Verified step-by-step solutions for your class notes.
Write a linear equation in two variables in which $a = 3$, $b = 0$ and $c = -\frac{1}{5}$.
Answer: The standard form is $ax + by + c = 0$.
Substituting the given values: $3x + 0y - \frac{1}{5} = 0$.
This can also be written simply as $3x - \frac{1}{5} = 0$, or multiplied by 5 to remove the fraction: $15x - 1 = 0$.
Complete the following table after expressing the given linear equations in standard form.
| Linear equation in two variables | Standard Form | Coefficient of x | Coefficient of y | Constant term |
|---|---|---|---|---|
| $y - 15 = \sqrt{2}x$ | ||||
| $3y - 2x = 0$ | ||||
| $5x = 3y$ | ||||
| $x = 8$ | ||||
| $3y = 1$ |
| Linear equation | Standard Form ($ax+by+c=0$) | Coefficient of x ($a$) | Coefficient of y ($b$) | Constant term ($c$) |
|---|---|---|---|---|
| $y - 15 = \sqrt{2}x$ | $-\sqrt{2}x + y - 15 = 0$ (or $\sqrt{2}x - y + 15 = 0$) | $-\sqrt{2}$ (or $\sqrt{2}$) | $1$ (or $-1$) | $-15$ (or $15$) |
| $3y - 2x = 0$ | $-2x + 3y + 0 = 0$ | $-2$ | $3$ | $0$ |
| $5x = 3y$ | $5x - 3y + 0 = 0$ | $5$ | $-3$ | $0$ |
| $x = 8$ | $1x + 0y - 8 = 0$ | $1$ | $0$ | $-8$ |
| $3y = 1$ | $0x + 3y - 1 = 0$ | $0$ | $3$ | $-1$ |
(i) The cost of a notebook is twice the cost of a pen. Consider the cost of a notebook to be ₹t and that of a pen to be ₹p. Charlie wrote $t = 2p$ whereas Meera wrote $p = 2t$. Which of these two representations is correct?
(ii) In a one-day International Cricket match... two Indian batsmen together scored 176 runs. Manisha expressed this situation as $x + y = 176$, where the number of runs scored by one batsman is x, and the number of runs scored by the other is y. Is this a correct representation?
Answer:
(i) Charlie's representation, $t = 2p$, is correct. Since the notebook ($t$) is twice as expensive as the pen ($p$), you must multiply the cost of the pen by 2 to equal the cost of the notebook.
(ii) Yes, this is a correct representation. The sum of the runs scored by the first batsman ($x$) and the second batsman ($y$) equals the total runs they scored together (176).
Verify if the ordered pair $(4, 3)$ is a solution of $5x - 6y = 2$. Explain your reasoning.
Answer: Substitute $x = 4$ and $y = 3$ into the equation $5x - 6y = 2$.
LHS = $5(4) - 6(3) = 20 - 18 = 2$.
RHS = 2.
Since LHS = RHS, the ordered pair $(4, 3)$ is a solution of the equation.
Find any two solutions for each of the following equations:
(i) $7x - 3y = 21$
(ii) $2x + 3y = 5$
Answer:
(i) $7x - 3y = 21$
Let $x = 0$: $7(0) - 3y = 21 \implies -3y = 21 \implies y = -7$. Solution 1: $(0, -7)$.
Let $y = 0$: $7x - 3(0) = 21 \implies 7x = 21 \implies x = 3$. Solution 2: $(3, 0)$.
(ii) $2x + 3y = 5$
Let $x = 1$: $2(1) + 3y = 5 \implies 3y = 3 \implies y = 1$. Solution 1: $(1, 1)$.
Let $y = 3$: $2x + 3(3) = 5 \implies 2x = -4 \implies x = -2$. Solution 2: $(-2, 3)$.
(Other valid solutions exist).
In the equations given below, m and n are unknown constants:
$2mx + 3y = 7$
$4x + ny = -10$
If $(2, 1)$ is the solution of both equations, find the values of m and n.
Answer: Substitute $x = 2$ and $y = 1$ into both equations.
First equation: $2m(2) + 3(1) = 7 \implies 4m + 3 = 7 \implies 4m = 4 \implies \mathbf{m = 1}$.
Second equation: $4(2) + n(1) = -10 \implies 8 + n = -10 \implies \mathbf{n = -18}$.
Consider the graph of the equation $3x - 7y = 21$ shown below. Does the point C $(2, 3)$ lie on the line? Does it satisfy the equation? Can points that do not lie on the line satisfy the equation?
Answer:
- Look at the graph: Point C $(2, 3)$ is clearly floating above the green line, so it does not lie on the line.
- Let's check algebraically: Substitute $x=2, y=3$ into $3x - 7y = 21$.
LHS = $3(2) - 7(3) = 6 - 21 = -15$. Since $-15 \neq 21$, it does not satisfy the equation.
- No, points that do not lie on the line can never satisfy the equation, because the line is the exact graphical representation of all the solutions to that equation.
The following diagrams represent ski hills. Rank the hills in order of their steepness, from least to greatest.
Answer: Steepness is determined by the slope (Rise / Run).
Slope of A = $60 / 70 \approx 0.857$
Slope of B = $60 / 110 \approx 0.545$
Slope of C = $80 / 100 = 0.800$
Ranking from least to greatest steepness: B, C, A.
The ramp at a loading dock rises 2.5 metres over a run of 4 metres. Find the slope of the ramp.
Answer: Slope = $\frac{\text{Rise}}{\text{Run}} = \frac{2.5}{4} = 0.625$.
An accessibility ramp for wheelchairs is to be made alongside the staircase. The guidelines given for the slope of the ramp is 1 cm vertical rise to 12 cm horizontal length. What should be the horizontal length of the ramp if the total height of the stairs is 18 cm?
Answer: The required slope is $\frac{\text{Rise}}{\text{Run}} = \frac{1}{12}$.
Given total rise = 18 cm.
$\frac{18}{\text{Run}} = \frac{1}{12} \implies \text{Run} = 18 \times 12 = 216$ cm.
The horizontal length should be 216 cm (or 2.16 metres).
Given below are some everyday situations. Construct a pair of linear equations in two variables for each situation:
Let the cost of one movie ticket be ₹x and the cost of one snack box be ₹y.
(i) On the first day, the family buys 2 movie tickets and 3 snack boxes for ₹850.
(ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹1100.
Answer:
(i) $2x + 3y = 850$
(ii) $4x + 1y = 1100$
Let the fixed charge for one taxi trip be ₹x and the additional charge per kilometre be ₹y.
(i) Sahil travelled 6 km and paid a total fare of ₹122.
(ii) Meena travelled 8 km and paid a total fare of ₹160.
Answer:
(i) $x + 6y = 122$
(ii) $x + 8y = 160$
In a sports meet, tickets for adults cost ₹150 each and tickets for children cost ₹100 each. A total of 200 people attended the meet, and the total amount collected from ticket sales was ₹25,000. (Let the number of adult tickets sold be x and the number of children's tickets sold be y).
Answer:
Total people equation: $x + y = 200$
Total amount equation: $150x + 100y = 25000$
Form a pair of linear equations for each of the following problems and find their solutions.
(i) The sum of two integers is +5 and their difference is -21. Find the two numbers.
Answer:
Equations: $x + y = 5$ (1), $x - y = -21$ (2)
Using elimination (add both): $2x = -16 \implies x = -8$
Substitute $x$ in (1): $-8 + y = 5 \implies y = 13$
The numbers are -8 and 13.
(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
Answer:
Equations: $x - y = 26$ (1), $x = 3y$ (2) (assuming $x > y$)
Substitute (2) into (1): $3y - y = 26 \implies 2y = 26 \implies y = 13$
Substitute $y$ in (2): $x = 3(13) = 39$
The numbers are 39 and 13.
(iv) Taxi charges consist of a fixed charge plus a per km charge. For 10 km, paid ₹155. For 15 km, paid ₹220. Find fixed charge and charge per km. Cost for 25 km?
Answer:
Let fixed charge = $x$, charge/km = $y$.
Equations: $x + 10y = 155$ (1), $x + 15y = 220$ (2)
Subtract (1) from (2): $5y = 65 \implies y = 13$ (Charge per km is ₹13)
Substitute $y$ in (1): $x + 10(13) = 155 \implies x = 155 - 130 = 25$ (Fixed charge is ₹25)
Cost for 25 km: $x + 25y = 25 + 25(13) = 25 + 325 = \mathbf{350}$ rupees.
(vii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Answer:
Let Nuri's current age = $N$, Sonu's = $S$.
5 years ago: $N - 5 = 3(S - 5) \implies N - 3S = -10$ (1)
10 years later: $N + 10 = 2(S + 10) \implies N - 2S = 10$ (2)
Subtract (1) from (2): $(N - 2S) - (N - 3S) = 10 - (-10) \implies S = 20$
Substitute $S=20$ in (2): $N - 2(20) = 10 \implies N = 50$
Nuri is 50, Sonu is 20.
Use the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$ to determine whether the lines representing the following pairs of linear equations intersect at a point, are parallel or are coincident.
(i) $5x - 4y + 8 = 0$; $7x + 6y - 9 = 0$
(ii) $9x + 3y + 12 = 0$; $18x + 6y + 24 = 0$
(iii) $6x - 3y + 10 = 0$; $2x - y + 9 = 0$
Answer:
(i) $\frac{a_1}{a_2} = \frac{5}{7}$, $\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, they intersect at a point (unique solution).
(ii) $\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, they are coincident (infinitely many solutions).
(iii) $\frac{a_1}{a_2} = \frac{6}{2} = 3$, $\frac{b_1}{b_2} = \frac{-3}{-1} = 3$, $\frac{c_1}{c_2} = \frac{10}{9}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, they are parallel (no solution).
5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of each pencil and pen.
Answer:
Let pencil cost = $x$, pen cost = $y$.
$5x + 7y = 50$ (1)
$7x + 5y = 46$ (2)
Add (1) and (2): $12x + 12y = 96 \implies x + y = 8$ (3)
Subtract (2) from (1): $-2x + 2y = 4 \implies -x + y = 2$ (4)
Add (3) and (4): $2y = 10 \implies y = 5$
Substitute $y=5$ in (3): $x + 5 = 8 \implies x = 3$
Cost of pencil = ₹3, Cost of pen = ₹5.
The graph of the line $y=3x$ passing through $(0, 0)$ and $(2, 6)$ is given below.
(i) Identify the slope of the line from the graph and explain how you calculated it.
(ii) What is the y-intercept of the line? Explain its significance.
(iv) Without drawing a new graph, determine whether the point $(4, 10)$ lies on this line. Justify your answer.
Answer:
(i) Slope = $\frac{\text{Rise}}{\text{Run}} = \frac{6 - 0}{2 - 0} = \frac{6}{2} = 3$.
(ii) The line passes through $(0,0)$, so the y-intercept is 0. This signifies that at the starting point ($x=0$), the value of $y$ is also 0.
(iv) No. Substitute $x=4$ into the equation $y=3x$. $y = 3(4) = 12$. Since $12 \neq 10$, the point $(4, 10)$ does not lie on the line.
The formula to convert Celsius to Fahrenheit is $F = \frac{9}{5}C + 32$.
(ii) If the temperature is $30^\circ C$, what is the temperature in Fahrenheit?
(iii) If the temperature is $95^\circ F$, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Answer:
(ii) $F = \frac{9}{5}(30) + 32 = 9(6) + 32 = 54 + 32 = \mathbf{86^\circ F}$.
(iii) $95 = \frac{9}{5}C + 32 \implies 63 = \frac{9}{5}C \implies C = \frac{63 \times 5}{9} = 7 \times 5 = \mathbf{35^\circ C}$.
(v) Yes. Set $F = C = x$.
$x = \frac{9}{5}x + 32 \implies x - \frac{9}{5}x = 32 \implies -\frac{4}{5}x = 32 \implies x = -\frac{32 \times 5}{4} = -4 \times 5 = \mathbf{-40}$. So, $-40^\circ C = -40^\circ F$.
Give a formula to find the x-intercept of the line $y = mx + c$.
Answer: The x-intercept occurs where the line crosses the x-axis, meaning $y = 0$.
Substitute $y=0$ into the equation: $0 = mx + c \implies mx = -c \implies x = -\frac{c}{m}$.
The formula for the x-intercept is $-\frac{c}{m}$ (assuming $m \neq 0$).
Plan A: ₹50 monthly fee + ₹0.20 per minute. Plan B: ₹30 monthly fee + ₹0.30 per minute.
Find the number of minutes at which both plans cost the same.
Answer: Let the number of minutes be $x$.
Cost of Plan A = $50 + 0.20x$
Cost of Plan B = $30 + 0.30x$
Equate them: $50 + 0.20x = 30 + 0.30x \implies 20 = 0.10x \implies x = \frac{20}{0.10} = 200$.
Both plans cost the same at 200 minutes. (Below 200 mins, Plan B is cheaper; above 200 mins, Plan A is cheaper).
For what values of $p$ does the pair of equations $4x + py + 8 = 0$ and $2x + 2y + 2 = 0$ have a unique solution?
Answer: For a unique solution, the ratio of coefficients must be $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
$\frac{4}{2} \neq \frac{p}{2} \implies 2 \neq \frac{p}{2} \implies p \neq 4$.
The equations have a unique solution for all real values of $p$ except $p = 4$.
Find the value of 'k' for which the following system of equations represents a pair of coincident lines:
$x + 2y = 3$; $(k-1)x + (k+1)y = k+3$.
Answer: For coincident lines (infinitely many solutions), $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$\frac{1}{k-1} = \frac{2}{k+1} = \frac{3}{k+3}$
Solve the first part: $\frac{1}{k-1} = \frac{2}{k+1} \implies k+1 = 2(k-1) \implies k+1 = 2k-2 \implies 3 = k$.
Verify with the third part: $\frac{2}{3+1} = \frac{2}{4} = \frac{1}{2}$ and $\frac{3}{3+3} = \frac{3}{6} = \frac{1}{2}$. It matches.
So, k = 3.
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