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Do the equations \(x+2y+2=0 \) and \(\dfrac{x}{2}+\dfrac{y}{4}1 \) represent a pair of intersecting lines? Justify your answer.
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It is not a intersecting line. It is parallel lines
Do the equations \(2x+4y=3\) and \(12+6y=6\) represent a pair of parallel lines ? Justify your answer.
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No
For the pair of equations \(\lambda x+3y=7,2x+6y=14 \) to have infinitely many solutions, the value of \(\lambda \) should be 1. Is this statement true ? Give reasons.
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Yes
Find the value of \(k \) for which the following system of equations have infinitely many solutions.
\( 2x – 3y = 7, (k + 2) x – (2k – 1)y = 3 (2k – 1)\)
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Solve the following pairs of linear equations by substitution method.
\( 3x – 5y – 4 = 0, 9x – 2y = 7\)
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…(1)
…(2)
eqn (1)
Solve the following pairs of linear equations by substitution method.
\( 3x – 5y = 20, 6x – 10y = 40\)
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…(1)
…(2)
Solve the following pairs of linear equations by elimination method.
\(\dfrac{3x}{2}\dfrac{5y}{3}=2,\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{13}{6} \)
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…(1)
…(2)
eqn (1)
eqn (2)
Solve the following pairs of linear equations by elimination method.
\(0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3 \)
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…(1)
….(2)
eqn (1)
Solve the following pairs of linear equations by cross multiplication method.
\(\sqrt2x+\sqrt3y=0,\ \sqrt3x\sqrt8y=0 \)
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Solve the following pairs of linear equations by cross multiplication method.
\(xy=3,\dfrac{x}{3}+\dfrac{y}{2}=6 \)
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Determine the value of \(k \) for which the following pair of linear equations is consistent :
\( 2x+(k1)\ y=5;\quad 3x+6y=5\)
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Determine the value of \(k \) for which the following pair of linear equations has no solution :
\( (k+1) x1(2k+1) y=4;\quad 6x10y=7\)
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Determine the value of \(k \) for which the following pair of linear equations represents a pair of parallel lines on the graph.
\((3k1) x + (k1) y= 5;\quad (K+1) x+y= 3 \)
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For which value (s) of \(k \), do the pair of linear equations : \(kx+y=k^2\) and \(x+ky=1\) have a unique solution?
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Find the relation between a,b,c and d for which the equations \(ax+by=c\) and \(cx+dy=a\) have a unique solution.
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For what value of \(k,\ 2x+3y=4\) and \((k+2) x + 6y = 3k +2\) will have infinitely many solutions.
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For which values of \(p \) does the pair of equations given below has unique solution?
\( 4x + py + 8 = 0;\ 2x + 2y + 2 = 0\)
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Solve for \(x \) and \(y \) : \(\dfrac{4}{x}+5y=7;\dfrac{3}{x}+4y=5 \)
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Let
…(1)
…(2)
Subtracting (2) from (1) we get
…(3)
Using (3) in eqn (1) we get
Solve for \(x \) and \(y \) : \(4x\dfrac{6}{y}=15;\ 3x\dfrac{4}{y}=7 \)
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Multiplying (1) by 4 and (2) by 6
…(3)
…(4)
Adding eqn (3) and (4)
Without drawing the graph, find out, whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident.
\(18x7y=24,\dfrac{9}{5}x\dfrac{7}{10}y=\dfrac{9}{10} \)
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It has no solution. It is parallel
Find the value of k for which the pair of linear equations kx + 3y = k – 3 and 12x + ky = k has no solution.
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Value of for which the equation has no solution is
Solve 2x + 3y = 11 and 2x – 4y = 24 and hence find the value of m for which y = mx + 3.
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The path of a Car I is given by the equation 3x + 4y – 12 = 0 and the path of another Car II is given by the equation 6x + 8y – 48 = 0. Represent this situation graphically.
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The paths of two car are giver by the following pair of linear equations.
3x + 4y – 12 ……..(1)
6x + 8y – 48 …….. (2)
Putting , we get
Putting , we get
Thus, two solution of equation are and
We have,
Putting , we get
Putting , we get
Thus, two solution of equation are and
Determine algebraically, the vaertices of the triangle formed by the lines.
\(3xy=3,\ 2x3y=2,\ x+2y=8 \)
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Multiplying eqn (1) by (3) and then subtract (2) from it
Putting the value of in 1 we get
so one vertex is (1, 0)
Equation eqn (3)
Subtracting in eqn (3) we get
Coordinate of point is (4, 2) on solving lines (3) and (1) we get
Point
There are two numbers such that 3 times the greater is 18 times their difference and 4 times the smaller is 4 less than twice the sum of the two. What are the numbers?
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..(1)
…(2)
After solving eqn (1)
..(3)
After solving equation (2)
…(4)
Substituting equation (4) into (3)
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?
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Let numerator be , denumerator be
…(1)
Also …(2)
on solving
Fraction
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?
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Let present age of nuri and sonu years
A/Q …(1)
Also …(2)
on solving we get
Present age of nuri is 50 years and sonu is 20 years
Solve the following pair of linear equations \(\dfrac{2x4}{4}=\dfrac{3y2}{2};\dfrac{4x3}{2}\dfrac{6y}{3}=2\dfrac{5}{6} \)
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Represent the following pairs of linear equations graphically.
\(2x – y = 2 ; 4x – y = 8 \)
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Solve graphically \(x + y = 5 ; 4x + 3y = 17 \)
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The given equations are : …(1)
…(2)
Consider (1)
when
when
when
Consider (2),
when
when
Solve graphically
If 2x + y = 23 and 4x – y = 19, find the value of 5y – 2x.
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Adding both the equations we get
Solve graphically \(xy = 0.8, \dfrac{20}{(x+y)} = 2. \)
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…(1)
…(2)
Subtracting (2) from (1) we get
Solve graphically \(\dfrac{1}{2x}\dfrac{1}{y}=1,\dfrac{1}{x}+\dfrac{1}{2y}=8,x,y\ne0 \).
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Let
…(1)
…(2)
Multiplying equation (1) by and than adding with (2) we get
\(\dfrac{22}{3x+2y}+\dfrac{7}{3x2y}=3,\dfrac{33}{3x+2y}\dfrac{14}{3x2y}=1 \).
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Let
..(1)
…(2)
equation (1)
on solving
\(\dfrac{2xy}{x+y}+\dfrac{3}{2},\dfrac{xy}{2xy}=\dfrac{3}{10},x+y\ \ne\ 0,2xy\ \ne0 \).
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…(3)
…(4)
on solving we get
A part of monthly expenses of a family is constant and the remaining varies with the price of wheat. When the rate of wheat is Rs.250 a quintal, the total monthly expenses of the family are Rs.1000 and when it is Rs. 240 a quintal, the total monthly expenses of the family are Rs. 980. Find the total monthly expenses of the family when the cost of wheat is Rs. 350 a quintal.
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Let the constant expenditure be ₹
consumption of wheat quintals
A/Q …(1)
…(2)
From (1) we get
substituting in eqn (2)
Total monthly expenss and total expenss when the price is per quintal
A lady has only 25 p and 50 p coins in her purse. If in all she has 40 coins totalling Rs. 12.50, find the number of coins of each type in her purse.
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Let the number of 25 p = x, 50 p = y
A/Q …(1)
..(2)
equation (1) – equation (2) we get
No. of 25 paise coins = 30 and no. of 50 paise coins = 10
Solve for \(x \) and \(y \) : \((a – b) x + (a + b) y = a^2 – 2ab – b^2 \)
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…(1)
…(2)
subtracting (2) from (1)
The taxi charge in a city consists of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs.105 and for journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charges per km ?
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Let fixed charge and charge per km
…(1)
Fixed charge and charge per km
A part of monthly hostel charges is fixed and the remaining depends on the number of days on has taken food in the mess. When a student A takes food for 20 days, she has to pay Rs.1000 as hostel charges whereas a student B, who takes food 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of the food per day.
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Let fixed charge and charge per day taken food per day
…(1)
..(2)
From (1)
Putting value of in eqn (2)
so fixed charge and charges per day per day
Nine times a twodigit number is the same as twice the number obtained by interchanging the digits of the number. If one digit of the number exceeds the other number by 7, find the number.
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Let ten’s digit be and unit’s digit be
A/Q
..(4)
Also
Number