Algebraic Expressions: 14 Applications, Worked Step by Step

A Temperature Rule Written as an Expression and Evaluated

methodWrite the Rule as an Expression, Then Substitute with Brackets Round a Negative Value

A weather station changes a temperature of c °C into degrees Fahrenheit by this rule: multiply the temperature by 95, then add 32. (a) Write the rule as an expression in c, and find the Fahrenheit temperature when c = 25. (b) On a winter night c = −10. Find the Fahrenheit temperature.

Multiply c by 95, then add 32: the expression is 95c + 32.

Multiplying c by 95 gives 95c, and adding 32 gives the expression 95c + 32.

step 1 of 5

The rule is a list of operations on c, and writing them in order gives an expression. Substituting a value for c turns the expression into a calculation. A negative value goes in brackets so that its sign stays with it.

Multiplying c by 95 gives 95c, and adding 32 gives the expression 95c + 32.
Substitute c = 25: 95 × 25 + 32. Multiply before adding: 95 × 25 = 9 × 5 = 45.
(a) The expression is 95c + 32. When c = 25 its value is 45 + 32 = 77, so the temperature is 77 °F.
Substitute c = −10 with brackets round the negative number: 95 × (−10) + 32. A positive number times a negative number is negative, so 95 × (−10) = −18.
(b) −18 + 32 = 14, so the temperature is 14 °F. Check: from 25 °C down to −10 °C is a fall of 35 °C, which is 95 × 35 = 63 °F, and 77 − 63 = 14.

answer(a) 95c + 32, which is 77 °F; (b) 14 °F

techniqueSubstitution

examsPSLE · O-Level · GCSE Higher

Common pitfalls

Adding before multiplying, as in 95 × (25 + 32). The rule multiplies the temperature first and adds 32 afterwards, so only c is multiplied by 95.
Writing 95 × (−10) as +18. A positive number times a negative number is negative, so the product is −18 and the answer is −18 + 32 = 14, not 50.

The Fence Round a Plot Whose Sides Are Given in x and y

methodAdd the Four Sides, Then Collect the x Terms and the y Terms Separately

A plot of land has four sides. In meters they are 3x + 2y, x + 4y, 2x − y and x + 4y. (a) Write the length of the fence round the plot as a simplified expression. (b) Find the length of the fence when x = 5 and y = 3.

The fence is the perimeter: (3x + 2y) + (x + 4y) + (2x − y) + (x + 4y).

The length of the fence is the sum of the four sides: (3x + 2y) + (x + 4y) + (2x − y) + (x + 4y).

step 1 of 6

The fence is the perimeter, which is the sum of the four sides. Terms with the same letter are like terms and can be added, so the x terms are collected together and the y terms are collected together.

The length of the fence is the sum of the four sides: (3x + 2y) + (x + 4y) + (2x − y) + (x + 4y).
Collect the x terms: 3x + x + 2x + x = 7x.
Collect the y terms, keeping the minus sign with the y it belongs to: 2y + 4y − y + 4y = 9y.
(a) The length of the fence is (7x + 9y) m. 7x and 9y are unlike terms, so the expression cannot be made any simpler.
Substitute x = 5 and y = 3: 7 × 5 + 9 × 3 = 35 + 27 = 62.
(b) The fence is 62 m long. Check: the four sides are 21 m, 17 m, 7 m and 17 m, and 21 + 17 + 7 + 17 = 62.

answer(a) (7x + 9y) m; (b) 62 m

techniqueLike Terms · Substitution

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Adding the x terms to the y terms to get 16xy. x and y stand for different lengths, so 7x and 9y are unlike terms and stay separate.
Counting −y as +y and writing 11y. The third side is 2x − y, so its y term is subtracted: 2y + 4y − y + 4y = 9y.

A Vegetable Bed Made Longer, Its Area Written Two Ways

methodSplit the Rectangle into Two Parts: the Factor Outside Multiplies Every Term Inside the Bracket

A vegetable bed is x m long and 5 m wide. It is made 3 m longer. (a) Write the area of the longer bed in two ways, one with a bracket and one without. (b) A second bed is (2x + 1) m long and 4 m wide. Find the total area of the two beds as a single simplified expression.

The longer bed is (x + 3) m by 5 m, so its area is 5(x + 3) m².

The longer bed is (x + 3) m long and 5 m wide, so its area is 5(x + 3) m².

step 1 of 6

The longer bed is one rectangle, and it is also the old part and the new part put together. Writing its area both ways shows what expanding a bracket means. The second bed is expanded in the same way, and like terms are then collected.

The longer bed is (x + 3) m long and 5 m wide, so its area is 5(x + 3) m².
Split the bed into the old part and the new part. The old part is 5 × x = 5x m² and the new part is 5 × 3 = 15 m².
(a) The two parts make up the whole bed, so 5(x + 3) = 5x + 15. The 5 multiplies every term inside the bracket.
The area of the second bed is 4(2x + 1) m². Expand it in the same way: 4 × 2x + 4 × 1 = 8x + 4.
Add the two areas and collect like terms: 5x + 15 + 8x + 4 = 13x + 19.
(b) The total area is (13x + 19) m². Check with x = 2: the beds are 5 × 5 = 25 m² and 4 × 5 = 20 m², which is 45 m² in all, and 13 × 2 + 19 = 45.

answer(a) 5(x + 3) = 5x + 15; (b) (13x + 19) m²

techniqueExpanding Brackets · Like Terms

examsO-Level · SAT · GCSE Higher

Common pitfalls

Expanding 5(x + 3) as 5x + 3. The 5 multiplies both parts of the bed, so the new part is 5 × 3 = 15 m², not 3 m².
Adding 5x + 15 + 8x + 4 to get 32x. Only like terms can be added: the x terms make 13x and the numbers make 19.

A Path of Constant Width Round a Rectangular Lawn

methodOuter Rectangle Minus Inner Rectangle, Expanded and Simplified

A rectangular lawn is x m wide and (x + 5) m long. A path 1 m wide runs all the way round it. (a) Find the area of the path as a simplified expression in x. (b) Find the area of the path when x = 8.

The path adds 1 m at both ends, so the outer rectangle is (x + 2) m by (x + 7) m.

The path adds 1 m at both ends of each side, so the outer rectangle is (x + 2) m wide and (x + 7) m long.

step 1 of 6

The path is what is left when the lawn is taken away from the rectangle that holds the lawn and the path together. Each area is a product of two sides, so the two products are expanded and the second is subtracted from the first.

The path adds 1 m at both ends of each side, so the outer rectangle is (x + 2) m wide and (x + 7) m long.
The area of the path is the outer rectangle minus the lawn: (x + 2)(x + 7) − x(x + 5).
Expand each product: (x + 2)(x + 7) = x² + 7x + 2x + 14 = x² + 9x + 14, and x(x + 5) = x² + 5x.
Subtract every term of the lawn, so both of its terms change sign: x² + 9x + 14 − x² − 5x.
(a) Collect like terms: x² − x² = 0 and 9x − 5x = 4x. The area of the path is (4x + 14) m².
(b) When x = 8 the area is 4 × 8 + 14 = 46 m². Check: the outer rectangle is 10 × 15 = 150 m² and the lawn is 8 × 13 = 104 m², and 150 − 104 = 46.

answer(a) (4x + 14) m²; (b) 46 m²

techniqueExpanding Two Brackets · Like Terms · Substitution

examsO-Level · SAT · GCSE Higher

Common pitfalls

Making the outer rectangle (x + 1) m by (x + 6) m. The path lies on both sides of the lawn, so each side of the outer rectangle is 2 m longer than the lawn, not 1 m.
Subtracting only the first term of the lawn, as in x² + 9x + 14 − x² + 5x. The whole area x² + 5x is taken away, so both of its terms change sign.

An Order Totaled Quickly by Taking Out a Common Factor

methodab + ac = a(b + c): Take the Common Factor Outside the Bracket

A school orders 37 sets. Each set is a calculator at $46 and a case of instruments at $54. (a) Without a calculator, find the total cost of the 37 sets. (b) The delivery bill for n boxes is (24n + 36) dollars, and 12 classes share it equally. Write one class's share as an expression in n, and find the share when n = 10.

The total cost is 37 × 46 + 37 × 54 dollars.

The calculators cost 37 × 46 dollars and the cases cost 37 × 54 dollars, so the total is 37 × 46 + 37 × 54.

step 1 of 5

When two products share a factor, the factor can be taken outside a bracket, and the bracket may then be easy to work out. Factorizing a bill by the number of classes shows one class's share inside the bracket.

The calculators cost 37 × 46 dollars and the cases cost 37 × 54 dollars, so the total is 37 × 46 + 37 × 54.
Both products have the factor 37. Take it outside a bracket: 37 × 46 + 37 × 54 = 37(46 + 54).
(a) 46 + 54 = 100, so the total cost is 37 × 100 = $3700.
The highest common factor of 24 and 36 is 12, so 24n + 36 = 12 × 2n + 12 × 3 = 12(2n + 3).
(b) The bill is 12 equal shares of (2n + 3) dollars, so one class pays (2n + 3) dollars. When n = 10 the share is 2 × 10 + 3 = $23. Check: the bill is 24 × 10 + 36 = $276, and 12 × 23 = 276.

answer(a) $3700; (b) (2n + 3) dollars, which is $23 when n = 10

techniqueFactoring by Common Factor

examsO-Level · SAT · GCSE Higher

Common pitfalls

Writing 37 × 46 + 37 × 54 as 37 × 37 × (46 + 54). The common factor is taken out once, because each of the 37 sets costs 46 + 54 dollars.
Taking out 6 and stopping at 6(4n + 6). The factorization is correct but not complete, and it shows 6 equal shares, not the 12 shares that the classes need.

A Rectangular Plot Whose Area Is a Quadratic Expression

methodRun the Area Model Backwards: Two Numbers with a Product of 12 and a Sum of 7

A rectangular plot has an area of (x² + 7x + 12) m². Each of its sides is x m plus a whole number of meters. (a) Factorize the area to find the two sides. (b) Find the length of the fencing round the plot when x = 6.

The x² is a square of side x and the 12 is a rectangle. The other two parts must make 7x.

In an area model the x² is a square of side x, and the 12 is a rectangle whose sides are two whole numbers. The other two parts are the x terms, and together they must make 7x.

step 1 of 6

Expanding two brackets fills an area model with four parts. Factorizing runs the model backwards: the x² and the number are two of the parts, and the two numbers on the sides are found from their product and their sum.

In an area model the x² is a square of side x, and the 12 is a rectangle whose sides are two whole numbers. The other two parts are the x terms, and together they must make 7x.
The two numbers have a product of 12 and a sum of 7. The pairs with a product of 12 are 1 × 12, 2 × 6 and 3 × 4, and only 3 + 4 = 7.
With 3 and 4 on the sides, the four parts are x², 3x, 4x and 12, and 3x + 4x = 7x.
(a) x² + 7x + 12 = (x + 3)(x + 4), so the sides are (x + 3) m and (x + 4) m.
When x = 6 the sides are 6 + 3 = 9 m and 6 + 4 = 10 m. The fencing is the perimeter: 2 × (9 + 10) = 38.
(b) The fencing is 38 m long. Check: the area is 9 × 10 = 90 m², and 6² + 7 × 6 + 12 = 36 + 42 + 12 = 90.

answer(a) (x + 3) m and (x + 4) m; (b) 38 m

techniqueFactoring Quadratics · Expanding Two Brackets

examsO-Level · SAT · GCSE Higher

Common pitfalls

Choosing 2 and 6 because 2 × 6 = 12. Their sum is 8, which belongs to x² + 8x + 12. The two numbers must have a sum of 7 as well as a product of 12.
Substituting x = 6 into the area and giving 90 as the fencing. 90 m² is the area of the plot. The fencing is the perimeter, 2 × (9 + 10) = 38 m.

The Area of a Square Frame from the Difference of Two Squares

methoda² − b² = (a − b)(a + b)

A square metal plate has a side of 103 mm. A square hole of side 97 mm is cut from its middle, which leaves a frame. (a) Without a calculator, find the area of the frame. (b) A second frame has an outer side of (x + 5) cm and a square hole of side (x − 5) cm. Find its area as a simplified expression, and find the area when x = 12.

The frame is the plate minus the hole: 103² − 97², a difference of two squares.

The area of the frame is the plate minus the hole: 103² − 97². This is a difference of two squares.

step 1 of 6

The area of a frame is a large square minus a small square, which is a difference of two squares. The identity a² − b² = (a − b)(a + b) turns it into one product. The same identity works when a and b are expressions.

The area of the frame is the plate minus the hole: 103² − 97². This is a difference of two squares.
Use a² − b² = (a − b)(a + b) with a = 103 and b = 97: 103² − 97² = (103 − 97)(103 + 97).
(a) 103 − 97 = 6 and 103 + 97 = 200, so the area of the frame is 6 × 200 = 1200 mm².
The area of the second frame is (x + 5)² − (x − 5)². Here a = x + 5 and b = x − 5, so a − b = x + 5 − x + 5 = 10 and a + b = 2x.
The area is (a − b)(a + b) = 10 × 2x = 20x.
(b) The area is 20x cm², which is 20 × 12 = 240 cm² when x = 12. Check: the two sides are 17 cm and 7 cm, and 17² − 7² = 289 − 49 = 240.

answer(a) 1200 mm²; (b) 20x cm², which is 240 cm² when x = 12

techniqueAlgebraic Identities

examsO-Level · SAT · GCSE Higher

Common pitfalls

Working out (103 − 97)² = 36. A difference of two squares is not the square of the difference: a² − b² = (a − b)(a + b), and the factor a + b has been left out.
Writing (x + 5) − (x − 5) as 0. Subtracting −5 adds 5, so (x + 5) − (x − 5) = x + 5 − x + 5 = 10.

A Square Floor Cut into Four Parts to Square a Number Mentally

method(a + b)² = a² + 2ab + b², and (a − b)² = a² − 2ab + b²

A square floor has 51 tiles along each side. (a) Cut the square into four parts, using 51 = 50 + 1, and find the number of tiles without a calculator. (b) A smaller square floor has 49 tiles along each side. Use 49 = 50 − 1 to find its number of tiles.

Write 51 as 50 + 1 and cut each side of the square into 50 and 1.

The floor holds 51² tiles. Write 51 as 50 + 1, and cut each side into a part of 50 tiles and a part of 1 tile.

step 1 of 6

A square of side a + b can be cut into a square of side a, a square of side b and two equal rectangles of a by b. That is the identity (a + b)² = a² + 2ab + b², and with a = 50 every part is easy to work out.

The floor holds 51² tiles. Write 51 as 50 + 1, and cut each side into a part of 50 tiles and a part of 1 tile.
The four parts are a square of 50 × 50 = 2500 tiles, two strips of 50 × 1 = 50 tiles each, and a corner of 1 × 1 = 1 tile.
This is the identity (a + b)² = a² + 2ab + b² with a = 50 and b = 1: 51² = 2500 + 2 × 50 + 1.
(a) 2500 + 100 + 1 = 2601 tiles.
For 49 use (a − b)² = a² − 2ab + b² with a = 50 and b = 1: 49² = 2500 − 2 × 50 + 1.
(b) 2500 − 100 + 1 = 2401 tiles. Check: 51² − 49² = (51 − 49)(51 + 49) = 2 × 100 = 200, and 2601 − 2401 = 200.

answer(a) 2601 tiles; (b) 2401 tiles

techniqueAlgebraic Identities · Expanding Two Brackets

examsO-Level · SAT · GCSE Higher

Common pitfalls

Writing 51² = 50² + 1² = 2501. That counts the large square and the corner only. The two strips of 50 tiles, which are the 2ab term, have been left out.
Writing 49² = 50² − 1² = 2499. (a − b)² = a² − 2ab + b², so the middle term −100 is needed and the last term is +1.

The Total Time for a Journey Out and Back at Two Speeds

methodWrite Each Time as an Algebraic Fraction, Then Add Them over a Common Denominator

A van drives 60 km to a depot at x km/h and returns by the same road at (x + 10) km/h. (a) Write the total driving time as a single fraction. (b) Find the total driving time when x = 20.

Time is distance divided by speed: 60x hours out and 60x + 10 hours back.

Time is distance divided by speed. The drive out takes 60x hours and the return takes 60x + 10 hours.

step 1 of 5

Time is distance divided by speed, so each part of the journey is an algebraic fraction. Algebraic fractions are added in the same way as numerical fractions: each is rewritten over a common denominator, and then the numerators are added.

Time is distance divided by speed. The drive out takes 60x hours and the return takes 60x + 10 hours.
The total time is 60x + 60x + 10. The two denominators have no common factor, so the common denominator is their product, x(x + 10).
Multiply the numerator and the denominator of each fraction by the factor its denominator lacks: 60(x + 10)x(x + 10) + 60xx(x + 10).
(a) Add the numerators: 60x + 600 + 60x = 120x + 600. The total time is 120x + 600x(x + 10) hours.
(b) When x = 20 the total time is 120 × 20 + 60020 × 30 = 3000600 = 5 hours. Check: 6020 = 3 hours out and 6030 = 2 hours back, and 3 + 2 = 5.

answer(a) 120x + 600x(x + 10) hours; (b) 5 hours

techniqueAdding and Multiplying Algebraic Fractions · Algebraic Fractions · Substitution

examsO-Level · SAT · GCSE Higher

Common pitfalls

Adding the numerators and adding the denominators to get 1202x + 10. When x = 20 that is 2.4 hours, which is less than the 3 hours of the drive out alone. Fractions are added over a common denominator.
Multiplying only the denominators, to get 60 + 60x(x + 10). A fraction keeps its value only when its numerator and its denominator are multiplied by the same expression.

Three Pupils' Expressions for the Tiles Round a Square Pond

methodExpand Both Expressions to Compare Them: One Agreeing Value Is Not a Proof

A square pond is n tiles long on each side, and one row of square tiles is laid all the way round it. Aisha says the border needs (n + 2)² − n² tiles. Ben says it needs 4(n + 1) tiles. (a) Show that the two expressions are equivalent. (b) Chen's expression is n² + 8. It gives 12 tiles when n = 2, and so do the other two. Is Chen's expression equivalent to them?

Aisha: the large square minus the pond. (n + 2)² − n² = n² + 4n + 4 − n² = 4n + 4.

Aisha takes the pond, n² tiles, away from the large square of side n + 2. Expand: (n + 2)² = n² + 4n + 4, so (n + 2)² − n² = 4n + 4.

step 1 of 6

Two expressions are equivalent when they are equal for every value of the letter. Expanding and simplifying both is a proof, because it holds for every value at once. Testing a value can show that two expressions are different, but it cannot show that they are the same.

Aisha takes the pond, n² tiles, away from the large square of side n + 2. Expand: (n + 2)² = n² + 4n + 4, so (n + 2)² − n² = 4n + 4.
Ben counts four runs of n + 1 tiles, each run being one side and one corner. Expand: 4(n + 1) = 4n + 4.
(a) Both expressions simplify to 4n + 4, so they are equal for every value of n. They are equivalent.
When n = 2, 4n + 4 = 12 and n² + 8 = 4 + 8 = 12. One value that agrees does not prove that two expressions are equivalent, because two different expressions can be equal at a single value.
Test another value. When n = 3, 4n + 4 = 16 but n² + 8 = 9 + 8 = 17.
(b) Chen's expression is not equivalent to the others: when n = 3 it gives 17 tiles and the border needs 16. One value that disagrees is enough to show this.

answer(a) Both simplify to 4n + 4; (b) No: when n = 3 Chen's expression gives 17 tiles and the border needs 16

techniqueRecognizing Equivalent Expressions · Expanding Two Brackets · Substitution

examsO-Level · SAT · GCSE Higher

Common pitfalls

Deciding that n² + 8 is equivalent because it agrees when n = 2. Two different expressions can be equal at one value. Equivalent expressions are equal at every value.
Expanding (n + 2)² as n² + 4. (n + 2)² = (n + 2)(n + 2) = n² + 2n + 2n + 4, so the 4n is part of it.

A Drawer Organizer with Four Compartments, and the Sides of the Tray from Its Area

methodFactor by Grouping: Take the Four Terms Two at a Time, Then Take Out the Shared Bracket

A drawer organizer is a rectangular tray divided into four rectangular compartments. Its largest compartment is x cm long and y cm wide, and the maker gives the floor area of the whole tray as (xy + 8x + 6y + 48) cm². (a) Factor the area by grouping to find the two sides of the tray in terms of x and y. (b) A customer orders a square organizer whose largest compartment is 20 cm long. Find the width of that compartment and the floor area of the tray.

No factor divides all four terms, so take them in pairs: (xy + 8x) + (6y + 48).

The four terms have no common factor, so group them in pairs: (xy + 8x) + (6y + 48).

step 1 of 5

No single factor divides all four terms of the area. Taken two at a time, though, each pair has a common factor, and both pairs leave the same bracket behind. That bracket is then a common factor of the whole expression, and the area becomes a product of two sides.

The four terms have no common factor, so group them in pairs: (xy + 8x) + (6y + 48).
Take x out of the first pair and 6 out of the second: x(y + 8) + 6(y + 8). Both pairs leave the same bracket, y + 8.
(a) Take out the common bracket: xy + 8x + 6y + 48 = (x + 6)(y + 8). The sides of the tray are (x + 6) cm and (y + 8) cm. Check by expanding: (x + 6)(y + 8) = xy + 8x + 6y + 48.
A square tray has two equal sides, so x + 6 = y + 8. The largest compartment is 20 cm long, so x = 20, which gives 26 = y + 8 and y = 18.
(b) The largest compartment is 18 cm wide. The tray is 26 cm by 26 cm, so its floor area is 26 × 26 = 676 cm². Check: 20 × 18 + 8 × 20 + 6 × 18 + 48 = 360 + 160 + 108 + 48 = 676.

answer(a) (x + 6) cm and (y + 8) cm; (b) the compartment is 18 cm wide, and the floor area of the tray is 676 cm²

techniqueFactoring by Grouping · Factoring by Common Factor · Substitution

examsO-Level

Common pitfalls

Pairing xy with 48 and 8x with 6y. Those pairs share no factor, so no common bracket appears. Pair the terms so that each pair has a factor in common: xy with 8x, and 6y with 48.
Writing the sides as (x + 8) and (y + 6). Expanding gives xy + 6x + 8y + 48, which is a different area. The 6 taken out of the second pair joins the x, so the sides are (x + 6) and (y + 8).

The Length and Width of a Loading Pallet from Its Recorded Top Area

methodSplit the Middle Term with Two Numbers Whose Product Is 6 × 10 and Whose Sum Is 19, Then Group

A warehouse loads square boxes with sides of x cm onto a rectangular pallet. Its records give the top area of the pallet as (6x² + 19x + 10) cm², and each side of the pallet is a whole number of box sides plus a whole number of centimeters. (a) Factor the area to find the length and the width of the pallet in terms of x. (b) The boxes have sides of 40 cm. Find the length and the width of the pallet, and check them against the recorded area.

Two numbers with a product of 6 × 10 = 60 and a sum of 19: only 4 and 15.

Look for two numbers with a product of 6 × 10 = 60 and a sum of 19. The pairs with a product of 60 are 1 and 60, 2 and 30, 3 and 20, 4 and 15, 5 and 12, and 6 and 10. Only 4 + 15 = 19.

step 1 of 6

When the x² term has a coefficient other than 1, the two numbers that split the middle term have a product equal to that coefficient times the number term, here 6 × 10 = 60, and a sum equal to the coefficient of x. Splitting the middle term gives four terms, and grouping them in pairs gives the two factors.

Look for two numbers with a product of 6 × 10 = 60 and a sum of 19. The pairs with a product of 60 are 1 and 60, 2 and 30, 3 and 20, 4 and 15, 5 and 12, and 6 and 10. Only 4 + 15 = 19.
Split the middle term with them: 6x² + 19x + 10 = 6x² + 4x + 15x + 10.
Group in pairs and take out a factor from each: 2x(3x + 2) + 5(3x + 2). Both pairs leave 3x + 2, so the area is (2x + 5)(3x + 2).
(a) The pallet is (3x + 2) cm long and (2x + 5) cm wide: three box sides and 2 cm along it, and two box sides and 5 cm across it. When a box side is more than 3 cm, 3x + 2 is the longer side.
Substitute x = 40: the length is 3 × 40 + 2 = 122 cm and the width is 2 × 40 + 5 = 85 cm.
(b) The pallet is 122 cm long and 85 cm wide. Check: 122 × 85 = 10370, and the recorded area is 6 × 40² + 19 × 40 + 10 = 9600 + 760 + 10 = 10370 cm².

answer(a) (3x + 2) cm long and (2x + 5) cm wide; (b) 122 cm long and 85 cm wide, and 122 × 85 = 10370 cm² agrees with the recorded area

techniqueFactoring Quadratics with a Leading Coefficient · Factoring by Grouping · Substitution

examsO-Level · SAT · GCSE Higher

Common pitfalls

Looking for two numbers with a product of 10 and a sum of 19, as for x² + 19x + 10. The 6 in front of x² changes the search: the product must be 6 × 10 = 60.
Writing (3x + 5)(2x + 2). It expands to 6x² + 16x + 10, so the middle term is wrong. Expanding a factorization shows whether each number is in the right bracket.

The Steel in the Walls of a Hollow Cube-Shaped Box, and the Side of a Larger Box

methoda³ − b³ = (a − b)(a² + ab + b²)

A closed box in the shape of a cube is made from steel sheet 1 mm thick, so its outside is a cube of side a mm and its inside is a cube of side b mm, where a − b = 2. (a) A box has an outer side of 101 mm. Factor a³ − b³ and use it to find the volume of steel in the box without a calculator. (b) A larger box made from the same sheet contains 135002 mm³ of steel. Let its outer side be (m + 1) mm. Show that the volume of steel is (6m² + 2) mm³, and find the outer side.

The steel is the outer cube minus the inner cube: a³ − b³ = (a − b)(a² + ab + b²).

The steel is the outer cube minus the inner cube, so its volume is a³ − b³ = (a − b)(a² + ab + b²). Here a = 101 and b = 99.

step 1 of 6

The steel fills the space between the outer cube and the inner cube, so its volume is a³ − b³. The identity a³ − b³ = (a − b)(a² + ab + b²) turns this into a product whose first factor is the small number a − b = 2. Writing the two sides as m + 1 and m − 1 makes the second factor simple as well.

The steel is the outer cube minus the inner cube, so its volume is a³ − b³ = (a − b)(a² + ab + b²). Here a = 101 and b = 99.
a − b = 2, and a² + ab + b² = 101² + 101 × 99 + 99² = 10201 + 9999 + 9801 = 30001.
(a) The volume of steel is 2 × 30001 = 60002 mm³. Check: 101³ − 99³ = 1030301 − 970299 = 60002.
For the larger box a = m + 1 and b = m − 1. Then a² + ab + b² = (m² + 2m + 1) + (m² − 1) + (m² − 2m + 1) = 3m² + 1, so the volume of steel is 2(3m² + 1) = 6m² + 2.
Set this equal to the steel in the box: 6m² + 2 = 135002, so 6m² = 135000 and m² = 22500. Then m = 150 or m = −150, and m = −150 would make the outer side negative, so m = 150.
(b) The outer side is 150 + 1 = 151 mm. Check: 151³ − 149³ = 3442951 − 3307949 = 135002.

answer(a) 60002 mm³; (b) the volume of steel is 2(3m² + 1) = 6m² + 2, and the outer side is 151 mm

techniqueThe Sum and Difference of Two Cubes · Expanding Two Brackets

examsO-Level

Common pitfalls

Writing a³ − b³ as (a − b)³, which gives 2³ = 8 mm³. The difference of two cubes is not the cube of the difference: the second factor a² + ab + b² is needed.
Using a² − ab + b² as the second factor. That factor belongs to a³ + b³. For a difference of two cubes the middle sign in the second factor is a plus.

A Party Price in Two Forms That Must Always Agree, and a Second Venue's Price

methodExpand, Then Match the Terms in n and the Numbers; Where Two Prices Agree at One Value, Solve an Equation

A party venue's website prices a party for n children, where n is 5 or more, at (14n + 60) dollars. Its booking form gives the same price as q dollars for the first 5 children together plus p dollars for each child after the fifth, which is p(n − 5) + q dollars. The two must give the same price for every party. (a) Find p and q. (b) A second venue charges $110 for the first 5 children and $16 for each child after the fifth. For what number of children do the two venues charge the same, and what is that price?

Expand the booking form: p(n − 5) + q = pn − 5p + q.

Expand the booking form: p(n − 5) + q = pn − 5p + q. Its term in n is pn and its number term is q − 5p.

step 1 of 6

The website and the booking form describe one price list, so the two expressions are equal for every number of children. Two expressions that are equal for every n have the same coefficient of n and the same number term, which gives one equation for each unknown. Two different price lists can still be equal at a single value of n, and that value is found by solving an equation.

Expand the booking form: p(n − 5) + q = pn − 5p + q. Its term in n is pn and its number term is q − 5p.
The two forms are equal for every n, so their terms in n match: pn = 14n, which gives p = 14.
(a) The number terms match too: q − 5p = 60, so q = 60 + 5 × 14 = 130. The first 5 children cost $130 together, and each child after the fifth costs $14.
The second venue's price is 16(n − 5) + 110 = 16n − 80 + 110 = 16n + 30 dollars. Its terms do not match 14n + 60, so the two prices are equal only where 14n + 60 = 16n + 30.
Take 14n and 30 from both sides: 30 = 2n, so n = 15.
(b) The two venues charge the same for 15 children, and the price is 14 × 15 + 60 = $270. Check with the booking forms: 130 + 14 × 10 = 270 and 110 + 16 × 10 = 270.

answer(a) p = 14 and q = 130; (b) 15 children, at $270

techniqueFinding Unknown Coefficients by Matching · Expanding Brackets

examsSAT

Common pitfalls

Matching q with 60 and writing q = 60. The booking form's number term is q − 5p, not q, because the bracket p(n − 5) also gives −5p. Expand before matching.
Matching coefficients for the second venue as well, finding 14 ≠ 16 and deciding that the venues never charge the same. Matching applies only to two forms that are equal for every n. Two different price lists are equal where an equation holds, and here that is at n = 15.