Chapter 13: Two Variables, One Line (Linear Equations in Two Variables)

In the standard form $ax+by+c=0$:

$$3x+0\cdot y-\frac{1}{5}=0$$

$3x+0y-\frac{1}{5}=0$ (i.e. $3x=\frac{1}{5}$, with y free).

Bring every term to one side to get $ax+by+c=0$:

(Multiplying an equation by −1, e.g. 2x − 3y = 0 for the second row, gives an equivalent standard form.)

See the table above.

(i) The notebook costs twice as much as the pen, so $t=2p$. Charlie is correct. (Meera's equation would mean the pen costs twice the notebook.) Check: if a pen costs ₹10, the notebook costs ₹20, and $20=2\times 10$ ✓.

(ii) Yes. Their total is 176, so $x+y=176$ is correct (with x and y whole numbers from 0 to 176).

(i) Charlie's t = 2p is correct. (ii) Yes, x + y = 176 is correct.

Substitute x = 4, y = 3: $5(4)-6(3)=20-18=2$ ✓

Yes, (4, 3) is a solution because it makes both sides equal to 2.

Choose a value of x and solve for y.

• (i) x = 0: −3y = 21, y = −7 → (0, −7). x = 3: 21 − 3y = 21, y = 0 → (3, 0). (Also (6, 7).)
• (ii) x = 1: 3y = 3, y = 1 → (1, 1). x = 4: 3y = −3, y = −1 → (4, −1). (Also (−2, 3).)

(i) (0, −7) and (3, 0) (ii) (1, 1) and (4, −1) (many other answers are possible)

Substitute x = 2, y = −1:

$$2m(2)+3(-1)=7\Rightarrow 4m=10\Rightarrow m=\frac{5}{2}$$

$$4(2)+n(-1)=-10\Rightarrow 8-n=-10\Rightarrow n=18$$

m = 5/2 and n = 18

Check (i): 5(−1) + 3(4) = 7 ✓ and 5(2) + 3(−1) = 7 ✓; the others are checked the same way.

(i) (−1, 4) in QII, (2, −1) in QIV (ii) (2, 1) in QI, (−1, −4) in QIII (iii) (1, 4) in QI, (−2, −1) in QIII (iv) (−2, 1) in QII, (1, −4) in QIV

C does not lie on the line, and it does not satisfy the equation: $3(2)-7(3)=6-21=-15\ne 21$.

No: a point satisfies the equation exactly when it lies on the graph. The graph of an equation is the set of all its solutions, so points off the line never satisfy it.

No, C is not on the line and 3(2) − 7(3) = −15 ≠ 21. Points not on the line can never satisfy the equation.

• (i) False. It has infinitely many solutions, one for each point on its line.
• (ii) False. Only when c = 0. For example, x + y = 1 does not pass through (0, 0).
• (iii) False. For example (2, 1) is a rational solution of 2x + 3y = 7.
• (iv) True. It is $1\cdot x+0\cdot y-3=0$, with a = 1, b = 0 (not both zero).
• (v) True. For every value of x there is a value $y=\frac{7-2x}{3}$.
• (vi) False. $2(1)+3(2)=8\ne 7$.

(i) F (ii) F (iii) F (iv) T (v) T (vi) F

(i) The second equation is the first multiplied by 2. If (x, y) satisfies 3x + 4y = 7, multiplying by 2 gives 6x + 8y = 14. If (x, y) satisfies 6x + 8y = 14, dividing by 2 gives 3x + 4y = 7. So both equations have exactly the same solutions (the same line).

(ii) If ax + by = c, multiply by k: kax + kby = kc. Conversely, if kax + kby = kc, then since k ≠ 0 we can divide by k to get ax + by = c. So every solution of one is a solution of the other. (k ≠ 0 is needed: with k = 0 the second equation becomes 0 = 0, which every pair satisfies.)

Multiplying by k (and, since k ≠ 0, dividing by k) turns each equation into the other, so they have the same solutions.

Slope = rise ÷ run:

$$A:\frac{60}{70}\approx 0.86,B:\frac{60}{110}\approx 0.55,C:\frac{80}{100}=0.80$$

From least to greatest steepness: B, C, A.

$$\text{Slope}=\frac{2.5}{4}=0.625=\frac{5}{8}$$

The slope is 5/8 (0.625).

• (i) l is horizontal: as x changes, y does not change. Slope = 0.
• (ii) l passes through (−2, 2) and (0, 0) (and on through (2, −2)): it falls 2 for every 2 to the right. Slope = $\frac{0-2}{0-(-2)}=-1$.
• (iii) l passes through (−2, −1) and (1, 1): slope = $\frac{1-(-1)}{1-(-2)}=\frac{2}{3}$.
• (iv) l is vertical: the run is 0, so the slope is not defined.

(i) 0 (ii) −1 (iii) 2/3 (iv) not defined (vertical line)

The slope must be $\frac{1}{12}$, so

$$\frac{18}{\text{length}}=\frac{1}{12}\Rightarrow \text{length}=18\times 12=216\text{cm}$$

The horizontal length must be 216 cm (2.16 m).

$$2x+3y=850,4x+y=1100$$

(Solving: y = 1100 − 4x, so 2x + 3300 − 12x = 850, x = 245 and y = 120.)

2x + 3y = 850 and 4x + y = 1100 (a ticket costs ₹245, a snack box ₹120).

$$x+6y=122,x+8y=160$$

(Subtracting: 2y = 38, y = 19 and x = 8.)

x + 6y = 122 and x + 8y = 160 (fixed charge ₹8, ₹19 per km).

$$x+y=200,150x+100y=25000$$

(Dividing the second by 50: 3x + 2y = 500; subtracting 2(x + y) = 400 gives x = 100, y = 100.)

x + y = 200 and 150x + 100y = 25000 (100 adults and 100 children).

$$x+y=5,x-y=-21$$

Adding: 2x = −16, x = −8; then y = 13. Check: −8 + 13 = 5 and −8 − 13 = −21 ✓

The numbers are −8 and 13.

$$x-y=26,x=3y$$

Substituting: 3y − y = 26, y = 13, x = 39.

The numbers are 39 and 13.

Let a bat cost ₹x and a ball ₹y:

$$7x+6y=8880,3x+5y=4000$$

Multiply the first by 5 and the second by 6: $35x+30y=44400$ and $18x+30y=24000$. Subtracting: 17x = 20400, x = 1200. Then 3(1200) + 5y = 4000 gives y = 80.

A bat costs ₹1200 and a ball ₹80.

$$f+10k=155,f+15k=220$$

Subtracting: 5k = 65, k = 13; f = 155 − 130 = 25. For 25 km: 25 + 25 × 13 = ₹350.

Fixed charge ₹25, ₹13 per km; 25 km costs ₹350.

Let the fraction be x/y.

$$\frac{x+2}{y+2}=\frac{9}{11}\Rightarrow 11x-9y=-4,\frac{x+3}{y+3}=\frac{5}{6}\Rightarrow 6x-5y=-3$$

Multiply the first by 5 and the second by 9: $55x-45y=-20$, $54x-45y=-27$. Subtracting: x = 7; then 6(7) − 5y = −3 gives y = 9. Check: 9/11 ✓ and 10/12 = 5/6 ✓

The fraction is 7/9.

$$\frac{x+1}{y-1}=1\Rightarrow x-y=-2,\frac{x}{y+1}=\frac{1}{2}\Rightarrow 2x-y=1$$

Subtracting the first from the second: x = 3; then y = 5. Check: 4/4 = 1 ✓, 3/6 = 1/2 ✓

The fraction is 3/5.

Let their present ages be N and S.

$$N-5=3(S-5)\Rightarrow N-3S=-10,N+10=2(S+10)\Rightarrow N-2S=10$$

Subtracting: S = 20, so N = 50. Check: 5 years ago 45 = 3 × 15 ✓; in 10 years 60 = 2 × 30 ✓

Nuri is 50 years old and Sonu is 20.

Let the number be 10x + y.

$$x+y=9,9(10x+y)=2(10y+x)\Rightarrow 88x=11y\Rightarrow y=8x$$

So 9x = 9, x = 1, y = 8. Check: 9 × 18 = 162 = 2 × 81 ✓

The number is 18.

$$x+y=25,50x+100y=2000\Rightarrow x+2y=40$$

Subtracting: y = 15, x = 10.

10 notes of ₹50 and 15 notes of ₹100.

Seven days = 3 days + 4 extra days; five days = 3 days + 2 extra days.

$$f+4d=27,f+2d=21$$

Subtracting: 2d = 6, d = 3; f = 15.

Fixed charge ₹15; ₹3 for each extra day.

Let x = number of girls, y = number of boys: $x+y=10$ and $x-y=4$.

Points: x + y = 10 through (10, 0), (0, 10), (5, 5); x − y = 4 through (4, 0), (6, 2), (8, 4).

The lines meet at (7, 3).

7 girls and 3 boys.

• (i) $\frac{5}{7}$ and $\frac{-4}{6}=-\frac{2}{3}$ are not equal: the lines intersect at one point.
• (ii) $\frac{9}{18}=\frac{3}{6}=\frac{12}{24}=\frac{1}{2}$: the lines are coincident.
• (iii) $\frac{6}{2}=\frac{-3}{-1}=3$ but $\frac{10}{9}\ne 3$: the lines are parallel.

(i) intersecting (ii) coincident (iii) parallel

• (i) The second equation is 2 × the first, so the lines coincide: infinitely many solutions, all points (x, 5 − x), such as (0, 5), (2, 3), (5, 0).
• (ii) $\frac{1}{3}=\frac{-1}{-3}$ but $\frac{8}{16}=\frac{1}{2}\ne \frac{1}{3}$: parallel lines, no solution.
• (iii) $\frac{2}{4}\ne \frac{1}{-2}$: one solution. From the graph (and by elimination) the lines meet at (2, 2).
• (iv) $\frac{2}{4}=\frac{-2}{-4}=\frac{1}{2}$ but $\frac{-2}{-5}=\frac{2}{5}\ne \frac{1}{2}$: parallel lines, no solution.

(i) infinitely many solutions (coincident) (ii) no solution (iii) one solution, (2, 2) (iv) no solution

$$l+w=36,l-w=4$$

Adding: 2l = 40, l = 20 and w = 16.

Length 20 m, width 16 m.

• (i) Change the ratio of the x- and y-coefficients, e.g. $3x-2y+1=0$ (since $\frac{2}{3}\ne \frac{3}{-2}$).
• (ii) Keep the same ratio of coefficients but change the constant, e.g. $4x+6y-5=0$ (since $\frac{2}{4}=\frac{3}{6}\ne \frac{-8}{-5}$).
• (iii) Multiply the whole equation by a non-zero number, e.g. $4x+6y-16=0$.

For example (i) 3x − 2y + 1 = 0 (ii) 4x + 6y − 5 = 0 (iii) 4x + 6y − 16 = 0

The coefficients are swapped, so adding and subtracting give very simple equations:

$$\text{Add: }16x+16y=208\Rightarrow x+y=13$$

$$\text{Subtract: }2x-2y=6\Rightarrow x-y=3$$

So 2x = 16, x = 8 and y = 5. Check: 9(8) + 7(5) = 72 + 35 = 107 ✓

A citron costs 8 and a fragrant wood-apple costs 5.

$$5x+7y=50,7x+5y=46$$

Adding: 12(x + y) = 96, so x + y = 8. Subtracting: 2y − 2x = 4, so y − x = 2. Hence y = 5, x = 3.

A pencil costs ₹3 and a pen ₹5.

Let the stool be s cm high and the cat c cm high.

$$s+c=85,s-c=25$$

Adding: 2s = 110, s = 55 (and c = 30).

The stool is 55 cm high (the cat is 30 cm).

(i) From (0, 0) to (2, 6) the rise is 6 and the run is 2, so slope $=\frac{6-0}{2-0}=3$.

(ii) The line passes through (0, 0), so the y-intercept is 0. For the tank, it means the water level was 0 cm at the start (the tank was empty).

(iii) (a) $y=3\times 5=15$ cm. (b) $21=3x$ gives x = 7 minutes.

(iv) At x = 4 the line has y = 12, not 10. So (4, 10) does not lie on the line.

(v) On the line x = 3 every point has x = 3; putting x = 3 into y = 3x gives y = 9. The lines meet at (3, 9).

(i) slope 3 (ii) y-intercept 0: the tank starts empty (iii) (a) 15 cm (b) 7 minutes (iv) No, since 3 × 4 = 12 ≠ 10 (v) (3, 9)

(i) Points: (0, 32), (10, 50), (30, 86), (−40, −40). (One square = 10 degrees.)

(ii) $F=\frac{9}{5}\times 30+32=54+32=86$°F.

(iii) $95=\frac{9}{5}C+32\Rightarrow \frac{9}{5}C=63\Rightarrow C=35$°C.

(iv) 0°C: F = 32°F. 0°F: $0=\frac{9}{5}C+32\Rightarrow C=-\frac{160}{9}\approx -17.8$°C.

(v) Put F = C: $C=\frac{9}{5}C+32\Rightarrow -\frac{4}{5}C=32\Rightarrow C=-40$. So −40°C = −40°F.

(ii) 86°F (iii) 35°C (iv) 32°F; about −17.8°C (v) Yes: −40° (−40°C = −40°F).

Points: 2x + y = 6 through (0, 6), (3, 0), (1, 4); 2x − y = 2 through (0, −2), (1, 0), (3, 4).

The lines meet at (2, 2). Check: 2(2) + 2 = 6 ✓ and 2(2) − 2 − 2 = 0 ✓

x = 2, y = 2

Method (for any two lines drawn on a grid):

• Read two clear grid points on each line.
• Find each line's equation y = mx + c: m = rise ÷ run between the two points, and c from one point.
• Solve the two equations together (substitution or elimination). The solution (x, y) is the point of intersection; check it against the picture.

For example, if one line passes through (0, 1) and (2, 5) (so y = 2x + 1) and the other through (0, 7) and (3, 4) (so y = −x + 7), then 2x + 1 = −x + 7 gives x = 2, y = 5: the lines meet at (2, 5).

Find each line's equation from two grid points on it, then solve the pair of equations; the common solution is the intersection point.

The line meets the x-axis where y = 0: $0=mx+c$, so $x=-\frac{c}{m}$ (for m ≠ 0). If m = 0 the line is horizontal and meets the x-axis only if c = 0.

x-intercept $=-\frac{c}{m}$, i.e. the line cuts the x-axis at $\left(-\frac{c}{m},0\right)$ (m ≠ 0).

For t minutes: A = 50 + 0.2t and B = 30 + 0.3t.

$$50+0.2t=30+0.3t\Rightarrow 0.1t=20\Rightarrow t=200$$

At 200 minutes both cost ₹90. For more than 200 minutes, Plan A is cheaper (its rate per minute is lower); for fewer than 200 minutes, Plan B is cheaper (its fixed fee is lower).

Same cost at 200 minutes (₹90). Plan A is cheaper for more than 200 minutes; Plan B for fewer than 200 minutes.

(i) Infinitely many: all lines y = mx + c with the given m and any c. They are all parallel.

(ii) Exactly one: with slope m through (x₁, y₁) the y-intercept must be $c=y_1-m{x}_1$.

(i) Infinitely many (all parallel) (ii) exactly one

A unique solution needs $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$:

$$\frac{4}{2}\ne \frac{p}{2}\Rightarrow p\ne 4$$

For every value of p except p = 4.

For infinitely many solutions all three ratios must be equal:

$$\frac{a+b}{3}=\frac{-2b}{-1}=\frac{5a+2b+1}{14}$$

From the first two: a + b = 6b, so a = 5b. From the second and third: $28b=5a+2b+1=25b+2b+1$, so b = 1 and a = 5.

Check: the first equation becomes 6x − 2y = 28, which is 2 × (3x − y = 14) ✓

a = 5, b = 1

$$\frac{1}{k-1}=\frac{2}{k+1}=\frac{3}{k+3}$$

From the first two: k + 1 = 2k − 2, so k = 3. Check the third: $\frac{2}{4}=\frac{3}{6}=\frac{1}{2}$ ✓ (the second equation becomes 2x + 4y = 6).

k = 3

(i) On average Robot 1 moves along the line $y=\frac{4}{3}x$ and Robot 2 along $y=2(x-10)$, which meet where $\frac{4}{3}x=2x-20$, i.e. at (30, 40). Robot 2's line is steeper, so its path catches up with Robot 1's path.

Tracing the actual staircase paths: Robot 1's corners are (3, 0), (3, 4), (6, 4), (6, 8), ..., (24, 28), (24, 32), ...; Robot 2's corners are (11, 0), (11, 2), (12, 2), ..., (24, 26), (24, 28), .... The paths first meet at (24, 28), where both robots are at a corner. After that they cross several more times (for example at (27, 32) and (30, 40)).

(ii) Robot 1's first move is from (3, 0) right to (8, 0); Robot 2's first move is from (7, 0) right to (12, 0). Both moves lie along the x-axis, so the paths overlap along the segment from (7, 0) to (8, 0). After that, Robot 1 goes up and Robot 2 goes down, so they never meet again.

(If each robot is instead imagined moving in a straight line, along y = x − 3 and $y=-\frac{3}{5}(x-7)$, those lines meet at (4.5, 1.5); but the actual paths are staircases.)

(i) Yes: the paths first meet at (24, 28) (and later also at (30, 40), among other points). (ii) Yes: they share the segment from (7, 0) to (8, 0) on the x-axis.

Fifteen minutes after 2:a it is 3:b, so a + 15 = 60 + b, i.e. a − b = 45. Also a = 6b. So 5b = 45, b = 9 and a = 54.

First look: 2:54; second look: 3:09.

3:09 (the first reading was 2:54).

Let the number be 10x + y.

$$x+y=15,(10y+x)-(10x+y)=9\Rightarrow y-x=1$$

So y = 8, x = 7. Check: 87 − 78 = 9 ✓

The number is 78.

Opposite angles of a cyclic quadrilateral add up to 180°:

$$(x+7)+(3y+23)=180\Rightarrow x+3y=150$$

$$(y+8)+(4x+12)=180\Rightarrow 4x+y=160$$

From the first, x = 150 − 3y; substituting, 600 − 12y + y = 160, so y = 40 and x = 30.

$$\angle A=37°,\angle B=48°,\angle C=143°,\angle D=132°$$

Check: 37 + 143 = 180 ✓, 48 + 132 = 180 ✓

∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°

Let the speed be v km/h and the time t hours, so the distance is vt.

$$vt=(v+6)(t-6)\Rightarrow 0=-6v+6t-36\Rightarrow t-v=6$$

$$vt=(v-4)(t+6)\Rightarrow 0=6v-4t-24\Rightarrow 3v-2t=12$$

Substituting t = v + 6: 3v − 2v − 12 = 12, so v = 24 and t = 30. Distance = 24 × 30 = 720 km.

Check: at 30 km/h it takes 24 h (6 less) ✓; at 20 km/h it takes 36 h (6 more) ✓

Speed 24 km/h; distance 720 km.

Let the father's age be F; the children's ages add up to F too. In 20 years each of the 4 children is 20 years older, so their sum is F + 80, and the father is F + 20.

$$F+80=2(F+20)\Rightarrow F=40$$

The father is 40 years old.