Question 1
Solve the following pairs of linear equations by the substitution method:
(i) $x + y = 14; \quad x - y = 4$
(ii) $s - t = 3; \quad \frac{s}{3} + \frac{t}{2} = 6$
(iii) $3x - y = 3; \quad 9x - 3y = 9$
(iv) $0.2x + 0.3y = 1.3; \quad 0.4x + 0.5y = 2.3$
(v) $\sqrt{2}x + \sqrt{3}y = 0; \quad \sqrt{3}x - \sqrt{8}y = 0$
(vi) $\frac{3x}{2} - \frac{5y}{3} = -2; \quad \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
(ii) $s - t = 3; \quad \frac{s}{3} + \frac{t}{2} = 6$
(iii) $3x - y = 3; \quad 9x - 3y = 9$
(iv) $0.2x + 0.3y = 1.3; \quad 0.4x + 0.5y = 2.3$
(v) $\sqrt{2}x + \sqrt{3}y = 0; \quad \sqrt{3}x - \sqrt{8}y = 0$
(vi) $\frac{3x}{2} - \frac{5y}{3} = -2; \quad \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
(i) $x + y = 14$ and $x - y = 4$
From first equation: $x = 14 - y \quad \text{--- (1)}$
Substitute in $x - y = 4$: $$(14 - y) - y = 4 \implies 14 - 2y = 4 \implies 2y = 10 \implies y = 5$$ Substitute $y = 5$ in (1): $x = 14 - 5 = 9$.
$$\therefore x = 9, \quad y = 5$$
Substitute in $x - y = 4$: $$(14 - y) - y = 4 \implies 14 - 2y = 4 \implies 2y = 10 \implies y = 5$$ Substitute $y = 5$ in (1): $x = 14 - 5 = 9$.
$$\therefore x = 9, \quad y = 5$$
(ii) $s - t = 3$ and $\dfrac{s}{3} + \dfrac{t}{2} = 6$
From $s - t = 3 \implies s = t + 3 \quad \text{--- (1)}$
Substitute in second equation: $$\frac{t + 3}{3} + \frac{t}{2} = 6 \implies \frac{2(t + 3) + 3t}{6} = 6 \implies 5t + 6 = 36 \implies 5t = 30 \implies t = 6$$ From (1): $s = 6 + 3 = 9$.
$$\therefore s = 9, \quad t = 6$$
Substitute in second equation: $$\frac{t + 3}{3} + \frac{t}{2} = 6 \implies \frac{2(t + 3) + 3t}{6} = 6 \implies 5t + 6 = 36 \implies 5t = 30 \implies t = 6$$ From (1): $s = 6 + 3 = 9$.
$$\therefore s = 9, \quad t = 6$$
(iii) $3x - y = 3$ and $9x - 3y = 9$
From $3x - y = 3 \implies y = 3x - 3 \quad \text{--- (1)}$
Substitute in $9x - 3y = 9$: $$9x - 3(3x - 3) = 9 \implies 9x - 9x + 9 = 9 \implies 9 = 9$$ This statement is always true for all real values of $x$.
$$\therefore \text{The pair of linear equations has infinitely many solutions.}$$
Substitute in $9x - 3y = 9$: $$9x - 3(3x - 3) = 9 \implies 9x - 9x + 9 = 9 \implies 9 = 9$$ This statement is always true for all real values of $x$.
$$\therefore \text{The pair of linear equations has infinitely many solutions.}$$
(iv) $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$
Multiply both equations by 10 to clear decimals:
$$2x + 3y = 13 \quad \text{--- (1)}$$
$$4x + 5y = 23 \quad \text{--- (2)}$$
From (1): $x = \dfrac{13 - 3y}{2}$. Substitute in (2):
$$4\left(\frac{13 - 3y}{2}\right) + 5y = 23 \implies 2(13 - 3y) + 5y = 23 \implies 26 - y = 23 \implies y = 3$$
Substitute $y = 3$: $x = \dfrac{13 - 3(3)}{2} = \dfrac{4}{2} = 2$.
$$\therefore x = 2, \quad y = 3$$
$$\therefore x = 2, \quad y = 3$$
(v) $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x - \sqrt{8}y = 0$
From first equation: $x = -\dfrac{\sqrt{3}}{\sqrt{2}}y \quad \text{--- (1)}$
Substitute in second equation: $$\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0 \implies -\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0 \implies y\left(-\frac{3}{\sqrt{2}} - 2\sqrt{2}\right) = 0 \implies y = 0$$ From (1): $x = 0$.
$$\therefore x = 0, \quad y = 0$$
Substitute in second equation: $$\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0 \implies -\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0 \implies y\left(-\frac{3}{\sqrt{2}} - 2\sqrt{2}\right) = 0 \implies y = 0$$ From (1): $x = 0$.
$$\therefore x = 0, \quad y = 0$$
(vi) $\dfrac{3x}{2} - \dfrac{5y}{3} = -2$ and $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$
Multiplying both equations by 6:
$$9x - 10y = -12 \quad \text{--- (1)}$$
$$2x + 3y = 13 \implies x = \frac{13 - 3y}{2} \quad \text{--- (2)}$$
Substitute in (1):
$$9\left(\frac{13 - 3y}{2}\right) - 10y = -12 \implies \frac{117 - 27y - 20y}{2} = -12 \implies 117 - 47y = -24 \implies 47y = 141 \implies y = 3$$
Substitute $y = 3$: $x = \dfrac{13 - 3(3)}{2} = 2$.
$$\therefore x = 2, \quad y = 3$$
$$\therefore x = 2, \quad y = 3$$
Answers: (i) $(9, 5)$; (ii) $(9, 6)$; (iii) Infinitely many; (iv) $(2, 3)$; (v) $(0, 0)$; (vi) $(2, 3)$.
Question 2
Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of '$m$' for which $y = mx + 3$.
Step 1: Solving for $x$ and $y$
$$2x + 3y = 11 \quad \text{--- (1)}$$
$$2x - 4y = -24 \implies 2x = 4y - 24 \implies x = 2y - 12 \quad \text{--- (2)}$$
Substitute (2) in (1):
$$2(2y - 12) + 3y = 11 \implies 4y - 24 + 3y = 11 \implies 7y = 35 \implies y = 5$$
Substitute $y = 5$ in (2):
$$x = 2(5) - 12 = 10 - 12 = -2$$
$$\text{Solution: } x = -2, \quad y = 5$$
Step 2: Finding $m$
Substitute $x = -2$ and $y = 5$ into $y = mx + 3$:
$$5 = m(-2) + 3$$
$$5 - 3 = -2m \implies 2 = -2m \implies m = -1$$
$$\therefore x = -2, \quad y = 5, \quad \mathbf{m = -1}$$
Question 3
Form the pair of linear equations for the following problems and find their solution by substitution method:
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
(i) Difference Between Two Numbers
Let the two numbers be $x$ and $y$ ($x > y$).
$$x - y = 26 \quad \text{--- (1)}$$ $$x = 3y \quad \text{--- (2)}$$ Substitute (2) in (1): $3y - y = 26 \implies 2y = 26 \implies y = 13$.
$$x = 3(13) = 39$$ $$\therefore \text{The two numbers are } 39 \text{ and } 13.$$
$$x - y = 26 \quad \text{--- (1)}$$ $$x = 3y \quad \text{--- (2)}$$ Substitute (2) in (1): $3y - y = 26 \implies 2y = 26 \implies y = 13$.
$$x = 3(13) = 39$$ $$\therefore \text{The two numbers are } 39 \text{ and } 13.$$
(ii) Supplementary Angles
Let the larger angle be $x^\circ$ and smaller be $y^\circ$.
$$x + y = 180^\circ \quad \text{--- (1)}$$ $$x - y = 18^\circ \implies x = y + 18^\circ \quad \text{--- (2)}$$ Substitute (2) in (1): $(y + 18) + y = 180 \implies 2y = 162 \implies y = 81^\circ$.
$$x = 81 + 18 = 99^\circ$$ $$\therefore \text{The angles are } 99^\circ \text{ and } 81^\circ.$$
$$x + y = 180^\circ \quad \text{--- (1)}$$ $$x - y = 18^\circ \implies x = y + 18^\circ \quad \text{--- (2)}$$ Substitute (2) in (1): $(y + 18) + y = 180 \implies 2y = 162 \implies y = 81^\circ$.
$$x = 81 + 18 = 99^\circ$$ $$\therefore \text{The angles are } 99^\circ \text{ and } 81^\circ.$$
(iii) Cost of Bats and Balls
Let cost of 1 bat be ₹$x$ and 1 ball be ₹$y$.
$$7x + 6y = 3800 \quad \text{--- (1)}$$ $$3x + 5y = 1750 \implies x = \frac{1750 - 5y}{3} \quad \text{--- (2)}$$ Substitute in (1): $$7\left(\frac{1750 - 5y}{3}\right) + 6y = 3800 \implies 12250 - 35y + 18y = 11400 \implies -17y = -850 \implies y = 50$$ From (2): $x = \dfrac{1750 - 5(50)}{3} = \dfrac{1500}{3} = 500$.
$$\therefore \text{Cost of each bat } = ₹500, \quad \text{Cost of each ball } = ₹50$$
$$7x + 6y = 3800 \quad \text{--- (1)}$$ $$3x + 5y = 1750 \implies x = \frac{1750 - 5y}{3} \quad \text{--- (2)}$$ Substitute in (1): $$7\left(\frac{1750 - 5y}{3}\right) + 6y = 3800 \implies 12250 - 35y + 18y = 11400 \implies -17y = -850 \implies y = 50$$ From (2): $x = \dfrac{1750 - 5(50)}{3} = \dfrac{1500}{3} = 500$.
$$\therefore \text{Cost of each bat } = ₹500, \quad \text{Cost of each ball } = ₹50$$
(iv) City Taxi Charges
Let fixed charge be ₹$x$ and rate per km be ₹$y$.
$$x + 10y = 105 \implies x = 105 - 10y \quad \text{--- (1)}$$ $$x + 15y = 155 \quad \text{--- (2)}$$ Substitute (1) into (2): $$(105 - 10y) + 15y = 155 \implies 5y = 50 \implies y = 10$$ From (1): $x = 105 - 10(10) = 5$.
For a journey of 25 km: $$\text{Total charge} = x + 25y = 5 + 25(10) = 5 + 250 = ₹255$$ $$\therefore \text{Fixed charge} = ₹5, \quad \text{Charge per km} = ₹10, \quad \text{Cost for 25 km} = ₹255$$
$$x + 10y = 105 \implies x = 105 - 10y \quad \text{--- (1)}$$ $$x + 15y = 155 \quad \text{--- (2)}$$ Substitute (1) into (2): $$(105 - 10y) + 15y = 155 \implies 5y = 50 \implies y = 10$$ From (1): $x = 105 - 10(10) = 5$.
For a journey of 25 km: $$\text{Total charge} = x + 25y = 5 + 25(10) = 5 + 250 = ₹255$$ $$\therefore \text{Fixed charge} = ₹5, \quad \text{Charge per km} = ₹10, \quad \text{Cost for 25 km} = ₹255$$
(v) Fraction Problem
Let the fraction be $\dfrac{x}{y}$.
Condition 1: $\dfrac{x + 2}{y + 2} = \dfrac{9}{11} \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4 \quad \text{--- (1)}$
Condition 2: $\dfrac{x + 3}{y + 3} = \dfrac{5}{6} \implies 6x + 18 = 5y + 15 \implies x = \dfrac{5y - 3}{6} \quad \text{--- (2)}$
Substitute (2) in (1): $$11\left(\frac{5y - 3}{6}\right) - 9y = -4 \implies 55y - 33 - 54y = -24 \implies y = 9$$ From (2): $x = \dfrac{5(9) - 3}{6} = \dfrac{42}{6} = 7$.
$$\therefore \text{The required fraction is } \frac{7}{9}$$
Condition 1: $\dfrac{x + 2}{y + 2} = \dfrac{9}{11} \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4 \quad \text{--- (1)}$
Condition 2: $\dfrac{x + 3}{y + 3} = \dfrac{5}{6} \implies 6x + 18 = 5y + 15 \implies x = \dfrac{5y - 3}{6} \quad \text{--- (2)}$
Substitute (2) in (1): $$11\left(\frac{5y - 3}{6}\right) - 9y = -4 \implies 55y - 33 - 54y = -24 \implies y = 9$$ From (2): $x = \dfrac{5(9) - 3}{6} = \dfrac{42}{6} = 7$.
$$\therefore \text{The required fraction is } \frac{7}{9}$$
(vi) Jacob Age Problem
Let Jacob's present age be $x$ years and son's present age be $y$ years.
5 years hence: $(x + 5) = 3(y + 5) \implies x - 3y = 10 \implies x = 3y + 10 \quad \text{--- (1)}$
5 years ago: $(x - 5) = 7(y - 5) \implies x - 7y = -30 \quad \text{--- (2)}$
Substitute (1) in (2): $$(3y + 10) - 7y = -30 \implies -4y = -40 \implies y = 10$$ From (1): $x = 3(10) + 10 = 40$.
$$\therefore \text{Jacob's present age} = 40\text{ years}, \quad \text{Son's present age} = 10\text{ years}$$
5 years hence: $(x + 5) = 3(y + 5) \implies x - 3y = 10 \implies x = 3y + 10 \quad \text{--- (1)}$
5 years ago: $(x - 5) = 7(y - 5) \implies x - 7y = -30 \quad \text{--- (2)}$
Substitute (1) in (2): $$(3y + 10) - 7y = -30 \implies -4y = -40 \implies y = 10$$ From (1): $x = 3(10) + 10 = 40$.
$$\therefore \text{Jacob's present age} = 40\text{ years}, \quad \text{Son's present age} = 10\text{ years}$$
Solutions: (i) $39, 13$; (ii) $99^\circ, 81^\circ$; (iii) ₹500, ₹50; (iv) ₹5, ₹10/km, ₹255; (v) $\frac{7}{9}$; (vi) 40 yrs, 10 yrs.