Class 10 RD Sharma Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4 | Set 2 (original) (raw)
Last Updated : 12 Sep, 2024
Chapter 3 of RD Sharmaβs Class 10 Mathematics book delves into the topic of Pair of Linear Equations in Two Variables. This chapter is fundamental in understanding how to solve systems of linear equations and their applications. Exercise 3.4 | Set 2 provides the practice problems to strengthen students' ability to solve these equations using the various methods including the graphical, substitution, and elimination methods.
What is a Pair of Linear Equations in Two Variables?
A pair of linear equations in the two variables consists of the two equations each of which is linear in two variables. The general form of these equations is:
π1π₯+π1π¦=π1
a2x+b2y=c2
where π1 , π1 , π1, π2, π2 and π2 are constants. Solving this pair of equations involves finding the values of the x and y that satisfy both equations simultaneously. This can be done using various methods such as:
- Graphical Method: Plotting both the lines on a graph to find their intersection.
- Substitution Method: Solving one equation for one variable and substituting it into the other equation.
- Elimination Method: Adding or subtracting the equations to eliminate one of the variables.
Solve each of the following systems of equations by the method of cross-multiplication.
Question 15. 2ax + 3by = a + 2b and 3ax + 2by = 2a + b
**Solution:
Given that,
2ax + 3by = a + 2b
3ax + 2by = 2a + b
On comparing both the equation with the general form we get
a1 = 2a, b1 = 3b, c1 = -(a + 2b), a2 = 3a, b2 = 2b, c2 = -(2a + b)
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/(-3b(2a + b) + 2b(a + 2b)) = y/(-3a(a + 2b) + 2a(2a + b)) = 1/(4ab - 9ab)
x/(b2 - 4ab) = y/(a2 - 4ab) = 1/-5ab
x/(-4ab + b2) = 1/-5ab
x/b(b - 4a) = 1/-5ab
x = (4a - b)/5a
and,
= -y/(-a2 + 4ab) = 1/-5ab
= -y/a(-a + 4b) = 1/-5ab
y = (4b - a)/5b
**Hence, x = (4a - b)/5a and y = (4b - a)/5b
Question 16. 5ax + 6by = 28 and 3ax + 4by = 18
**Solution:-
Given that,
5ax + 6by = 28
3ax + 4by = 18
Or, 5ax + 6by - 28 = 0
3ax + 4by - 18 = 0
Here, a1 = 5a, b1 = 6b, c1=-28
aβ= 3a, bβ = 4b, cβ =-18
On comparing both the equation with the general form we get
a1 = 5a, b1 = 6b, c1 = -28, a2 = 3a, b2 = 4b, c2 = -18
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/4b = -y/-6a = 1/2ab
x/4b = 1/2b
x = 2/a
and,
-y/-6a = 1/2ab
y = 3/b
**Hence, x = 2/a and y = 3/b
Question 17. (a + 2b)x + (2a - b)y = 2 and (a β 2b)x + (2a + b)y = 3
**Solution:
Given that,
(a + 2b)x + (2a - b)y = 2
(a β 2b)x + (2a + b)y = 3
Or
(a + 2b) x + (2a β b)y - 2 = 0
(a β 2b) x + (2a + b)y - 3 = 0
On comparing both the equation with the general form we get
a1 = a + 2b, b1 = 2a - b, c1 = -2, a2 = a - 2b, b2 = 2a + b, c2 = -3
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/(-3(2a - b)) + (2(2a + b)) = y/(-2(a - 2b)) + (3(a + 2b)) = 1/((a + 2b)(2a + b) - (a - 2b)(2a - b))
x/(5b - 2a) = y/(a + 10b) = 1/(2a2 + 5ab + 2b2 - 2a2 + 5ab - 2b2)
x/(5b - 2a) = y/(a + 10b) = 1/10ab
So,
x/(5b - 2a) = 1/10ab
x= (5b - 2a)/10ab
and,
y/(a + 10b) = 1/10ab
y = (a + 10b)/10ab
**Hence, x = (5b - 2a)/10ab and y= (a + 10b)/10ab
Question 18. x((a - b) + (ab/(a - b))) = y((a + b) - (ab/(a + b))) and x + y = 2a2
**Solution:
Given that,
x((a - b) + (ab/(a - b))) = y((a + b) - (ab/(a + b)))
Or on solving we get
x((a2 + b2 - 2ab + ab)/(a - b)) = y((a2 + b2 + 2ab - ab)/(a + b))
= x((a2 + b2 - 2ab + ab)/(a - b)) - y((a2 + b2 + 2ab - ab)/(a + b)) = 0
and,
x + y = 2a2
On comparing both the equation with the general form we get
a1 = (a2 + b2 - 2ab + ab)/(a - b), b1 = (a2 + b2 + 2ab - ab)/(a + b), c1 = 0,
a2 = 1, b2 = 1, c2 = 2a2
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
β x/(2a2((a2 + b2 - 2ab + ab)/(a - b)) - 0) = y/(0 + 2a2((a2 + b2 + 2ab - ab)/(a + b)))
= 1/(((a2 + bΒ²)/(a - b) + (a2 + b2 + ab)/(a + b)))
β x/(2a2((a2 + b2 - 2ab + ab)/(a - b))) = y/(-2a2)((a2 + b2 + 2ab - ab)/(a + b)) = 1/(2a3/(a2 - b2))
Now,
x/(2a2((a2 + b2 - 2ab + ab)/(a - b)) = 1/(2a3/(a2 - b2))
x = (2a2(a2 + ab + b2)(a2 - b2)) / 2a3(a + b)
x = (a3 - b3)/a
and,
y/(-2a2)((a2 + b2 + 2ab - ab)/(a + b)) = 1/(2a3/(a2 - b2))
y = (2a2(a2 - ab + b2)(a2 - b2))/2a3(a - b)
y = a3 + b3/a
**Hence, x = (a 3 **- b 3 )/a and y = a 3 + b 3 /a
Question 19. bx + cy = a + b and ax[(1/(a - b)) - (1/(a + b))] + cy[(1/(b - a)) - (1/(b + a))] = 2a/(a + b)
**Solution:
Given that,
bx + cy = a + b
ax[(1/(a - b)) - (1/(a + b))] + cy[(1/(b - a)) - (1/(b + a))] = 2a/(a + b)
Or
bx + cy -(a + b) = 0
ax((1/(a - b)) - (1/(a + b))) + cy(((1/(b - a)) - (1/(b + a))) - 2a/(a + b) = 0
ax(2b/(a2 - b2))) + cy(2a/(b2 - a2)) - 2a/(a + b) = 0
On comparing both the equation with the general form we get
a1 = b, b1 = c, c1 = -(a + b),
a2 = 2b/(a2 - b2), b2 = 2a/(b2 - a2), c2 = 2a/(a + b)
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
β x / ((-2ac/(a + b)) + ((2ac(a + b)/(b2 - a2)))) = y/((-(a + b)2ab)/(a2 - b2)) + (2ab/(a + b))
= 1/((2abc/(b2- a2)) - (2abc/(a2 - b2)))
β x / ((-2ac/(a + b)) + ((2ac(a + b)/(b2 - a2)))) = y/((-(a + b)2ab)/(a2 - b2)) + (2ab/(a + b))
= 1/(-4abc/(a2 - b2))
β x/(-2ac((1/(a + b)) + (1/(a - b))) = y/(2ab((-1/(a - b)) + (1/(a + b))) = 1/(-4abc/(a2 - b2))
β x/(-4a2c/(a2 - b2)) = y/(4ab2/(a2 - b2)) = 1/1/(-4abc/(a2 - b2))
So,
x/(-4a2c/(a2 - b2)) = 1/(-4abc/(a2 - b2))
x = a/b
and,
y/(4ab2/(a2 - b2)) = 1/(-4abc/(a2 - b2))
y = b/c
**Hence, x = a/b, y = b/c
Question 20. (a β b) x + (a + b) y = 2a2 β 2b2 and (a + b) (x + y) = 4ab
**Solution:
Given that,
(a β b) x + (a + b) y = 2a2 β 2b2
(a + b) (x + y) = 4ab
(a β b) x + (a + b) y - 2(a2 β b2) = 0
(a + b)x + (a + b)y - 4ab = 0
On comparing both the equation with the general form we get
a1 = a - b, b1 = a + b, c1 = -2,
a2 = a + b, b2 = a + b, c2 = -4ab
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
β x/(-(a + b)4ab + 2(a + b) (a2 - b2)) = y/(β 2(a2 β b2)(a + b) + 4ab(a β b))
= 1/((a β b)(a + b) β (a + b)(a + b))
β x/(2(a + b)(a2 - b2 + 2ab)) = 1/-2b(a + b)
x = (2ab - a2 + b2)/b
and,
= -y/(2(a - b) (a2 + b2) -2b (a + b)) = 1/ -2b(a + b)
y = (a - b)(a2 + b2)/ b(a + b)
**Hence, x = (2ab - a 2 + b 2 )/b and y = (a - b)(a 2 + b 2 )/ b(a + b)
Question 21. a2x + b2y = c2 and b2x + a2y = d2
**Solution:
Given that,
a2x + b2y = c2
b2x + a2y = d2
Or
a2x + b2y - c2 = 0
b2x + a2y - d2 = 0
On comparing both the equation with the general form we get
a1 = a2, b1 = b2, c1 = -c2,
a2 = b2, b2 = a2, c2 = -d2
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/(-b2d2 + a2c2) = y/(-c2b2 + a2d2) = 1/(a4-b4)
x/(a2c2 - b2d2) = y/(a2d2 - c2b2) = 1/(a4-b4)
Therefore,
= x/(a2c2 - b2d2) = 1/(a4 - b4)
x = (a2c2 - b2d2)/(a4 - b4)
and,
= y/(a2d2 - c2b2) = 1/(a4-b4)
y = (a2c2 - b2d2) / (a4-b4)
**Hence, x = (a 2 c 2 **- b 2 d 2 )/(a 4 **- b 4 ), y = (a 2 c 2 **- b 2 d 2 ) / (a 4 **- b 4 )
Question 22. ax + by = (a + b)/2 and 3x + 5y = 4
**Solution:
Given that,
ax + by = (a+b)/2
3x + 5y = 4
Or
ax + by - (a + b)/2 = 0
3x + 5y - 4 = 0
On comparing both the equation with the general form we get
a1 = a, b1 = b, c1 = -(a + b)/2,
a2 = 3, b2 = 5, c2 = -4
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
= x/(-4b + 5((a + b)/2)) = y/(-3((a + b)/2) + 4a) = 1/(5a - 3b)
= x/((5a - 3b)/2) = y/((5a - 3b)/2) = 1/(5a - 3b)
Now,
x/((5a - 3b)/2) = 1/(5a - 3b)
x = (5a - 3b)/(2(5a - 3b))
x = 1/2
and,
y/((5a - 3b)/2) = 1/(5a - 3b)
y = (5a - 3b)/(2(5a - 3b))
y = 1/2
**Hence, x = 1/2, y = 1/2
Question 23. 2 (ax β by) + a + 4b = 0 and 2 (bx + ay) + b β 4a = 0
**Solution:
Given that,
2 (ax β by) + (a + 4b) = 0
2 (bx + ay) + (b β 4a) = 0
On comparing both the equation with the general form we get
a1 = 2a, b1 = -2b, c1 = a + 4b,
a2 = 2b, b2 = 2a, c2 = b - 4a
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
= x/((-2b(a + 4b)) - (2a(b - 4a ))) = y/((2b(a + 4b)) - (2a(b - 4a))) = 1/(4a2 + 4b2)
= x/(-2b2 + 8ab - 2ab + 8a2) = y/(2ab + 8b2 - 2ab + 8a2) = 1/4(a2 + b2)
= x/-2(a2 + b2) = y/8(a2 + b2) = 1/4(a2 + b2)
So,
= x/-2(a2 + b2) = 1/4(a2 + b2)
x = -1/2
and,
= y/8(a2 + b2) = 1/4(a2 + b2)
y = 2
**Hence, x = -1/2 and y = 2
Question 24. 6 (ax + by) = 3a + 2b and 6 (bx β ay) = 3b β 2a
**Solution:
given that,
6 (ax + by) = 3a + 2b
6 (bx β ay) = 3b β 2a
6 (ax + by) -(3a + 2b)=0.... (1)
6 (bx β ay) -(3b β 2a) =0..... (2)
On comparing both the equation with the general form we get
a1 = 6a, b1 = 6b, c1 = -(3a - 2b),
a2 = 6b, b2 = 66a, c2 = -(3b - 2a)
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
= x/(-6b(3b - 2a) - 6a(3a - 2b)) = y/(-6b(3a - 2b) + 6a(3b - 2a)) = 1/(-36a2 - 36b2)
= x/(-18(a2 + b2)) = y/(-12(a2 + b2)) = 1/(-36(a2 + b2))
Therefore,
x/(-18(a2 + b2)) = 1/(-36(a2 + b2))
x = 1/2
and,
y/(-12(a2 + b2)) = 1/(-36(a2 + b2))
y = 1/3
**Hence, x = 1/2 and y = 1/3
Question 25. (a2/x) β (b2/y) = 0 and (a2b/x) β (b2a/y) = a + b, x, y β 0
**Solution:
Given that,
(a2/x) β (b2/y) = 0
(a2b/x) β (b2a/y) = a + b
Or
(a2b/x) β (b2a/y) - (a + b) = 0
On comparing both the equation with the general form we get
a1 = a2, b1 = -b2, c1 = 0,
a2 = a2b, b2 = b2a, c2 = -(a + b)
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
= (1/x)/(b2(a + b) - 0) = (1/y)/(0 + (a2(a + b))) = 1/(a3b2 - a2b3)
= (1/x)/(b2(a + b)) = (1/y)/(a2(a + b)) = 1/a2b2(a + b)
So,
= (1/x)/(b2(a + b)) = 1/a2b2(a + b)
x = a2
and,
= (1/y)/(a2(a + b)) = 1/a2b2(a + b)
y = b2
**Hence, x = a 2 **and y = b 2
Question 26. mx β ny = m2 + n2 and x + y = 2m
**Solution:
Given that,
mx β ny = m2 + n2
x + y = 2m
Or
mx β ny -(m2 + n2) = 0
x + y - 2m = 0
On comparing both the equation with the general form we get
a1 = m, b1 = -n, c1 = -(m2 + n2),
a2 = 1, b2 = 1, c2 = -2m
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
= x/(2mn + (m2 + n2)) = y/(-(m2 + n2) + 2m2) = 1/(m + n)
= x/(m + n)2 = y/(m2 - n2) = 1/(m + n)
Therefore,
x/(m + n)2 = 1/(m + n)
x = m + n
and,
y/(m2 - n2) = 1/(m + n)
y = m - n
**Hence, x = m + n, y = m - n
Question 27. (ax/b) - (by/a) = a + b and ax - by = 2ab
**Solution:
Given that,
(ax/b) - (by/a) = a + b
ax - by = 2ab
Or
(ax/b) - (by/a) - (a + b) = 0
ax - by - 2ab = 0
On comparing both the equation with the general form we get
a1 = a/b, b1 = -b/a, c1 = -(a + b),
a2 = a, b2 = b, c2 = -2ab
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
= x/b(b - a) = -y/a(-a + b) = 1/(b - a)
So,
x/b(b - a) = 1/(b - a)
x = b
and,
-y/a(-a + b) = 1/(b - a)
y = -a
**Hence, x = b, y = -a
Question 28. (b/a)x + (a/b)y = a2 + b2 and x + y = 2ab
**Solution:
Given that,
(b/a)x + (a/b)y = a2 + b2
x + y = 2ab
Or
(b/a)x + (a/b)y - (a2 + b2) = 0
x + y - 2ab = 0
On comparing both the equation with the general form we get
a1 = b/a, b1 = a/b, c1 = -(a2 + b2),
a2 = 1, b2 = 1, c2 = -2ab
Now by using cross multiplication we get
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
= x/(b2 - a2) = y/(-b2 + a2) = 1/((b2 - a2)/ab)
Therefore,
x/(b2 - a2) = 1/((b2 - a2)/ab)
x = ab
y/(-b2 + a2) = 1/((b2 - a2)/ab)
y = ab
**Hence, x = ab, y = ab
Conclusion
Mastering the concepts of solving pairs of linear equations in the two variables is crucial for the tackling more complex algebraic problems. Exercise 3.4 | Set 2 of RD Sharma's textbook offers valuable practice in the applying different solving techniques. By solving these exercises, students will enhance their problem-solving skills and build a strong foundation in algebra.