Check graphically whether the pair of equations \(x+3y=6\) and \(2x-3y=12\) is consistent. If so, solve them graphically.
To draw a line we only need two points on it.
For \(x+3y=6\): taking \(x=0\) gives \(y=2\), point A(0, 2); taking \(x=6\) gives \(y=0\), point B(6, 0).
For \(2x-3y=12\): taking \(x=0\) gives \(y=-4\), point P(0, −4); taking \(x=3\) gives \(y=-2\), point Q(3, −2).
Plotting A, B, P, Q and drawing the two lines, they meet only at the single point B(6, 0). So the pair of equations is consistent, with the unique solution x = 6, y = 0.
Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions: \(5x-8y+1=0\) and \(3x-\dfrac{24}{5}y+\dfrac{3}{5}=0\).
Multiply the second equation throughout by \(\dfrac{5}{3}\):
This is exactly the first equation. So the two equations describe the same line — the lines are coincident, and the pair has infinitely many solutions (every point on the line is a solution). Plotting a few points on the graph confirms both equations trace the identical line.
Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased.” Help her friends to find how many pants and skirts Champa bought.
Let the number of pants be x and the number of skirts be y. The two conditions give:
Two points on each line: for \(y=2x-2\), \(x=2\Rightarrow y=2\) (A(2, 2)) and \(x=0\Rightarrow y=-2\) (B(0, −2)). For \(y=4x-4\), \(x=0\Rightarrow y=-4\) (P(0, −4)) and \(x=1\Rightarrow y=0\) (Q(1, 0)).
The two lines intersect at (1, 0). So x = 1, y = 0 is the required solution, i.e. the number of pants Champa purchased is 1 and she did not buy any skirt (0 skirts).
Solve the following pair of equations by substitution method: \(7x-15y=2\) and \(x+2y=3\).
Step 1: From \(x+2y=3\), write x in terms of y:
Step 2: Substitute this value of x in \(7x-15y=2\):
Step 3: Substitute this value of y back into \(x=3-2y\):
The solution is \(x=\dfrac{49}{29},\ y=\dfrac{19}{29}\).
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and solve by the method of substitution.
Let s and t be the present ages (in years) of Aftab and his daughter, respectively.
Seven years ago: \(s-7=7(t-7)\), i.e.
Three years from now: \(s+3=3(t+3)\), i.e.
From the second equation, \(s=3t+6\). Substituting in the first:
Substituting \(t=12\) back into \(s=3t+6\):
Aftab is 42 years old and his daughter is 12 years old.
In a shop the cost of 2 pencils and 3 erasers is ₹9, and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each pencil and each eraser.
Let pencil = x, eraser = y. The equations formed are:
Express x from the first equation:
Substituting into the second equation:
This statement is true for every value of y — there is no unique value. This happens because the second equation is just twice the first, so the two lines are coincident. The pair has infinitely many solutions, and no unique cost can be found for a pencil or an eraser from this information alone.
Two rails are represented by the equations \(x+2y-4=0\) and \(2x+4y-12=0\). Will the rails cross each other?
Express x from the first equation:
Substituting into the second equation:
This is a false statement — it holds for no value of y at all. So the equations have no common solution: the lines are parallel, and the two rails will not cross each other.
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹2000 per month, find their monthly incomes.
Let the incomes be ₹9x and ₹7x, and the expenditures be ₹4y and ₹3y. Since savings = income − expenditure = ₹2000 for each person:
Step 1: Multiply the first equation by 3 and the second by 4, to make the coefficients of y numerically equal:
Step 2: Subtract the first of these from the second, eliminating y:
Step 3: Substitute \(x=2000\) into \(9x-4y=2000\):
So the monthly incomes are \(9x=9(2000)=\)₹18,000 and \(7x=7(2000)=\)₹14,000.
Verification: \(18000:14000=9:7\ \checkmark\). Expenditures \(=18000-2000:14000-2000=16000:12000=4:3\ \checkmark\).
Use the elimination method to find all possible solutions of the following pair of linear equations: \(2x+3y=8\) and \(4x+6y=7\).
Step 1: Multiply the first equation by 2, to make the coefficients of x equal:
Step 2: Subtracting the second from the first:
This is a false statement. Therefore the pair of equations has no solution (the lines are parallel — note that \(\dfrac{2}{4}=\dfrac{3}{6}\ne\dfrac{8}{7}\)).
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Let the ten’s digit be x and the unit’s digit be y, so the number is \(10x+y\). Reversing the digits gives \(10y+x\).
The digits differ by 2, so either \(x-y=2\) or \(y-x=2\).
Case 1: \(x+y=6\) and \(x-y=2\). Adding: \(2x=8\Rightarrow x=4,\ y=2\). Number = 42.
Case 2: \(x+y=6\) and \(y-x=2\). Adding: \(2y=8\Rightarrow y=4,\ x=2\). Number = 24.
So there are two such numbers: 42 and 24.
Verification: \(42+24=66\), and \(4-2=2\ \checkmark\). Also \(24+42=66\), and \(4-2=2\ \checkmark\).
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
(i) Let the number of boys be x and the number of girls be y. Total students give \(x+y=10\); girls being 4 more than boys gives \(y=x+4\).
The lines meet at (3, 7). So there are 3 boys and 7 girls.
(ii) Let pencil = x, pen = y. \(5x+7y=50\) and \(7x+5y=46\).
The lines meet at (3, 5). So one pencil costs ₹3 and one pen costs ₹5.
On comparing the ratios \(\dfrac{a_{1}}{a_{2}},\dfrac{b_{1}}{b_{2}}\) and \(\dfrac{c_{1}}{c_{2}}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) \(5x-4y+8=0\) ; \(7x+6y-9=0\) (ii) \(9x+3y+12=0\) ; \(18x+6y+24=0\) (iii) \(6x-3y+10=0\) ; \(2x-y+9=0\)
(i)
Since \(\dfrac{5}{7}\ne-\dfrac{2}{3}\), the lines intersect at a point (a unique solution).
(ii)
All three ratios are equal (\(=\tfrac{1}{2}\)), so the lines are coincident.
(iii)
Here \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=3\) but \(\dfrac{c_{1}}{c_{2}}=\dfrac{10}{9}\ne3\), so the lines are parallel.
On comparing the ratios \(\dfrac{a_{1}}{a_{2}},\dfrac{b_{1}}{b_{2}}\) and \(\dfrac{c_{1}}{c_{2}}\), find out whether the following pairs of linear equations are consistent, or inconsistent:
(i) \(3x+2y=5\) ; \(2x-3y=7\) (ii) \(2x-3y=8\) ; \(4x-6y=9\)
(iii) \(\dfrac{3}{2}x+\dfrac{5}{3}y=7\) ; \(9x-10y=14\) (iv) \(5x-3y=11\) ; \(-10x+6y=-22\)
(v) \(\dfrac{4}{3}x+2y=8\) ; \(2x+3y=12\)
(i) In the standard form: \(3x+2y-5=0\), \(2x-3y-7=0\).
\(\dfrac{a_{1}}{a_{2}}\ne\dfrac{b_{1}}{b_{2}}\) — the lines intersect, so the pair is consistent (unique solution).
(ii) Standard form: \(2x-3y-8=0\), \(4x-6y-9=0\).
\(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\ne\dfrac{c_{1}}{c_{2}}\) — the lines are parallel, so the pair is inconsistent (no solution).
(iii) Standard form: \(\dfrac{3}{2}x+\dfrac{5}{3}y-7=0\), \(9x-10y-14=0\).
\(\dfrac{a_{1}}{a_{2}}\ne\dfrac{b_{1}}{b_{2}}\) — the lines intersect, so the pair is consistent (unique solution).
(iv) Standard form: \(5x-3y-11=0\), \(-10x+6y+22=0\).
All three ratios are equal — the lines coincide, so the pair is consistent (dependent, infinitely many solutions).
(v) Standard form: \(\dfrac{4}{3}x+2y-8=0\), \(2x+3y-12=0\).
All three ratios are equal — the lines coincide, so the pair is consistent (dependent, infinitely many solutions).
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) \(x+y=5\), \(2x+2y=10\) (ii) \(x-y=8\), \(3x-3y=16\)
(iii) \(2x+y-6=0\), \(4x-2y-4=0\) (iv) \(2x-2y-2=0\), \(4x-4y-5=0\)
(i) \(\dfrac{a_{1}}{a_{2}}=\dfrac{1}{2},\ \dfrac{b_{1}}{b_{2}}=\dfrac{1}{2},\ \dfrac{c_{1}}{c_{2}}=\dfrac{-5}{-10}=\dfrac{1}{2}\). All equal — lines coincide. Consistent, with infinitely many solutions (every point on the line \(x+y=5\)).
(ii) \(\dfrac{a_{1}}{a_{2}}=\dfrac{1}{3},\ \dfrac{b_{1}}{b_{2}}=\dfrac{-1}{-3}=\dfrac{1}{3},\ \dfrac{c_{1}}{c_{2}}=\dfrac{-8}{-16}=\dfrac{1}{2}\). \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\ne\dfrac{c_{1}}{c_{2}}\) — lines parallel. Inconsistent (no solution).
(iii) \(\dfrac{a_{1}}{a_{2}}=\dfrac{2}{4}=\dfrac{1}{2},\ \dfrac{b_{1}}{b_{2}}=\dfrac{1}{-2}=-\dfrac{1}{2}\). These are unequal — lines intersect. Consistent, unique solution. Solving: from \(2x+y-6=0\), \(y=6-2x\). Substituting into \(4x-2y-4=0\): \(4x-2(6-2x)-4=0\Rightarrow 4x-12+4x-4=0\Rightarrow 8x=16\Rightarrow x=2,\ y=2\).
(iv) \(\dfrac{a_{1}}{a_{2}}=\dfrac{2}{4}=\dfrac{1}{2},\ \dfrac{b_{1}}{b_{2}}=\dfrac{-2}{-4}=\dfrac{1}{2},\ \dfrac{c_{1}}{c_{2}}=\dfrac{-2}{-5}=\dfrac{2}{5}\). \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\ne\dfrac{c_{1}}{c_{2}}\) — lines parallel. Inconsistent (no solution).
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Let the width be w metres, so the length is \(w+4\). Half the perimeter is length + width:
So the width is 16 m and the length is \(16+4=\)20 m.
Given the linear equation \(2x+3y-8=0\), write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines
For \(2x+3y-8=0\): \(a_{1}=2,\ b_{1}=3,\ c_{1}=-8\).
(i) Intersecting: any line with a different \(\dfrac{a}{b}\) ratio works, e.g. \(x+y-5=0\). Check: \(\dfrac{a_{1}}{a_{2}}=\dfrac{2}{1}=2\), \(\dfrac{b_{1}}{b_{2}}=\dfrac{3}{1}=3\) — unequal, so the lines intersect.
(ii) Parallel: keep the same \(\dfrac{a}{b}\) ratio but change the constant, e.g. \(4x+6y+7=0\) (double the coefficients of \(x,y\) but not \(-8\)). Check: \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{1}{2}\), but \(\dfrac{c_{1}}{c_{2}}=\dfrac{-8}{7}\ne\dfrac{1}{2}\) — the lines are parallel.
(iii) Coincident: scale the whole equation by a constant, e.g. multiply by 2 to get \(4x+6y-16=0\). Check: \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}=\dfrac{1}{2}\) — the lines are coincident.
Draw the graphs of the equations \(x-y+1=0\) and \(3x+2y-12=0\). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solving the two equations for their intersection: from \(x-y+1=0\), \(y=x+1\). Substituting into \(3x+2y-12=0\): \(3x+2(x+1)-12=0\Rightarrow 5x-10=0\Rightarrow x=2,\ y=3\). Intersection: (2, 3).
The x-intercept of \(x-y+1=0\) (set \(y=0\)) is \(x=-1\), giving (−1, 0). The x-intercept of \(3x+2y-12=0\) (set \(y=0\)) is \(x=4\), giving (4, 0).
The shaded triangle has vertices (2, 3), (−1, 0) and (4, 0). (As a bonus check: its base along the x-axis is \(4-(-1)=5\) units and its height is 3 units, so its area is \(\tfrac{1}{2}\times5\times3=\dfrac{15}{2}\) sq units.)
Solve the following pair of linear equations by the substitution method.
(i) \(x+y=14\) ; \(x-y=4\) (ii) \(s-t=3\) ; \(\dfrac{s}{3}+\dfrac{t}{2}=6\)
(iii) \(3x-y=3\) ; \(9x-3y=9\) (iv) \(0.2x+0.3y=1.3\) ; \(0.4x+0.5y=2.3\)
(v) \(\sqrt{2}\,x+\sqrt{3}\,y=0\) ; \(\sqrt{3}\,x-\sqrt{8}\,y=0\) (vi) \(\dfrac{3x}{2}-\dfrac{5y}{3}=-2\) ; \(\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{13}{6}\)
(i) From \(x+y=14\): \(x=14-y\). Substituting into \(x-y=4\):
Then \(x=14-5=9\). Solution: x = 9, y = 5.
(ii) From \(s-t=3\): \(s=t+3\). Substituting into \(\dfrac{s}{3}+\dfrac{t}{2}=6\):
Multiplying throughout by 6:
Then \(s=6+3=9\). Solution: s = 9, t = 6.
(iii) From \(3x-y=3\): \(y=3x-3\). Substituting into \(9x-3y=9\):
This gives \(9=9\), true for every value of x. The second equation is exactly 3 times the first, so the lines are coincident: the pair has infinitely many solutions (every point satisfying \(y=3x-3\)).
(iv) From \(0.2x+0.3y=1.3\): \(x=\dfrac{1.3-0.3y}{0.2}\). It is cleaner to first multiply both equations by 10: \(2x+3y=13\) and \(4x+5y=23\). From the first, \(x=\dfrac{13-3y}{2}\). Substituting into the second:
Then \(x=\dfrac{13-9}{2}=2\). Solution: x = 2, y = 3.
(v) From \(\sqrt{2}\,x+\sqrt{3}\,y=0\): \(x=-\dfrac{\sqrt{3}}{\sqrt{2}}y\). Substituting into \(\sqrt{3}\,x-\sqrt{8}\,y=0\) (note \(\sqrt{8}=2\sqrt{2}\)):
Multiplying throughout by \(\sqrt{2}\):
Then \(x=0\) as well. Solution: x = 0, y = 0 (the only solution, since the two lines both pass through the origin but are not parallel).
(vi) Multiplying the first equation by 6: \(9x-10y=-12\). Multiplying the second by 6: \(2x+3y=13\), so \(x=\dfrac{13-3y}{2}\). Substituting into \(9x-10y=-12\):
Then \(x=\dfrac{13-9}{2}=2\). Solution: x = 2, y = 3.
Solve \(2x+3y=11\) and \(2x-4y=-24\) and hence find the value of ‘m’ for which \(y=mx+3\).
From \(2x+3y=11\): \(x=\dfrac{11-3y}{2}\). Substituting into \(2x-4y=-24\):
Then \(x=\dfrac{11-15}{2}=-2\). Solution: x = −2, y = 5.
Substituting into \(y=mx+3\):
m = −1.
Form the pair of linear equations for the following problems and find their solution by the 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 \(\dfrac{9}{11}\), if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \(\dfrac{5}{6}\). 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) Let the numbers be x (larger) and y (smaller), with \(x=3y\) and \(x-y=26\). Substituting:
The numbers are 39 and 13.
(ii) Let the angles be x (larger) and y (smaller). Supplementary: \(x+y=180\); condition: \(x-y=18\), so \(x=y+18\). Substituting:
The angles are 99° and 81°.
(iii) Let bat = x, ball = y. \(7x+6y=3800\) and \(3x+5y=1750\). From the second, \(x=\dfrac{1750-5y}{3}\). Substituting into the first:
Then \(x=\dfrac{1750-250}{3}=500\). Each bat costs ₹500 and each ball costs ₹50.
(iv) Let the fixed charge be x and the per-km charge be y. \(x+10y=105\) and \(x+15y=155\). From the first, \(x=105-10y\). Substituting into the second:
Then \(x=105-100=5\). Fixed charge = ₹5, charge per km = ₹10. For 25 km: \(5+25(10)=\)₹255.
(v) Let the fraction be \(\dfrac{a}{b}\). The conditions give:
From the first, \(a=\dfrac{9b-4}{11}\). Substituting into the second:
Then \(a=\dfrac{9(9)-4}{11}=\dfrac{77}{11}=7\). The fraction is \(\dfrac{7}{9}\). Check: \(\dfrac{7+2}{9+2}=\dfrac{9}{11}\ \checkmark\), \(\dfrac{7+3}{9+3}=\dfrac{10}{12}=\dfrac{5}{6}\ \checkmark\).
(vi) Let Jacob’s present age be x and his son’s be y. Five years hence: \(x+5=3(y+5)\Rightarrow x-3y=10\). Five years ago: \(x-5=7(y-5)\Rightarrow x-7y=-30\). From the first, \(x=3y+10\). Substituting into the second:
Then \(x=3(10)+10=40\). Jacob is 40 years old and his son is 10 years old.
Solve the following pair of linear equations by the elimination method and the substitution method:
(i) \(x+y=5\) and \(2x-3y=4\) (ii) \(3x+4y=10\) and \(2x-2y=2\)
(iii) \(3x-5y-4=0\) and \(9x=2y+7\) (iv) \(\dfrac{x}{2}+\dfrac{2y}{3}=-1\) and \(x-\dfrac{y}{3}=3\)
(i) Elimination: Multiply \(x+y=5\) by 3: \(3x+3y=15\). Adding to \(2x-3y=4\):
Substitution: \(y=5-x\). Substituting into \(2x-3y=4\): \(2x-3(5-x)=4\Rightarrow5x-15=4\Rightarrow x=\dfrac{19}{5}\), \(y=\dfrac{6}{5}\) — same answer.
Solution: \(x=\dfrac{19}{5},\ y=\dfrac{6}{5}\).
(ii) Simplify \(2x-2y=2\) to \(x-y=1\). Elimination: multiply \(x-y=1\) by 4: \(4x-4y=4\). Adding to \(3x+4y=10\):
Substitution: \(x=y+1\). Substituting into \(3x+4y=10\): \(3(y+1)+4y=10\Rightarrow7y=7\Rightarrow y=1,\ x=2\) — same answer.
Solution: x = 2, y = 1.
(iii) Rewrite as \(3x-5y=4\) and \(9x-2y=7\). Elimination: multiply the first by 3: \(9x-15y=12\). Subtracting \(9x-2y=7\):
Substitution: from \(3x-5y=4\), \(x=\dfrac{4+5y}{3}\). Substituting into \(9x-2y=7\): \(3(4+5y)-2y=7\Rightarrow13y=-5\Rightarrow y=-\dfrac{5}{13}\), \(x=\dfrac{9}{13}\) — same answer.
Solution: \(x=\dfrac{9}{13},\ y=-\dfrac{5}{13}\).
(iv) Multiply \(\dfrac{x}{2}+\dfrac{2y}{3}=-1\) by 6: \(3x+4y=-6\). Multiply \(x-\dfrac{y}{3}=3\) by 3: \(3x-y=9\). Elimination: subtracting the second from the first:
Substitution: from \(3x-y=9\), \(x=\dfrac{9+y}{3}\). Substituting into \(3x+4y=-6\): \((9+y)+4y=-6\Rightarrow5y=-15\Rightarrow y=-3,\ x=2\) — same answer.
Solution: x = 2, y = −3.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\dfrac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
(ii) 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?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
(i) Let the fraction be \(\dfrac{a}{b}\). The conditions give:
Subtracting the first from the second eliminates b:
Then \(b=a+2=5\). The fraction is \(\dfrac{3}{5}\). Check: \(\dfrac{3+1}{5-1}=1\ \checkmark\), \(\dfrac{3}{5+1}=\dfrac{1}{2}\ \checkmark\).
(ii) Let Nuri’s present age be x and Sonu’s be y. Five years ago: \(x-5=3(y-5)\Rightarrow x-3y=-10\). Ten years later: \(x+10=2(y+10)\Rightarrow x-2y=10\). Subtracting the first from the second eliminates x:
Then \(x=10+2(20)=50\). Nuri is 50 years old and Sonu is 20 years old.
(iii) Let the ten’s digit be x and the unit’s digit be y, so the number is \(10x+y\). \(x+y=9\). Also, \(9(10x+y)=2(10y+x)\), which simplifies to \(88x-11y=0\), i.e. \(8x-y=0\). Adding this to \(x+y=9\) eliminates y:
Then \(y=9-1=8\). The number is 18. Check: \(9\times18=162\), and twice the reversed number \(81\) is \(2\times81=162\ \checkmark\).
(iv) Let the number of ₹50 notes be x and ₹100 notes be y. \(x+y=25\) and \(50x+100y=2000\), i.e. \(x+2y=40\). Subtracting the first from this eliminates x:
Then \(x=25-15=10\). Meena received 10 notes of ₹50 and 15 notes of ₹100.
(v) Let the fixed charge for the first 3 days be x and the extra charge per day thereafter be y. Saritha (7 days = 3 fixed + 4 extra): \(x+4y=27\). Susy (5 days = 3 fixed + 2 extra): \(x+2y=21\). Subtracting the second from the first eliminates x:
Then \(x=21-2(3)=15\). The fixed charge is ₹15 and the extra charge per day is ₹3.
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