject.

  1. A top-up of p dollars leaves p − 2 dollars after the fee. Each of those dollars buys 5 minutes, so m = 5(p − 2).
  2. (a) The formula is m = 5(p − 2). For p = 20, m = 5 × (20 − 2) = 5 × 18 = 90 minutes.
  3. To make p the subject, divide both sides by 5 so that the bracket stands alone: m5 = p − 2.
  4. Add 2 to both sides: m5 + 2 = p, so p = m5 + 2.
  5. (b) For m = 150, p = 1505 + 2 = 30 + 2 = 32. The top-up is $32. Check: 5 × (32 − 2) = 5 × 30 = 150 minutes.

answer(a) m = 5(p − 2), 90 minutes; (b) p = m5 + 2, $32

techniqueWriting Formulas from Words · Changing the Subject of a Formula · Treating a Bracket as a Single Quantity

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Writing m = 5p − 2. That takes 2 minutes off at the end. The rule takes $2 off the payment before multiplying by 5, so the bracket is needed: m = 5(p − 2).
  • Inverting the formula as p = m − 25. The operations are undone in the reverse order: the last thing done was multiplying by 5, so divide by 5 first and then add 2.
04

Adult and Child Tickets with a Total Count and a Total Value

methodWrite Two Equations, Then Substitute One into the Other

For a school concert an adult ticket costs $12 and a child ticket costs $7. Altogether 150 tickets are sold for a total of $1360. (a) How many tickets of each kind are sold? (b) How much more money comes from the adult tickets than from the child tickets?

a + c=150(1) the tickets12a + 7c=1360(2) the dollars
With a adult tickets and c child tickets, a + c = 150 and 12a + 7c = 1360.
Let a be the number of adult tickets and c the number of child tickets. The count gives a + c = 150, and the money gives 12a + 7c = 1360.
step 1 of 6

There are two unknowns, so two equations are needed: one for the number of tickets and one for the money. The first equation gives one letter in terms of the other, and substituting it into the second leaves an equation in one letter.

  1. Let a be the number of adult tickets and c the number of child tickets. The count gives a + c = 150, and the money gives 12a + 7c = 1360.
  2. From the first equation, a = 150 − c. Substitute this for a in the second equation: 12(150 − c) + 7c = 1360.
  3. Expand the bracket: 1800 − 12c + 7c = 1360. Collect the c terms: 1800 − 5c = 1360.
  4. Add 5c to both sides: 1800 = 1360 + 5c. Subtract 1360 from both sides: 440 = 5c. Divide both sides by 5: c = 88.
  5. Then a = 150 − 88 = 62. (a) 62 adult tickets and 88 child tickets are sold. Check in the second equation: 12 × 62 + 7 × 88 = 744 + 616 = 1360.
  6. (b) The adult tickets bring in $744 and the child tickets $616, so the adult tickets bring in 744 − 616 = $128 more.

answer(a) 62 adult tickets and 88 child tickets; (b) $128

techniqueSimultaneous by Substitution · Writing a Pair of Equations from a Word Problem · Equations with Brackets

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Expanding 12(150 − c) as 1800 − c. The 12 multiplies both terms inside the bracket, so the expansion is 1800 − 12c.
  • Stopping at c = 88 and giving it as the number of adult tickets. The letter c was chosen for the child tickets, and a = 150 − c still has to be worked out.
05

Two Orders at a Cafe with Two Unknown Prices

methodScale Both Equations to Match One Coefficient, Then Subtract

At a cafe every coffee costs the same and every tea costs the same. One table orders 3 coffees and 2 teas and pays $21. Another table orders 2 coffees and 5 teas and pays $25. (a) Find the price of a coffee and the price of a tea. (b) Find the cost of an order of 5 coffees and 4 teas.

3x + 2y=21(1)2x + 5y=25(2)
A coffee costs x dollars and a tea costs y dollars. Each order is one equation.
Let a coffee cost x dollars and a tea cost y dollars. The first order gives equation (1), 3x + 2y = 21. The second order gives equation (2), 2x + 5y = 25.
step 1 of 6

Each order is one equation in the two prices. Neither letter has the same coefficient in both equations, so each equation is multiplied by a number that makes the coffee terms equal. Subtracting one equation from the other then eliminates the coffee price.

  1. Let a coffee cost x dollars and a tea cost y dollars. The first order gives equation (1), 3x + 2y = 21. The second order gives equation (2), 2x + 5y = 25.
  2. Multiply equation (1) by 2 and equation (2) by 3, so that both have 6x: 6x + 4y = 42 and 6x + 15y = 75.
  3. Subtract the first of these from the second. The 6x terms cancel: 15y − 4y = 75 − 42, so 11y = 33 and y = 3.
  4. Substitute y = 3 into equation (1): 3x + 6 = 21, so 3x = 15 and x = 5.
  5. (a) A coffee costs $5 and a tea costs $3. Check in equation (2): 2 × 5 + 5 × 3 = 10 + 15 = 25.
  6. (b) 5 coffees and 4 teas cost 5 × 5 + 4 × 3 = 25 + 12 = $37.

answer(a) a coffee costs $5 and a tea costs $3; (b) $37

techniqueSolving Simultaneous Equations by Scaling · Simultaneous by Elimination · Writing a Pair of Equations from a Word Problem

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Multiplying only the left-hand side and writing 6x + 4y = 21. Every term on both sides is multiplied, so equation (1) becomes 6x + 4y = 42.
  • Checking the prices in equation (1) only, the equation they were found from. A slip made earlier can still satisfy that equation, so the check uses the other equation.
06

Two Data Plans with the Same Rate That Never Cost the Same

methodEqual Gradients: Different Intercepts Give No Solution, the Same Intercept Gives Infinitely Many

Plan A for mobile data costs $18 a month and then $2 for each gigabyte used. Plan B costs $25 a month and then $2 for each gigabyte used. (a) Show that there is no amount of data for which the two plans cost the same, and find how much more Plan B always costs. (b) A leaflet describes Plan C by the equation 2y − 4x = 36, where y dollars is the monthly cost for x gigabytes. How many pairs of values (x, y) satisfy the equations of both Plan A and Plan C?

010203040500246810gigabytes, xcost in dollars, yABA: y = 2x + 18B: y = 2x + 25
For x gigabytes, Plan A costs y = 2x + 18 dollars and Plan B costs y = 2x + 25 dollars.
Let x be the number of gigabytes used and y the monthly cost in dollars. Plan A is y = 2x + 18 and Plan B is y = 2x + 25.
step 1 of 5

Write each plan as an equation y = mx + c for the cost y of x gigabytes. A pair of values that satisfies both equations is a point where the two lines meet. Lines with the same gradient either never meet or are the same line.

  1. Let x be the number of gigabytes used and y the monthly cost in dollars. Plan A is y = 2x + 18 and Plan B is y = 2x + 25.
  2. If the two plans cost the same, then 2x + 18 = 2x + 25. Subtract 2x from both sides: 18 = 25. This statement is false, whatever the value of x.
  3. (a) No value of x satisfies both equations, so the pair of equations has no solution. Both lines have gradient 2, and their intercepts, 18 and 25, are different, so the lines are parallel. Plan B always costs 25 − 18 = $7 more.
  4. For Plan C, divide both sides of 2y − 4x = 36 by 2: y − 2x = 18. Add 2x to both sides: y = 2x + 18. This is exactly the equation of Plan A.
  5. (b) The two equations describe the same line, so every point on it satisfies both: there are infinitely many solutions. Setting the costs equal gives 18 = 18, which is true for every value of x.

answer(a) no solution: the lines are parallel, and Plan B always costs $7 more; (b) infinitely many

techniqueHow Many Solutions a Pair of Equations Has · How Many Solutions an Equation Has

examsSAT

Common pitfalls

  • Reading 18 = 25 as a slip in the algebra and starting again. A false statement with no x left in it is the algebra showing that the equations have no solution.
  • Saying that Plan A and Plan C have no solution because 2y − 4x = 36 looks different from y = 2x + 18. Dividing every term by 2 shows that they are the same equation, so there are infinitely many solutions.
07

The Greatest Number of Rides a Budget Allows

methodForm a Two-Step Inequality, Solve It, Then Take the Greatest Whole Number

Wei Ling has $40 to spend at a fair. It costs $8 to enter and $3 for each ride. (a) Find the greatest number of rides she can afford. (b) She decides to keep at least $10 for food. Find the greatest number of rides she can afford now.

n rides: $8 to enter and $3 for each ride8 + 3n≤40
The entry and n rides cost 8 + 3n dollars, which cannot be more than 40: 8 + 3n ≤ 40.
Let n be the number of rides. The entry and the rides cost 8 + 3n dollars, and this cannot be more than 40: 8 + 3n ≤ 40.
step 1 of 6

She may spend up to $40 and not more, so the total cost is less than or equal to 40. That is an inequality in the number of rides. The number of rides is a whole number, so the answer is the greatest whole number that satisfies the inequality.

  1. Let n be the number of rides. The entry and the rides cost 8 + 3n dollars, and this cannot be more than 40: 8 + 3n ≤ 40.
  2. Subtract 8 from both sides: 3n ≤ 32. Divide both sides by 3: n ≤ 1023. Dividing by a positive number keeps the inequality sign as it is.
  3. The number of rides is a whole number, and the greatest whole number that is not more than 1023 is 10. (a) She can afford 10 rides. Check: 8 + 3 × 10 = 38 ≤ 40, but 11 rides cost 8 + 33 = 41 dollars.
  4. Keeping $10 for food adds 10 to what must fit into the $40: 8 + 3n + 10 ≤ 40, which is 3n + 18 ≤ 40.
  5. Subtract 18 from both sides: 3n ≤ 22. Divide both sides by 3: n ≤ 713.
  6. (b) She can afford 7 rides. Check: 18 + 3 × 7 = 39 ≤ 40, but 8 rides would need 18 + 24 = 42 dollars.

answer(a) 10 rides; (b) 7 rides

techniqueTwo-Step Inequalities · Integer Solutions of Inequalities · Inequalities

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Rounding 1023 up to 11 because it is nearer. Eleven rides cost $41, which is more than she has. The answer is the greatest whole number that satisfies the inequality, so it is always rounded down here.
  • Dividing 40 by 3 and answering 13 rides. The entry charge of $8 is paid first, so only $32 is left for the rides.
08

Chairs and Tables That Must Fit a Floor Area

methodForm an Inequality in Two Variables, Draw Its Boundary Line and Test a Point

A hall has 60 m2 of floor for furniture. Each chair needs 2 m2 of floor and each table needs 6 m2. There are x chairs and y tables. (a) Write an inequality in x and y for the furniture to fit, and describe the region of the graph that it gives. (b) Use the inequality to decide whether 12 chairs and 5 tables fit, and whether 18 chairs and 5 tables fit.

0246810120102030chairs, xtables, y2x + 6y ≤ 60
The chairs use 2x m2 and the tables use 6y m2, so 2x + 6y ≤ 60.
The chairs use 2x m2 and the tables use 6y m2. Together they cannot use more than 60 m2, so 2x + 6y ≤ 60.
step 1 of 5

The floor used by the furniture cannot be more than 60 m2, which gives an inequality in two variables. Its solutions are all the points on one side of a straight boundary line. A test point shows which side.

  1. The chairs use 2x m2 and the tables use 6y m2. Together they cannot use more than 60 m2, so 2x + 6y ≤ 60.
  2. The boundary is the line 2x + 6y = 60. When x = 0, 6y = 60 and y = 10. When y = 0, 2x = 60 and x = 30. The line joins (0, 10) and (30, 0), and it is drawn solid because the sign ≤ includes the points on the line.
  3. Test the point (0, 0): 2 × 0 + 6 × 0 = 0, and 0 ≤ 60 is true. (a) The inequality is 2x + 6y ≤ 60. Its region is the side of the line that contains (0, 0), with x and y not negative.
  4. For 12 chairs and 5 tables, 2 × 12 + 6 × 5 = 24 + 30 = 54, and 54 ≤ 60 is true. The point (12, 5) lies inside the region.
  5. (b) For 18 chairs and 5 tables, 2 × 18 + 6 × 5 = 36 + 30 = 66, and 66 ≤ 60 is false. So 12 chairs and 5 tables fit, using 54 m2, but 18 chairs and 5 tables do not, because they need 66 m2.

answer(a) 2x + 6y ≤ 60: the region on and below the line through (0, 10) and (30, 0); (b) 12 chairs and 5 tables fit (54 m2), 18 chairs and 5 tables do not (66 m2)

techniqueGraphing an Inequality in Two Variables · Forming an Inequality in Two Variables

examsSAT · GCSE Higher

Common pitfalls

  • Writing x + y ≤ 60. That counts pieces of furniture. The limit is on floor area, so each chair counts as 2 and each table as 6.
  • Shading the side of the line away from (0, 0) without testing a point. An empty hall uses 0 m2, which certainly fits, so (0, 0) must be in the shaded region.
09

Kites Limited by Paper and by Time at Once

methodShade Both Regions, Then Count the Integer Points in the Overlap

Farid makes large kites and small kites for a stall. A large kite uses 3 sheets of paper and takes 1 hour. A small kite uses 2 sheets of paper and takes 2 hours. He has 12 sheets of paper and 8 hours, and he wants to make at least one kite of each size. He makes x large kites and y small kites. (a) In how many different ways can he choose x and y? (b) Which choice gives the greatest total number of kites?

012345670123456789large kites, xsmall kites, ypaper: 3x + 2y ≤ 12time: x + 2y ≤ 8
The paper gives 3x + 2y ≤ 12 and the time gives x + 2y ≤ 8.
The paper used is 3x + 2y sheets, so 3x + 2y ≤ 12. The time taken is x + 2y hours, so x + 2y ≤ 8.
step 1 of 6

The paper gives one inequality and the time gives another. A choice is possible only when it satisfies both, so it lies in the overlap of the two regions. The numbers of kites are whole numbers, so the choices are the points with integer coordinates in that overlap.

  1. The paper used is 3x + 2y sheets, so 3x + 2y ≤ 12. The time taken is x + 2y hours, so x + 2y ≤ 8.
  2. The boundary 3x + 2y = 12 joins (4, 0) and (0, 6). The point (0, 0) gives 0 ≤ 12, which is true, so the region is the side of the line that contains (0, 0).
  3. The boundary x + 2y = 8 joins (8, 0) and (0, 4), and its region also contains (0, 0). Subtracting the second equation from the first gives 2x = 4, so the lines meet at (2, 3). The overlap is the region that lies under both lines.
  4. List the integer points in the overlap with x ≥ 1 and y ≥ 1. For x = 1: y = 1, 2, 3. For x = 2: y = 1, 2, 3. For x = 3: only y = 1, because 9 + 2y ≤ 12. For x = 4 the paper is used up, so y = 0, which is not allowed.
  5. (a) There are 3 + 3 + 1 = 7 different ways.
  6. (b) The total x + y is 5 at (2, 3) and at most 4 at the other six points, so he should make 2 large kites and 3 small kites, 5 kites in all. Check: 3 × 2 + 2 × 3 = 12 ≤ 12 and 2 + 2 × 3 = 8 ≤ 8.

answer(a) 7 ways; (b) 2 large kites and 3 small kites, 5 kites in all

techniqueThe Overlap of Two Inequality Regions · Forming an Inequality in Two Variables · Integer Solutions of Inequalities

examsSAT · GCSE Higher

Common pitfalls

  • Counting the points that satisfy only the paper inequality. The point (1, 4) uses 11 sheets, which is allowed, but it takes 1 + 8 = 9 hours, so it is outside the overlap.
  • Leaving out the points on the boundary lines, such as (2, 3). Both signs are ≤, so using exactly 12 sheets or exactly 8 hours is allowed.
10

A Machined Rod Accepted Within a Tolerance

methodWrite the Tolerance as an Absolute Value Inequality, Then Split It into Two Inequalities

A machine cuts steel rods that should be 50 mm long. A rod of length x mm is accepted when its length differs from 50 mm by at most 0.5 mm. (a) Write this condition as an absolute value inequality and solve it. (b) Five rods measure 49.4 mm, 49.5 mm, 50.2 mm, 50.6 mm and 50.5 mm. How many of them are accepted?

4949.55050.551|x − 50| ≤ 0.5
The distance between x and 50 on the number line is |x − 50|, and it must be at most 0.5.
The distance between x and 50 on the number line is |x − 50|. It must be at most 0.5, so |x − 50| ≤ 0.5.
step 1 of 5

The difference between the length and 50, without its sign, is the absolute value |x − 50|. An absolute value that is at most 0.5 means a number between −0.5 and 0.5, which is two inequalities that must both hold.

  1. The distance between x and 50 on the number line is |x − 50|. It must be at most 0.5, so |x − 50| ≤ 0.5.
  2. An absolute value is at most 0.5 when the number inside is between −0.5 and 0.5. This gives two inequalities: x − 50 ≥ −0.5 and x − 50 ≤ 0.5.
  3. Add 50 to both sides of each: x ≥ 49.5 and x ≤ 50.5. (a) |x − 50| ≤ 0.5, and the solution is 49.5 ≤ x ≤ 50.5. On the number line both ends have closed dots, because both end values are accepted.
  4. Test each rod. |49.4 − 50| = 0.6 and |50.6 − 50| = 0.6, and these are more than 0.5. |49.5 − 50| = 0.5, |50.2 − 50| = 0.2 and |50.5 − 50| = 0.5, and these are at most 0.5.
  5. (b) 3 of the five rods are accepted: the rods of 49.5 mm, 50.2 mm and 50.5 mm.

answer(a) |x − 50| ≤ 0.5, so 49.5 ≤ x ≤ 50.5; (b) 3 rods

techniqueSolving Absolute Value Inequalities · Solution Sets of Inequalities

examsH2

Common pitfalls

  • Writing only x − 50 ≤ 0.5 and accepting every rod shorter than 50.5 mm. A rod can also be too short. The absolute value gives a second inequality, x − 50 ≥ −0.5.
  • Rejecting the rods of 49.5 mm and 50.5 mm. The sign is ≤, so a difference of exactly 0.5 mm is accepted and both end values belong to the solution.
11

A Coffee Machine at Home or Coffee Bought at a Cafe, Month by Month

methodWrite Each Total in the Same Letter, Set the Totals Equal, Then Clear the Smaller Letter Term

Nadia spends $66 a month on coffee bought at a cafe. She could instead buy a coffee machine for $180 and then spend $30 a month on coffee pods. (a) After how many months would the two ways have cost her the same in total? (b) Over the first year, which way is cheaper, and by how much?

0200400600800024681012months, mtotal cost in dollarscafemachinecafe: 66m dollarsmachine: 180 + 30m dollars
For m months the cafe costs 66m dollars and the machine costs 180 + 30m dollars.
Let m be the number of months. Buying at the cafe costs 66m dollars in total. The machine costs 180 + 30m dollars in total: $180 once and $30 each month.
step 1 of 5

Let one letter stand for the number of months. Each way of buying coffee then has a total cost in that letter, and the totals are equal when the two expressions are equal. That equation has the unknown on both sides, so the smaller letter term is cleared from both sides first.

  1. Let m be the number of months. Buying at the cafe costs 66m dollars in total. The machine costs 180 + 30m dollars in total: $180 once and $30 each month.
  2. The totals are the same when 66m = 180 + 30m. Subtract 30m, the smaller letter term, from both sides: 36m = 180.
  3. Divide both sides by 36: m = 5. (a) The two ways have cost the same after 5 months, $330 each. Check: 66 × 5 = 330 and 180 + 30 × 5 = 330.
  4. For the first year put m = 12. The cafe costs 66 × 12 = $792 and the machine costs 180 + 30 × 12 = 180 + 360 = $540.
  5. (b) The machine is cheaper over the first year, by 792 − 540 = $252. This agrees with (a): after month 5 the cafe costs 66 − 30 = $36 a month more, for 7 more months, and 36 × 7 = 252.

answer(a) 5 months; (b) the machine, by $252

techniqueEquations with the Unknown on Both Sides · Forming Equations

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding 30m to the left side instead of subtracting it from both sides, which gives 96m = 180. Whatever is done to one side is done to the other, so the 30m comes off both sides: 36m = 180.
  • Comparing only the monthly amounts and saying that the machine saves 36 × 12 = $432 over the year. That leaves out the $180 the machine costs at the start, which is why the saving is only $252.
12

A Car Rental Price at the Airport Branch Set Against the City Branch

methodMatch the Letter Terms on Both Sides, Then Let the Constants Decide

At its city branch a car rental firm charges $30 a day plus a cleaning fee of $45, so a rental of d days costs 30d + 45 dollars. At its airport branch it will charge k dollars a day and add an airport fee equal to two days' rent, so a rental of d days costs k(d + 2) dollars. (a) The firm wants the airport branch to charge the same amount more than the city branch on every rental, however long. Find k, and show that the equation k(d + 2) = 30d + 45 then has no solution. (b) The firm later decides that the two branches should charge the same for every rental. The airport keeps its price rule from (a). What cleaning fee should the city branch charge instead of $45?

airport: k(d + 2) = kd + 2kcity: 30d + 45airport − city = (k − 30)d + 2k − 45
The airport charges k(d + 2) = kd + 2k dollars, which is (k − 30)d + 2k − 45 dollars more than the city.
Expand the airport's rule: k(d + 2) = kd + 2k. The airport price minus the city price is (kd + 2k) − (30d + 45) = (k − 30)d + 2k − 45 dollars.
step 1 of 6

An equation ad + b = cd + e has exactly one solution when the letter terms differ. When the letter terms match, they cancel, and the constants decide: different constants leave a false statement and no solution, and equal constants leave a true statement and infinitely many solutions.

  1. Expand the airport's rule: k(d + 2) = kd + 2k. The airport price minus the city price is (kd + 2k) − (30d + 45) = (k − 30)d + 2k − 45 dollars.
  2. For this difference to be the same on every rental, it must not change with d, so the d terms must match: k − 30 = 0, and k = 30. For any other k the difference changes by k − 30 dollars with each extra day.
  3. With k = 30 the equation is 30(d + 2) = 30d + 45, that is 30d + 60 = 30d + 45. Subtract 30d from both sides: 60 = 45, which is false whatever d is.
  4. (a) k = 30. The equation has no solution: the two prices are never equal, and the airport always charges 60 − 45 = $15 more.
  5. With a city fee of f dollars the equation is 30d + 60 = 30d + f. The d terms already match, so the constants decide: subtracting 30d from both sides leaves 60 = f.
  6. (b) The city branch should charge a cleaning fee of $60. The equation then reads 60 = 60 once the d terms cancel, which is true for every d: there are infinitely many solutions. Check with d = 3: 30 × 5 = 150 and 90 + 60 = 150.

answer(a) k = 30: the prices are never equal, and the airport always charges $15 more; (b) $60

techniqueChoosing a Coefficient to Fix the Solution Count · How Many Solutions an Equation Has · Equations with Brackets

examsSAT

Common pitfalls

  • Treating d as a single number and solving k(d + 2) = 30d + 45 for k. The firm's condition is about every length of rental, so it fixes the coefficient of d, not one value of k for one rental.
  • Writing the airport price as kd + 2. The fee is two days' rent, which is 2k dollars, so the bracket multiplies both terms: k(d + 2) = kd + 2k.
13

Two Resistors in Parallel and the Second Resistor a Circuit Needs

methodClear the Fraction, Gather the Terms in the Wanted Letter, Factor It Out, Then Divide

When two resistors of R1 ohms and R2 ohms are connected in parallel, their combined resistance R ohms is given by R = R1 R2R1 + R2. (a) Make R2 the subject of the formula. (b) A technician has a 60 ohm resistor and needs a combined resistance of 24 ohms. Find the resistance of the second resistor she must connect in parallel with it. Then use your formula to explain why no resistor connected in parallel with the 60 ohm one gives a combined resistance of 75 ohms.

R=R1 × R2/(R1 + R2)R(R1 + R2)=R1 × R2multiply both sides by R1 + R2R × R1 + R × R2=R1 × R2expand the bracket
Multiply both sides by R1 + R2 to clear the fraction, then expand: R R1 + R R2 = R1 R2.
Multiply both sides by R1 + R2 to clear the fraction: R(R1 + R2) = R1 R2. Expand the bracket: R R1 + R R2 = R1 R2.
step 1 of 6

The wanted letter R2 appears twice, on the top and the bottom of the fraction. Clear the fraction, gather every term in R2 on one side, factor R2 out, and divide by what is left in the bracket.

  1. Multiply both sides by R1 + R2 to clear the fraction: R(R1 + R2) = R1 R2. Expand the bracket: R R1 + R R2 = R1 R2.
  2. R2 is in two terms. Gather them on one side by subtracting R R2 from both sides: R R1 = R1 R2 − R R2.
  3. Factor out R2: R R1 = R2(R1 − R). Divide both sides by R1 − R. (a) R2 = R R1R1 − R.
  4. Substitute R1 = 60 and R = 24: R2 = 24 × 6060 − 24 = 144036 = 40. Check in the original formula: 60 × 4060 + 40 = 2400100 = 24 ohms.
  5. Substitute R = 75: R2 = 75 × 6060 − 75 = 4500−15 = −300. A resistance cannot be negative, so this value is rejected.
  6. (b) The second resistor is 40 ohms. No resistor gives 75 ohms: R2 is positive only when R1 − R is positive, so the combined resistance is always less than 60 ohms.

answer(a) R2 = R R1R1 − R; (b) 40 ohms; 75 ohms would need R2 = −300 ohms, which is impossible

techniqueMaking a Twice-Appearing Letter the Subject · Changing the Subject of a Formula

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Stopping at R2 = R(R1 + R2)R1. That is not a formula for R2, because R2 is still on the right. The terms in R2 must be gathered on one side and R2 factored out.
  • Taking −300 ohms as the resistor to use, or as a slip in the arithmetic. The formula gives a negative number because the request is impossible: two resistors in parallel always have less resistance than either one alone.
14

An Order Recovered from Two Suppliers' Quotes

methodScale One Equation to Match the Letters, Then Compare the Numbers

A grocer ordered x crates of apples and y crates of oranges, and then lost the order sheet. Two suppliers had priced the same order. Supplier P charges $8 a crate of apples and $5 a crate of oranges, and quoted $260. Supplier Q charges $12 a crate of apples and k dollars a crate of oranges, and quoted $404. (a) Write an equation for each quote. Find the value of k for which this pair of equations has no solution, and explain why. (b) Supplier Q in fact charges $8 a crate of oranges. Find the order.

8x + 5y=260(1) Supplier P12x + ky=404(2) Supplier Q
Each quote is one equation: 8x + 5y = 260 for Supplier P and 12x + ky = 404 for Supplier Q.
Supplier P's quote gives 8x + 5y = 260 (1). Supplier Q's quote gives 12x + ky = 404 (2).
step 1 of 6

Two equations have no solution when their letter terms can be made to match exactly while their numbers still differ: then one expression would have to equal two different numbers. So scale one equation until its x term matches the other's, and look at what remains.

  1. Supplier P's quote gives 8x + 5y = 260 (1). Supplier Q's quote gives 12x + ky = 404 (2).
  2. Match the x terms: multiply equation (1) by 1.5, which gives 12x + 7.5y = 390 (3).
  3. If k = 7.5, equations (2) and (3) have the same left side, 12x + 7.5y, but different right sides, 404 and 390. No pair (x, y) can make one expression equal two different numbers, so there is no solution. On a graph the two lines are parallel.
  4. (a) k = 7.5. At that price every price at Supplier Q would be 1.5 times the price at Supplier P, so Q's quote for any order would be 1.5 × 260 = $390, never $404. For any other k the y terms differ and the pair has exactly one solution.
  5. With k = 8, subtract equation (3) from equation (2): 0.5y = 14, so y = 28. Substitute into (1): 8x + 140 = 260, so 8x = 120 and x = 15.
  6. (b) The order was 15 crates of apples and 28 crates of oranges. Check with Supplier Q: 12 × 15 + 8 × 28 = 180 + 224 = 404.

answer(a) 8x + 5y = 260 and 12x + ky = 404; k = 7.5, since Q's quote would then always be 1.5 times P's; (b) 15 crates of apples and 28 crates of oranges

techniqueChoosing a Coefficient in a Pair of Equations · How Many Solutions a Pair of Equations Has · Solving Simultaneous Equations by Scaling

examsSAT

Common pitfalls

  • Comparing the y coefficients 5 and k directly and answering k = 5. The letters must be matched first: once the x terms agree, the y coefficient of equation (1) has become 7.5.
  • Saying that k = 7.5 gives infinitely many solutions because the left sides match. That happens only when the right sides match too. Here 390 and 404 differ, so there is no solution.
15

Two Garden Center Receipts and the Cost of a Border in Pairs

methodAdd the Equations to Read Off x + y, and Subtract Them to Read Off x − y

A garden center sells rose bushes at one price and lavender plants at another. One receipt shows 7 rose bushes and 3 lavender plants for $101. Another shows 3 rose bushes and 7 lavender plants for $89. A gardener plans a border of 12 pairs, each pair one rose bush and one lavender plant. (a) Without finding either price, find the cost of one pair, and of the border. (b) The owner then asks for 5 of the 12 rose bushes to be replaced by lavender plants. Without finding either price, find how much the border costs now.

7x + 3y=101(1)3x + 7y=89(2)
A rose bush costs x dollars and a lavender plant y dollars. Each receipt is one equation.
Let a rose bush cost x dollars and a lavender plant y dollars. The receipts give 7x + 3y = 101 (1) and 3x + 7y = 89 (2).
step 1 of 5

The two receipts swap their numbers of rose bushes and lavender plants, so adding them gives the same number of each, and subtracting them leaves the same number of each with opposite signs. One step then gives the sum of the prices, or their difference, without either price.

  1. Let a rose bush cost x dollars and a lavender plant y dollars. The receipts give 7x + 3y = 101 (1) and 3x + 7y = 89 (2).
  2. Add the two equations: 10x + 10y = 190. Divide both sides by 10: x + y = 19.
  3. (a) One pair costs $19, and the border of 12 pairs costs 12 × 19 = $228.
  4. Replacing a rose bush by a lavender plant lowers the cost by x − y. Subtract equation (2) from equation (1): 4x − 4y = 12, so x − y = 3. Each replacement saves $3.
  5. (b) Five replacements save 5 × 3 = $15, so the border now costs 228 − 15 = $213. Check: x + y = 19 and x − y = 3 give x = 11 and y = 8, and 7 rose bushes and 17 lavender plants cost 77 + 136 = 213.

answer(a) $19 a pair, $228 for the border; (b) $213

techniqueSolving for x + y Without Finding x and y · Simultaneous by Elimination

examsSAT

Common pitfalls

  • Solving for x and y by elimination before answering (a). It reaches the same total by a longer road. Because the receipts swap their numbers, adding them gives 10x + 10y at once, and the sum of the prices follows in one line.
  • Working out (b) as 228 − 5 × 19, as if five whole pairs were removed. The rose bushes are replaced, not removed, so each one lowers the cost by x − y = $3.
16

The Two Alarm Temperatures of a Medicine Fridge

methodSplit the Absolute Value into Two Equations: the Inside Is b or −b

A pharmacy keeps a medicine fridge at 5 °C. Its alarm is set to sound at the two temperatures that are exactly 3 °C away from 5 °C. (a) Write an absolute value equation for the alarm temperatures, x °C, and solve it. (b) The pharmacy's freezer has its alarm set to sound at −25 °C and at −15 °C. Write these two temperatures as the solutions of one equation |x − a| = b, giving the values of a and b.

051033the distance of x from 5 is 3
The distance between x and 5 is |x − 5|, and the alarm sounds when it is 3: |x − 5| = 3.
The distance between x and 5 on the temperature scale is |x − 5|. The alarm sounds when this distance is 3, so |x − 5| = 3.
step 1 of 5

The distance between x and a on a number line is |x − a|, whichever side of a the number x is on. So |x − a| = b has two solutions, one on each side: x = a − b and x = a + b. Read backwards, a is halfway between the two solutions and b is half the gap between them.

  1. The distance between x and 5 on the temperature scale is |x − 5|. The alarm sounds when this distance is 3, so |x − 5| = 3.
  2. The number inside the bars is either 3 or −3: x − 5 = 3 or x − 5 = −3.
  3. Add 5 to both sides of each: x = 8 or x = 2. (a) |x − 5| = 3, and the alarm sounds at 2 °C and at 8 °C. Check: |8 − 5| = 3 and |2 − 5| = |−3| = 3.
  4. The solutions of |x − a| = b are a − b and a + b, so a is halfway between them: a = −25 + (−15)2 = −402 = −20.
  5. b is half the gap between them: b = −15 − (−25)2 = 102 = 5. (b) a = −20 and b = 5, so the equation is |x − (−20)| = 5, that is |x + 20| = 5. Check: |−25 + 20| = 5 and |−15 + 20| = 5.

answer(a) |x − 5| = 3, so x = 2 or x = 8: the alarm sounds at 2 °C and at 8 °C; (b) a = −20 and b = 5, so |x + 20| = 5

techniqueSolving Absolute Value Equations

examsSAT · H2

Common pitfalls

  • Solving only x − 5 = 3 and giving the single answer 8 °C. The bars hide the sign of x − 5: a temperature 3 °C below 5 °C is also 3 °C away, so x = 2 is a second solution.
  • Writing the freezer's equation as |x − 20| = 5. Its solutions are 15 and 25, both above freezing. The middle of −25 and −15 is −20, and x − (−20) is x + 20.
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