Chapter 4: Expressions using Letter-Numbers (Algebra)

Shabnam $=a+3=18+3=$ 21 years. Aftab $=s-3=20-3=$ 17 years.

21 years; 17 years

$8\times 35+9\times 60=280+540=$ ₹820

$c=7,j=4$: $7\times 35+4\times 60=245+240=$ ₹485

Square: $4\times q=4\times 7=$ 28 cm

₹820; ₹485; 28 cm

Let each side be $s$.

(a) $P=3\times s=3s$ (b) $P=5s$ (c) $P=6s$

(a) 3s (b) 5s (c) 6s, where s is the side length.

Combined length $=(20+k)$ m

20 + k metres

₹430; ₹695; 880 + 5z; 100x + 20y + 5z

Starting time (once) + 8 seconds for each of the $y$ kg: (d) $10+8\times y$ seconds.

(d) 10 + 8 × y

Let the number be $n$.

(a) $n+5$ (b) $n-4$ (c) $13n-2$ (d) $2n-13$

(a) n + 5 (b) n − 4 (c) 13n − 2 (d) 2n − 13

(a) A pencil costs ₹x and an eraser ₹y. Riya buys 8 pencils and 3 erasers. Total cost $=8\times x+3\times y$ rupees.

(b) Raju earns ₹15 for every jar of pickle he sells and pays ₹2 for every empty jar he buys. He sells $j$ jars and buys $k$ empty jars. His profit $=15\times j-2\times k$ rupees.

E.g. (a) cost of 8 pencils at ₹x and 3 erasers at ₹y (b) earning ₹15 on each of j items minus ₹2 spent on each of k items.

Moving one place right adds 1; moving one row up subtracts 7.

(Check with the marked dates 12, 13, 14 / 19, 20, 21: $w=20$.)

Top row: w − 8, w − 7, w − 6; bottom right: w + 1.

(1) $23+(-20)=3$ (2) $(83-13)+(28+32)=70+60=130$ (3) $20+20=40$ (4) $57-1=56$ (5) $68-31=37$ (6) $28+54=82$ (7) $20+8\times 10=100$

3, 130, 40, 56, 37, 82, 100

3rd term $=4\times 3=12$; 29th term $=4\times 29=116$; nth term $=4n$.

$5m+3=5\times 2+3=13$.

12, 116, 4n; 13

Correct values: 14, 18, 19, 17, 10 (correct), −15, 4, 11, 8

Day 2: $3c$; Day 3: $10c$. Pencils: $5c+3c+10c=18c$; for $c=50$: $18\times 50=$ ₹900.

Erasers: $4d+6d+1d=(4+6+1)d=11d$. Total money $=18c+11d$.

Check $5c+3c+10c=18c$ for $c=2$: $10+6+20=36=18\times 2$ ✓.

3c, 10c; ₹900; 4d + 6d + d = 11d

Yes: $(40x+75y)-(6x+10y)=40x-6x+75y-10y=34x+65y$.

And yes, subtracting $(6x+10y)$ is the same as adding its inverse $(-6x-10y)$, so $(40x+75y)+(-6x-10y)$ is the same expression and also gives $34x+65y$.

34x + 65y; yes, subtracting (6x + 10y) = adding (−6x − 10y).

$7p-3q$ means 7 correct answers and 3 wrong answers in round 1 (similarly for the others).

Round 2: $8\times 4-4\times 1=28$; Round 3: $6\times 4-2\times 1=22$. If there is no penalty, $q=0$.

Krishita's rounds could be, e.g., $8p-2q$, $7p-3q$ and $8p-2q$ (sum $23p-7q$).

Difference: $23p-7q-(21p-9q)=23p-7q-21p+9q=2p+2q$. Since $p$ and $q$ are positive, Krishita scored more, by $2p+2q$ (by 10 when $p=4$, $q=1$). She had 2 more correct and 2 fewer wrong answers.

28 and 22; q = 0; Krishita scored more, by 2p + 2q.

Neither pair is equal. $10(y-3)=10y-30$, which is always 27 less than $10y-3$.

No: 5u ≠ 5 + u, and 10y − 3 ≠ 10(y − 3) = 10y − 30.

Picture 1. Row-wise: $(5y-6+x)+(x+2+5y)$. Like terms: $(5y+5y)+(x+x)+(-6+2)=$ $10y+2x-4$.

Picture 2. Like terms: $4\times 2p+4\times 3q+2\times 3+2\times (-2)=$ $8p+12q+2$. (Column-wise also gives $(2p+3q)+(3q+2p)+(2p+3q+1)+(3q+2p+1)=8p+12q+2$.)

Picture 3. Like terms: $4\times (-5g)+12\times 5k=$ $-20g+60k$. (Column-wise: $(10k-10g)+20k+20k+(10k-10g)=60k-20g$.)

(1) 10y + 2x − 4 (2) 8p + 12q + 2 (3) 60k − 20g

(a) $4p$; $3p+q$ (b) $2p$ (c) $2p-2q$ (d) $2q$ (e) $0$ (f) $-2q$ (g) $-d$ (h) $-c$ (i) $2d-d-d+c=c$ (j) $2d-0-c=2d-c$ (k) $d-2c$

(a) 4p; 3p + q (b) 2p (c) 2p − 2q (d) 2q (e) 0 (f) −2q (g) −d (h) −c (i) c (j) 2d − c (k) d − 2c

3a + 2b; 0; 6p + 12; x − y; 3 + 6z; x + 5; 5y − 6; 6p + 3q; 30w + 15x; 3j + 6k + 9h + 12; 8r + 12s + 20

In the simplest form there is one term for each different letter-number, plus at most one term that is only a number. So:

number of terms = number of letter-numbers, or number of letter-numbers + 1 (when there is a constant term).

Examples: $3a+2b$ (2 letters, 2 terms); $6p+12$ (1 letter, 2 terms); $8r+12s+20$ (2 letters, 3 terms). (In $0$ there are no letters and no terms left.)

Terms = number of different letter-numbers, plus 1 if there is a number-only term; so terms are never more than letters + 1.

$2\times 5-2=8$ ✓; $2\times 8-1=15$ ✓; $2\times 9-11=7$ ✓; $2\times 10-10=10$ ✓; $2\times 6-4=$ 8.

Holds for all; (6, 4) gives 8.

Machine 1: "add the two numbers and subtract 2": $a+b-2$.

$5+2-2=5$; $8+1-2=7$; $9+11-2=18$; $10+10-2=18$.

Machine 2: "multiply the two numbers and add 1": $ab+1$.

$4\times 1+1=5$; $6\times 0+1=1$; $3\times 2+1=7$; $10\times 3+1=$ 31.

Make your own: e.g. "3 times the first plus the second", $3a+b$: (2, 5) → 11, (4, 0) → 12, (1, 7) → 10.

Machine 1: a + b − 2. Machine 2: ab + 1, so (10, 3) → 31.

A: $3n-2$; B: $3n-1$; C: $3n$.

A at 3n − 2, B at 3n − 1, C at 3n; 99 → C, 122 → B, 148 → A.

Let the centre be $a$. The number above is 7 less and the number below 7 more; left and right are 1 less and 1 more:

Sum $=(a-7)+(a-1)+a+(a+1)+(a+7)=5a$. So it always happens.

Other shapes:

• 3 × 3 square with centre $a$: the sum is $9a$.
• "X" shape (a − 8, a − 6, a, a + 6, a + 8): sum $=5a$.
• Three numbers in a column (a − 7, a, a + 7) or in a row (a − 1, a, a + 1): sum $=3a$.

Yes, the sum is (a − 7) + (a − 1) + a + (a + 1) + (a + 7) = 5a; a 3 × 3 square gives 9a, an X gives 5a, three in a line give 3a.

Step $y$ has $3+2(y-1)=2y+1$ matchsticks.

Step 33: $2\times 33+1=$ 67; Step 84: 169; Step 108: 217.

$y+(y+1)=2y+1$ ✓

67, 169, 217; Step y has y horizontal and y + 1 diagonal sticks, total 2y + 1.

Only (a) $30x+20y$.

(a) 30x + 20y

The customers are $p+q+r$ (those who bought both are counted once): (a) $p+q+r$.

(a) p + q + r

(a) Each day-and-night it rises $u-d$ cm, so after 10: $10(u-d)=10u-10d$ cm above the start.

(b) If $d>u$, then $u-d$ is negative: the snail goes down overall, ending $10(d-u)$ cm below its starting position.

(a) 10(u − d) cm (b) It moves down overall, ending 10(d − u) cm below the start.

Week 1: $7\times 5$; Week 2: $7\times (5+z)$; Week 3: $7\times (5+2z)$.

Total $=35+35+7z+35+14z=$ $105+21z$ km

105 + 21z km

• Top left: $w+2$ → −5 → $w-3$ → ×3 → $3w-9$
• Bottom left: $w+2$ → −8 → $w-6$ → −4 → $w-10$
• Bottom right: $w+2$ → −4 → $w-2$ → ×3 → $3w-6$

3w − 9; w − 6 then w − 10; w − 2.

Yahapur → S1 → S2 → S3 → Vahapur: 4 stretches of $t$ minutes and 3 stops of 2 minutes.

(b) Time $=4t+3\times 2=$ $4t+6$ minutes. (a) $t=4$: $16+6=$ 22 minutes.

(a) 22 minutes (b) 4t + 6 minutes

(a) $(3+8-7)a+(9-4)b+(-6+16)=4a+5b+10$

(b) $9a-9b-8a-4b-16=a-13b-16$

(c) $4x-6+8x+12=12x+6$

(d) $8x-2x+3+12=6x+15$

(e) $8h-5-7h+9=h+4$

(f) $23+24m-12n-8n-3m-18=21m-20n+5$

(a) 4a + 5b + 10 (b) a − 13b − 16 (c) 12x + 6 (d) 6x + 15 (e) h + 4 (f) 21m − 20n + 5

(a) $13d+c-2$ (b) $7f+4s-4$ (c) $8d-14c+9+16c-11-9d=-d+2c-2$ (d) $-7f-4s+3$ (e) $0$ (f) $0$

(a) 13d + c − 2 (b) 7f + 4s − 4 (c) 2c − d − 2 (d) −7f − 4s + 3 (e) 0 (f) 0

"Subtract A from B" means $B-A$.

(a) $6a+9b-18-9a+6b-14=-3a+15b-32$

(b) $7y-10+3x+15x-13+9y=18x+16y-23$

(c) $11-10g+3h-17g-9+7h=-27g+10h+2$

(d) $6a-9b-18-9a+6b-14=-3a-3b-32$

(e) $-3y+8-3x-10x-2-10y=-13x-13y+6$

(f) $7h-8g+20-8g-4h+10=-16g+3h+30$

(a) −3a + 15b − 32 (b) 18x + 16y − 23 (c) −27g + 10h + 2 (d) −3a − 3b − 32 (e) −13x − 13y + 6 (f) −16g + 3h + 30

(a) Pens cost ₹8 each and pencils ₹3 each. The total cost of $x$ pens and $y$ pencils is $8x+3y$ rupees.

(b) Meena earns ₹x per hour and works 15 hours, but 2 hours' pay is cut for coming late: $15x-2x=13x$ rupees.

E.g. (a) x pens at ₹8 and y pencils at ₹3 (b) 15 hours' pay at ₹x per hour minus 2 hours' pay = 13x.

Each fold adds one more strand at the cut, and so one more piece: 0 folds → 2, 1 fold → 3, 2 folds → 4, …

Folded 10 times: $10+2=$ 12 pieces. Folded $r$ times: $r+2$ pieces.

12 pieces; r + 2

The first square needs 4; each new square needs 3 more. $w$ squares: $4+3(w-1)=3w+1$.

10 squares: $3\times 10+1=$ 31 matchsticks.

31; 3w + 1

The pattern repeats every 4: red (1), yellow (2), green (3), yellow (4).

• Red: $4n-3$ (1, 5, 9, …)
• Yellow: $2n$ (every even position)
• Green: $4n-1$ (3, 7, 11, …)

90 and 190 are even → yellow. $343=4\times 86-1$ → green.

90 yellow, 190 yellow, 343 green; red 4n − 3, yellow 2n, green 4n − 1.

Each step adds 4 squares (one at the end of each arm): Step $n$ has $4n+1$ squares.

Step 4: 17; Step 10: 41; Step 50: 201.

Vertices: each square has 4 corners, but neighbouring squares in an arm touch at a corner, so one corner is shared for each of the $4n$ touching pairs:

Vertices $=4(4n+1)-4n=$ $12n+4$ (Step 1: 16). (If shared corners are counted separately for each square, the count is $4(4n+1)=16n+4$.)

17, 41, 201; 4n + 1 squares; 12n + 4 distinct vertices (16n + 4 if each square's corners are counted separately).

(a) Row $r$: column 1: $4r-3$; column 2: $4r-2$; column 3: $4r-1$; column 4: $4r$.

(b) Divide by 4:

• $124=4\times 31$ → row 31, column 4
• $147=4\times 36+3$ → row 37, column 3
• $201=4\times 50+1$ → row 51, column 1

(c) $4(r-1)+c$, i.e. $4r-4+c$.

(d) Multiples of 3 (3, 6, 9, 12, 15, 18, …) fall in columns 3, 2, 1, 4, 3, 2, 1, 4, … repeating every 12 numbers. Other patterns: column 4 holds the multiples of 4; columns 2 and 4 hold the even numbers and columns 1 and 3 the odd numbers; each number is 4 more than the one above it.

(a) 4r − 3, 4r − 2, 4r − 1, 4r (b) 124: row 31 col 4; 147: row 37 col 3; 201: row 51 col 1 (c) 4(r − 1) + c (d) multiples of 3 cycle through columns 3, 2, 1, 4.