Introduction to Pair of Linear Equations in Two Variables
In this chapter, we learn how to solve a pair of linear equations in two variables. A linear equation in two variables is an equation of the form:ax+by=cax + by = cax+by=c
where aaa, bbb, and ccc are constants, and xxx and yyy are the variables.
A pair of linear equations in two variables can be written as:a1x+b1y=c1a_1x + b_1y = c_1a1x+b1y=c1 a2x+b2y=c2a_2x + b_2y = c_2a2x+b2y=c2
There are various methods to solve such pairs of linear equations:
- Graphical Method
- Substitution Method
- Elimination Method
We will go through these methods and solve the problems step by step.
Exercise 3.1: Solving Linear Equations
Q1. Solve the following pair of linear equations by substitution method:
3x+4y=53x + 4y = 53x+4y=5 x+2y=3x + 2y = 3x+2y=3
Solution:
We will solve the second equation for xxx and then substitute it into the first equation.
From the second equation:x+2y=3⇒x=3−2yx + 2y = 3 \quad \Rightarrow \quad x = 3 – 2yx+2y=3⇒x=3−2y
Now, substitute x=3−2yx = 3 – 2yx=3−2y into the first equation:3(3−2y)+4y=53(3 – 2y) + 4y = 53(3−2y)+4y=5 9−6y+4y=59 – 6y + 4y = 59−6y+4y=5 9−2y=59 – 2y = 59−2y=5 −2y=5−9-2y = 5 – 9−2y=5−9 −2y=−4-2y = -4−2y=−4 y=2y = 2y=2
Now substitute y=2y = 2y=2 back into x=3−2yx = 3 – 2yx=3−2y:x=3−2(2)=3−4=−1x = 3 – 2(2) = 3 – 4 = -1x=3−2(2)=3−4=−1
Thus, the solution is x=−1x = -1x=−1 and y=2y = 2y=2.
Q2. Solve the following pair of linear equations by elimination method:
5x+3y=145x + 3y = 145x+3y=14 3x+2y=103x + 2y = 103x+2y=10
Solution:
We will multiply both equations by suitable numbers to eliminate one variable.
First, multiply the first equation by 2 and the second by 3:2(5x+3y)=2(14)⇒10x+6y=282(5x + 3y) = 2(14) \quad \Rightarrow \quad 10x + 6y = 282(5x+3y)=2(14)⇒10x+6y=28 3(3x+2y)=3(10)⇒9x+6y=303(3x + 2y) = 3(10) \quad \Rightarrow \quad 9x + 6y = 303(3x+2y)=3(10)⇒9x+6y=30
Now, subtract the second equation from the first:(10x+6y)−(9x+6y)=28−30(10x + 6y) – (9x + 6y) = 28 – 30(10x+6y)−(9x+6y)=28−30 x=−2x = -2x=−2
Now substitute x=−2x = -2x=−2 into the first equation 5x+3y=145x + 3y = 145x+3y=14:5(−2)+3y=145(-2) + 3y = 145(−2)+3y=14 −10+3y=14-10 + 3y = 14−10+3y=14 3y=243y = 243y=24 y=8y = 8y=8
Thus, the solution is x=−2x = -2x=−2 and y=8y = 8y=8.
Q3. Solve the following pair of linear equations by graphical method:
x+y=5x + y = 5x+y=5 x−y=1x – y = 1x−y=1
Solution:
To solve graphically, we first plot the two equations on a graph.
For the equation x+y=5x + y = 5x+y=5, rewrite it as y=5−xy = 5 – xy=5−x. For various values of xxx, calculate yyy:
- When x=0x = 0x=0, y=5y = 5y=5.
- When x=5x = 5x=5, y=0y = 0y=0.
For the equation x−y=1x – y = 1x−y=1, rewrite it as y=x−1y = x – 1y=x−1. For various values of xxx, calculate yyy:
- When x=0x = 0x=0, y=−1y = -1y=−1.
- When x=2x = 2x=2, y=1y = 1y=1.
Now, plot these points on a graph and find the point of intersection. The point of intersection is the solution to the system of equations. By plotting the two lines, we find that the point of intersection is (3,2)(3, 2)(3,2).
Thus, the solution is x=3x = 3x=3 and y=2y = 2y=2.
Exercise 3.2: Word Problems
Q1. A shopkeeper sells 3 pens and 5 pencils for Rs 35. He sells 5 pens and 2 pencils for Rs 30. Find the price of one pen and one pencil.
Let the price of one pen be xxx and the price of one pencil be yyy.
We can form the following system of equations:3x+5y=353x + 5y = 353x+5y=35 5x+2y=305x + 2y = 305x+2y=30
Now, we can solve these equations using either substitution or elimination. Let’s use the elimination method.
Multiply the first equation by 5 and the second by 3:5(3x+5y)=5(35)⇒15x+25y=1755(3x + 5y) = 5(35) \quad \Rightarrow \quad 15x + 25y = 1755(3x+5y)=5(35)⇒15x+25y=175 3(5x+2y)=3(30)⇒15x+6y=903(5x + 2y) = 3(30) \quad \Rightarrow \quad 15x + 6y = 903(5x+2y)=3(30)⇒15x+6y=90
Now subtract the second equation from the first:(15x+25y)−(15x+6y)=175−90(15x + 25y) – (15x + 6y) = 175 – 90(15x+25y)−(15x+6y)=175−90 19y=8519y = 8519y=85 y=8519=5y = \frac{85}{19} = 5y=1985=5
Now substitute y=5y = 5y=5 into 3x+5y=353x + 5y = 353x+5y=35:3x+5(5)=353x + 5(5) = 353x+5(5)=35 3x+25=353x + 25 = 353x+25=35 3x=103x = 103x=10 x=103=3.33x = \frac{10}{3} = 3.33x=310=3.33
Thus, the price of one pen is Rs 3.33 and the price of one pencil is Rs 5.
Q2. The sum of two numbers is 26. If one number exceeds the other by 6, find the numbers.
Let the numbers be xxx and yyy.
We can form the following system of equations:x+y=26x + y = 26x+y=26 x−y=6x – y = 6x−y=6
Now solve these equations. Add both equations:(x+y)+(x−y)=26+6(x + y) + (x – y) = 26 + 6(x+y)+(x−y)=26+6 2x=322x = 322x=32 x=16x = 16x=16
Substitute x=16x = 16x=16 into x+y=26x + y = 26x+y=26:16+y=2616 + y = 2616+y=26 y=10y = 10y=10
Thus, the two numbers are 16 and 10.
Additional Questions
1. Solve the pair of equations 2x+3y=72x + 3y = 72x+3y=7 and 4x−y=34x – y = 34x−y=3 using the substitution method.
Solution:
From the second equation 4x−y=34x – y = 34x−y=3, we can solve for yyy:y=4x−3y = 4x – 3y=4x−3
Now, substitute y=4x−3y = 4x – 3y=4x−3 into the first equation 2x+3y=72x + 3y = 72x+3y=7:2x+3(4x−3)=72x + 3(4x – 3) = 72x+3(4x−3)=7 2x+12x−9=72x + 12x – 9 = 72x+12x−9=7 14x=1614x = 1614x=16 x=1614=87x = \frac{16}{14} = \frac{8}{7}x=1416=78
Now, substitute x=87x = \frac{8}{7}x=78 into y=4x−3y = 4x – 3y=4x−3:y=4(87)−3=327−3=327−217=117y = 4\left(\frac{8}{7}\right) – 3 = \frac{32}{7} – 3 = \frac{32}{7} – \frac{21}{7} = \frac{11}{7}y=4(78)−3=732−3=732−721=711
Thus, the solution is x=87x = \frac{8}{7}x=78 and y=117y = \frac{11}{7}y=711.
2. Solve the system of equations 3x+4y=113x + 4y = 113x+4y=11 and 2x−y=−12x – y = -12x−y=−1 using the elimination method.
Solution:
We multiply the second equation by 4 to eliminate yyy:4(2x−y)=4(−1)⇒8x−4y=−44(2x – y) = 4(-1) \quad \Rightarrow \quad 8x – 4y = -44(2x−y)=4(−1)⇒8x−4y=−4
Now, add the first equation 3x+4y=113x + 4y = 113x+4y=11 to this:(3x+4y)+(8x−4y)=11+(−4)(3x + 4y) + (8x – 4y) = 11 + (-4)(3x+4y)+(8x−4y)=11+(−4) 3x+8x=73x + 8x = 73x+8x=7 11x=711x = 711x=7 x=711x = \frac{7}{11}x=117
Substitute x=711x = \frac{7}{11}x=117 into 2x−y=−12x – y = -12x−y=−1:2(711)−y=−12\left(\frac{7}{11}\right) – y = -12(117)−y=−1 1411−y=−1\frac{14}{11} – y = -11114−y=−1 −y=−1−1411=−11−1411=−2511-y = -1 – \frac{14}{11} = \frac{-11 – 14}{11} = \frac{-25}{11}−y=−1−1114=11−11−14=11−25 y=2511y = \frac{25}{11}y=1125
Thus, the solution is x=711x = \frac{7}{11}x=117 and y=2511y = \frac{25}{11}y=1125.
3. If x+y=10x + y = 10x+y=10 and x−y=2x – y = 2x−y=2, find the values of xxx and yyy.
Solution:
Add both equations:(x+y)+(x−y)=10+2(x + y) + (x – y) = 10 + 2(x+y)+(x−y)=10+2 2x=122x = 122x=12 x=6x = 6x=6
Substitute x=6x = 6x=6 into x+y=10x + y = 10x+y=10:6+y=106 + y = 106+y=10 y=4y = 4y=4
Thus, the values are x=6x = 6x=6 and y=4y = 4y=4.
4. A train travels 60 km/hr for 4 hours and then 50 km/hr for 3 hours. Find the total distance traveled.
Solution:
Distance = Speed × Time
For the first part, the distance is:Distance=60 km/hr×4 hours=240 km\text{Distance} = 60 \, \text{km/hr} \times 4 \, \text{hours} = 240 \, \text{km}Distance=60km/hr×4hours=240km
For the second part, the distance is:Distance=50 km/hr×3 hours=150 km\text{Distance} = 50 \, \text{km/hr} \times 3 \, \text{hours} = 150 \, \text{km}Distance=50km/hr×3hours=150km
Thus, the total distance is:240+150=390 km240 + 150 = 390 \, \text{km}240+150=390km
5. Two numbers are in the ratio 3:4. If their sum is 63, find the numbers.
Solution:
Let the two numbers be 3x3x3x and 4x4x4x, where xxx is the common factor.
The sum of the numbers is:3x+4x=633x + 4x = 633x+4x=63 7x=637x = 637x=63 x=9x = 9x=9
Thus, the numbers are:3x=3×9=273x = 3 \times 9 = 273x=3×9=27 4x=4×9=364x = 4 \times 9 = 364x=4×9=36
The two numbers are 27 and 36.
6. Solve the pair of equations 5x+2y=125x + 2y = 125x+2y=12 and 3x−y=43x – y = 43x−y=4 by the graphical method.
Solution:
To solve this by the graphical method, rewrite both equations in slope-intercept form:
- 5x+2y=12⇒y=−52x+65x + 2y = 12 \quad \Rightarrow \quad y = -\frac{5}{2}x + 65x+2y=12⇒y=−25x+6
- 3x−y=4⇒y=3x−43x – y = 4 \quad \Rightarrow \quad y = 3x – 43x−y=4⇒y=3x−4
Now, plot these equations on a graph. The point where both lines intersect gives the solution. After plotting the lines, we find the intersection point is (2,2)(2, 2)(2,2).
Thus, the solution is x=2x = 2x=2 and y=2y = 2y=2.
7. Solve the system of equations x+2y=7x + 2y = 7x+2y=7 and 3x−y=43x – y = 43x−y=4 using substitution.
Solution:
From the first equation x+2y=7x + 2y = 7x+2y=7, solve for xxx:x=7−2yx = 7 – 2yx=7−2y
Now, substitute x=7−2yx = 7 – 2yx=7−2y into the second equation 3x−y=43x – y = 43x−y=4:3(7−2y)−y=43(7 – 2y) – y = 43(7−2y)−y=4 21−6y−y=421 – 6y – y = 421−6y−y=4 21−7y=421 – 7y = 421−7y=4 −7y=4−21-7y = 4 – 21−7y=4−21 −7y=−17-7y = -17−7y=−17 y=177y = \frac{17}{7}y=717
Now, substitute y=177y = \frac{17}{7}y=717 back into x=7−2yx = 7 – 2yx=7−2y:x=7−2(177)=7−347=497−347=157x = 7 – 2\left(\frac{17}{7}\right) = 7 – \frac{34}{7} = \frac{49}{7} – \frac{34}{7} = \frac{15}{7}x=7−2(717)=7−734=749−734=715
Thus, the solution is x=157x = \frac{15}{7}x=715 and y=177y = \frac{17}{7}y=717.
8. The sum of two numbers is 30, and one number exceeds the other by 6. Find the numbers.
Solution:
Let the two numbers be xxx and yyy.
We can form the following system of equations:x+y=30x + y = 30x+y=30 x−y=6x – y = 6x−y=6
Add the two equations:(x+y)+(x−y)=30+6(x + y) + (x – y) = 30 + 6(x+y)+(x−y)=30+6 2x=362x = 362x=36 x=18x = 18x=18
Substitute x=18x = 18x=18 into x+y=30x + y = 30x+y=30:18+y=3018 + y = 3018+y=30 y=12y = 12y=12
Thus, the two numbers are 18 and 12.
9. Solve the pair of equations x+y=8x + y = 8x+y=8 and 2x−y=32x – y = 32x−y=3 using the substitution method.
Solution:
From the first equation x+y=8x + y = 8x+y=8, solve for xxx:x=8−yx = 8 – yx=8−y
Now, substitute x=8−yx = 8 – yx=8−y into the second equation 2x−y=32x – y = 32x−y=3:2(8−y)−y=32(8 – y) – y = 32(8−y)−y=3 16−2y−y=316 – 2y – y = 316−2y−y=3 16−3y=316 – 3y = 316−3y=3 −3y=3−16-3y = 3 – 16−3y=3−16 −3y=−13-3y = -13−3y=−13 y=133y = \frac{13}{3}y=313
Now substitute y=133y = \frac{13}{3}y=313 into x=8−yx = 8 – yx=8−y:x=8−133=243−133=113x = 8 – \frac{13}{3} = \frac{24}{3} – \frac{13}{3} = \frac{11}{3}x=8−313=324−313=311
Thus, the solution is x=113x = \frac{11}{3}x=311 and y=133y = \frac{13}{3}y=313.
10. A man can swim 12 km/hr in still water. If the stream flows at 4 km/hr, find the speed of the man in the stream.
Solution:
The speed of the man in the stream is the sum of the speed of the man in still water and the speed of the stream.Speed in the stream=12+4=16 km/hr\text{Speed in the stream} = 12 + 4 = 16 \, \text{km/hr}Speed in the stream=12+4=16km/hr
Thus, the speed of the man in the stream is 16 km/hr.
11. A rectangular plot has a length that is 5 meters more than twice its width. If the area of the rectangle is 50 square meters, find the dimensions of the rectangle.
Solution:
Let the width of the rectangle be xxx meters. Then, the length of the rectangle is 2x+52x + 52x+5 meters.
The area of the rectangle is given by:Area=Length×Width\text{Area} = \text{Length} \times \text{Width}Area=Length×Width 50=(2x+5)×x50 = (2x + 5) \times x50=(2x+5)×x 50=2×2+5×50 = 2x^2 + 5×50=2×2+5x
Rearrange the equation:2×2+5x−50=02x^2 + 5x – 50 = 02×2+5x−50=0
We now solve this quadratic equation using the quadratic formula:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac
For the equation 2×2+5x−50=02x^2 + 5x – 50 = 02×2+5x−50=0, a=2a = 2a=2, b=5b = 5b=5, and c=−50c = -50c=−50.x=−5±52−4(2)(−50)2(2)x = \frac{-5 \pm \sqrt{5^2 – 4(2)(-50)}}{2(2)}x=2(2)−5±52−4(2)(−50) x=−5±25+4004x = \frac{-5 \pm \sqrt{25 + 400}}{4}x=4−5±25+400 x=−5±4254x = \frac{-5 \pm \sqrt{425}}{4}x=4−5±425 x=−5±20.6154x = \frac{-5 \pm 20.615}{4}x=4−5±20.615
Now, let’s calculate the two possible values of xxx:x=−5+20.6154=15.6154≈3.90x = \frac{-5 + 20.615}{4} = \frac{15.615}{4} \approx 3.90x=4−5+20.615=415.615≈3.90 x=−5−20.6154=−25.6154≈−6.40x = \frac{-5 – 20.615}{4} = \frac{-25.615}{4} \approx -6.40x=4−5−20.615=4−25.615≈−6.40
Since width cannot be negative, the width is approximately 3.90 meters.
Now, find the length:Length=2x+5=2(3.90)+5=7.80+5=12.80 meters\text{Length} = 2x + 5 = 2(3.90) + 5 = 7.80 + 5 = 12.80 \, \text{meters}Length=2x+5=2(3.90)+5=7.80+5=12.80meters
Thus, the dimensions of the rectangle are approximately 3.90 meters (width) and 12.80 meters (length).
12. Solve the system of equations 4x+5y=204x + 5y = 204x+5y=20 and 2x−y=42x – y = 42x−y=4 using the elimination method.
Solution:
We multiply the second equation by 5 to align the coefficients of yyy:5(2x−y)=5(4)⇒10x−5y=205(2x – y) = 5(4) \quad \Rightarrow \quad 10x – 5y = 205(2x−y)=5(4)⇒10x−5y=20
Now, add the first equation 4x+5y=204x + 5y = 204x+5y=20 to the new equation 10x−5y=2010x – 5y = 2010x−5y=20:(4x+5y)+(10x−5y)=20+20(4x + 5y) + (10x – 5y) = 20 + 20(4x+5y)+(10x−5y)=20+20 4x+10x=404x + 10x = 404x+10x=40 14x=4014x = 4014x=40 x=4014=207x = \frac{40}{14} = \frac{20}{7}x=1440=720
Now substitute x=207x = \frac{20}{7}x=720 into the second equation 2x−y=42x – y = 42x−y=4:2(207)−y=42\left(\frac{20}{7}\right) – y = 42(720)−y=4 407−y=4\frac{40}{7} – y = 4740−y=4 −y=4−407=287−407=−127-y = 4 – \frac{40}{7} = \frac{28}{7} – \frac{40}{7} = \frac{-12}{7}−y=4−740=728−740=7−12 y=127y = \frac{12}{7}y=712
Thus, the solution is x=207x = \frac{20}{7}x=720 and y=127y = \frac{12}{7}y=712.
13. The sum of two numbers is 50, and their difference is 10. Find the numbers.
Solution:
Let the two numbers be xxx and yyy.
We can form the following system of equations:x+y=50x + y = 50x+y=50 x−y=10x – y = 10x−y=10
Add the two equations:(x+y)+(x−y)=50+10(x + y) + (x – y) = 50 + 10(x+y)+(x−y)=50+10 2x=602x = 602x=60 x=30x = 30x=30
Substitute x=30x = 30x=30 into x+y=50x + y = 50x+y=50:30+y=5030 + y = 5030+y=50 y=20y = 20y=20
Thus, the two numbers are 30 and 20.
14. Solve the pair of linear equations 2x+3y=182x + 3y = 182x+3y=18 and 5x−2y=45x – 2y = 45x−2y=4 using the elimination method.
Solution:
Multiply the first equation by 2 and the second equation by 3 to align the coefficients of yyy:2(2x+3y)=2(18)⇒4x+6y=362(2x + 3y) = 2(18) \quad \Rightarrow \quad 4x + 6y = 362(2x+3y)=2(18)⇒4x+6y=36 3(5x−2y)=3(4)⇒15x−6y=123(5x – 2y) = 3(4) \quad \Rightarrow \quad 15x – 6y = 123(5x−2y)=3(4)⇒15x−6y=12
Now, add the two equations:(4x+6y)+(15x−6y)=36+12(4x + 6y) + (15x – 6y) = 36 + 12(4x+6y)+(15x−6y)=36+12 4x+15x=484x + 15x = 484x+15x=48 19x=4819x = 4819x=48 x=4819x = \frac{48}{19}x=1948
Now substitute x=4819x = \frac{48}{19}x=1948 into 2x+3y=182x + 3y = 182x+3y=18:2(4819)+3y=182\left(\frac{48}{19}\right) + 3y = 182(1948)+3y=18 9619+3y=18\frac{96}{19} + 3y = 181996+3y=18 3y=18−9619=34219−9619=246193y = 18 – \frac{96}{19} = \frac{342}{19} – \frac{96}{19} = \frac{246}{19}3y=18−1996=19342−1996=19246 y=24619×13=24657=8219y = \frac{246}{19} \times \frac{1}{3} = \frac{246}{57} = \frac{82}{19}y=19246×31=57246=1982
Thus, the solution is x=4819x = \frac{48}{19}x=1948 and y=8219y = \frac{82}{19}y=1982.
15. The cost of 5 pencils and 7 pens is Rs 35, and the cost of 3 pencils and 4 pens is Rs 20. Find the cost of each pencil and each pen.
Solution:
Let the cost of one pencil be xxx and the cost of one pen be yyy.
We can form the following system of equations:5x+7y=355x + 7y = 355x+7y=35 3x+4y=203x + 4y = 203x+4y=20
Multiply the first equation by 4 and the second by 7 to align the coefficients of yyy:4(5x+7y)=4(35)⇒20x+28y=1404(5x + 7y) = 4(35) \quad \Rightarrow \quad 20x + 28y = 1404(5x+7y)=4(35)⇒20x+28y=140 7(3x+4y)=7(20)⇒21x+28y=1407(3x + 4y) = 7(20) \quad \Rightarrow \quad 21x + 28y = 1407(3x+4y)=7(20)⇒21x+28y=140
Now, subtract the second equation from the first:(20x+28y)−(21x+28y)=140−140(20x + 28y) – (21x + 28y) = 140 – 140(20x+28y)−(21x+28y)=140−140 −x=0-x = 0−x=0 x=0x = 0x=0
Now substitute x=0x = 0x=0 into 3x+4y=203x + 4y = 203x+4y=20:3(0)+4y=203(0) + 4y = 203(0)+4y=20 4y=204y = 204y=20 y=5y = 5y=5
Thus, the cost of one pencil is Rs 0 and the cost of one pen is Rs 5.
16. The sum of three consecutive numbers is 72. Find the numbers.
Solution:
Let the three consecutive numbers be xxx, x+1x + 1x+1, and x+2x + 2x+2.
The sum of the numbers is:x+(x+1)+(x+2)=72x + (x + 1) + (x + 2) = 72x+(x+1)+(x+2)=72 3x+3=723x + 3 = 723x+3=72 3x=693x = 693x=69 x=23x = 23x=23
Thus, the three consecutive numbers are 23, 24, and 25.
Class 10 Chapter 2 Maths Polynomials
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