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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.
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