RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3:
In Chapter 3, Exercise 3.3 of RD Sharma's Class 10 Maths, students explore solving pairs of linear equations in two variables using graphical and algebraic methods. This exercise delves into three main algebraic methods: substitution, elimination, and cross-multiplication.
Each method is applied to find the point of intersection, or the solution, of the two equations, which represents the values of the variables satisfying both equations simultaneously. Exercise 3.3 also includes practical problems, enhancing students' problem-solving skills and helping them understand the real-life application of linear equations in various contexts.
RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Overview
Exercise 3.3 in RD Sharma’s Class 10 Maths on Pair of Linear Equations in Two Variables is crucial for developing a strong foundation in algebra. Mastering methods like substitution, elimination, and cross-multiplication empowers students to solve real-world problems involving two variables, such as calculating rates, distances, and finances. These techniques are fundamental for higher mathematics, physics, and economics, were systems of equations frequently model complex relationships.
Understanding these concepts enhances logical thinking and problem-solving skills, essential for competitive exams and advanced studies. This exercise builds confidence in handling equations, laying groundwork for more advanced algebraic and analytical concepts.
RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 PDF
Below, we have provided a PDF of RD Sharma Solutions for Class 10 Maths Chapter 3, Exercise 3.3, focused on Pair of Linear Equations in Two Variables. This resource includes step-by-step solutions to each question, helping students understand methods like substitution, elimination, and cross-multiplication. These solutions simplify complex problems, offering a clear approach to mastering linear equations in two variables.
RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 PDF
RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Pair of Linear Equations in Two Variables
Below is the RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Pair of Linear Equations in Two Variables -
Solve the following system of equations:
1. 11x + 15y + 23 = 0
7x – 2y – 20 = 0
Solution:
The given pair of equations are
11x +15y + 23 = 0 …………………………. (i)
7x – 2y – 20 = 0 …………………………….. (ii)
From (ii),
2y = 7x – 20
⇒ y = (7x −20)/2 ……………………………… (iii)
Now, substituting y in equation (i), we get
⇒ 11x + 15((7x−20)/2) + 23 = 0
⇒ 11x + (105x − 300)/2 + 23 = 0
⇒ (22x + 105x – 300 + 46) = 0
⇒ 127x – 254 = 0
⇒ x = 2
Next, putting the value of x in equation (iii), we get,
⇒ y = (7(2) − 20)/2
∴ y= -3
Thus, the value of x and y is found to be 2 and -3, respectively.
2. 3x – 7y + 10 = 0
y – 2x – 3 = 0
Solution:
The given pair of equations are
3x – 7y + 10 = 0 …………………………. (i)
y – 2x – 3 = 0 ……………………………….. (ii)
From (ii),
y – 2x – 3 = 0
y = 2x+3 ……………………………… (iii)
Now, substituting y in equation (i), we get
⇒ 3x – 7(2x+3) + 10 = 0
⇒ 3x – 14x – 21 + 10 = 0
⇒ -11x = 11
⇒ x = -1
Next, putting the value of x in equation (iii), we get
⇒ y = 2(-1) + 3
∴ y= 1
Thus, the value of x and y is found to be -1 and 1, respectively.
3. 0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Solution:
The given pair of equations are
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Let’s, multiply LHS and RHS by 10 to make the coefficients an integer.
4x + 3y = 17 ……………………….. (i)
7x – 2y = 8 …………………………… (ii)
From (ii),
7x – 2y = 8
x = (8 + 2y)/7……………………………… (iii)
Now, substituting x in equation (i), we get
⇒ 4[(8 + 2y)/7] + 3y = 17
⇒ 32 + 8y + 21y = (17 x 7)
⇒ 29y = 87
⇒ y = 3
Next, putting the value of y in equation (iii), we get
⇒ x = (8 + 2(3))/ 7
⇒ x = 14/7
∴ x = 2
Thus, the value of x and y is found to be 2 and 3, respectively.
4. x/2 + y = 0.8
7/(x+ y/2) = 10
Solution:
The given pair of equations are
x/2 + y = 0.8 ⇒ x + 2y = 1.6…… (a)
7/(x + y/2) = 10 ⇒7 = 10(x + y/2) ⇒7 = 10x + 5y
Let’s, multiply the LHS and RHS of equation (a) by 10 for easy calculation.
So, we finally get
10x + 20y = 16 ……………………….. (i) And,
10x + 5y = 7 …………………………… (ii)
Now, subtracting two equations, we get
⇒ (i) – (ii)
15y = 9
⇒ y = 3/5
Next, putting the value of y in the equation (i), we get
x = [16 − 20(3/5)]/10
⇒ (16 – 12)/10 = 4/10
∴ x = 2/5
Thus, the value of x and y obtained are 2/5 and 3/5, respectively.
5. 7(y + 3) – 2(x + 2) = 14
4(y – 2) + 3(x – 3) = 2
Solution:
The given pair of equations are
7(y+3) – 2(x+2) = 14…………………………. (i)
4(y-2) + 3(x-3) = 2……………………………….. (ii)
From (i), we get
7y + 21 – 2x – 4 = 14
7y = 14 + 4 – 21 + 2x
⇒ y = (2x – 3)/7
From (ii), we get
4y – 8 + 3x – 9 = 2
4y + 3x – 17 – 2 = 0
⇒ 4y + 3x – 19 = 0 …………….. (iii)
Now, substituting y in equation (iii),
4[(2x − 3)/7] + 3x – 19=0
8x – 12 + 21x – (19 x 17) = 0 [after taking LCM]
29x = 145
⇒ x = 5
Now, putting the value of x and in equation (ii),
4(y-2) + 3(5-3) = 2
⇒ 4y -8 + 6 = 2
⇒ 4y = 4
∴ y = 1
Thus, the value of x and y obtained are 5 and 1, respectively.
6. x/7 + y/3 = 5
x/2 – y/9 = 6
Solution:
The given pair of equations are
x/7 + y/3 = 5…………………………. (i)
x/2 – y/9 = 6………………………………..(ii)
From (i), we get
x/7 + y/3 = 5
⇒3x + 7y = (5×21) [After taking LCM]
⇒ 3x =105 – 7y
⇒ x = (105 – 7y)/3……. (iv)
From (ii), we get
x/2 – y/9 = 6
⇒ 9x – 2y = 108 ……………………… (iii) [After taking LCM]
Now, substituting x in equation (iii), we get
9[(105 − 7y)/3] – 2y = 108
⇒ 945 – 63y – 6y = 324 [After taking LCM]
⇒ 945 – 324 = 69y
⇒ 69y = 621
⇒ y = 9
Now, putting the value of y in equation (iv),
x = (105 − 7(9))/3
⇒ x = (105 − 63)/3 = 42/3
∴ x = 14
Thus, the value of x and y obtained are 14 and 9, respectively.
7. x/3 + y/4 = 11
5x/6 − y/3 = −7
Solution:
The given pair of equations are
x/3 + y/4 = 11…………………………. (i)
5x/6 − y/3 = −7……………………………….. (ii)
From (i), we get
x/3 + y/4 = 11
⇒4x + 3y = (11×12) [After taking LCM]
⇒ 4x =132 – 3y
⇒ x = (132 – 3y)/4……. (iv)
From (ii), we get
5x/6 − y/3 = −7
⇒ 5x – 2y = -42 ……………………… (iii) [After taking LCM]
Now, substituting x in equation (iii), we get
5[(132 − 3y)/4] – 2y = -42
⇒ 660 – 15y – 8y = -42 x 4 [After taking LCM]
⇒ 660 + 168 = 23y
⇒ 23y = 828
⇒ y = 36
Now, putting the value of y in equation (iv),
x = (132 – 3(36))/4
⇒ x = (132 − 108)/4 = 24/4
∴ x = 6
Thus, the value of x and y obtained are 6 and 36, respectively.
8. 4/x + 3y = 8
6/x −4y = −5
Solution:
Taking 1/x = u
Then, the two equations become
4u + 3y = 8…………………… (i)
6u – 4y = -5……………………. (ii)
From (i), we get
4u = 8 – 3y
⇒ u = (8 − 3y)/4 …….. (iii)
Substituting u in (ii),
[6(8 − 3y)/4] – 4y = -5
⇒ [3(8−3y)/2] − 4y = −5
⇒ 24 − 9y −8y = −5 x 2 [After taking LCM]
⇒ 24 – 17y = -10
⇒ -17y =- 34
⇒ y = 2
Putting y=2 in (iii), we get
u = (8 − 3(2))/4
⇒ u = (8 − 6)/4
⇒ u = 2/4 = 1/2
⇒ x = 1/u = 2
∴ x = 2
So, the solution of the pair of equations given is x=2 and y =2.
9. x + y/2 = 4
2y + x/3 = 5
Solution:
The given pair of equations are
x + y/2 = 4 ……………………. (i)
2y + x/3 = 5……………………. (ii)
From (i), we get
x + y/2 = 4
⇒ 2x + y = 8 [After taking LCM]
⇒ y = 8 – 2x …..(iv)
From (ii), we get
x + 6y = 15 ……………… (iii) [After taking LCM]
Substituting y in (iii), we get
x + 6(8 – 2x) = 15
⇒ x + 48 – 12x = 15
⇒ -11x = 15 – 48
⇒ -11x = -33
⇒ x = 3
Putting x = 3 in (iv), we get
y = 8 – (2×3)
∴ y = 8 – 6 = 2
Hence, the solutions of the given system of the equation are x = 3 and y = 2, respectively.
10. x + 2y = 3/2
2x + y = 3/2
Solution:
The given pair of equations are
x + 2y = 3/2 …………………. (i)
2x + y = 3/2…………………… (ii)
Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1, respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.
Multiplying equation (i)x1 and (ii)x2 ⇒
x + 2y = 3/2 ………………………. (iii)
4x + 2y = 3 ……………………………………………………. (iv)
Subtracting equation (iii) from (iv),
(4x – x) + (2y-2y) = 3x = 3 – (3/2)
⇒ 3x = 3/2
⇒ x = 1/2
Putting x = 1/2 in equation (iv),
4(1/2) + 2y = 3
⇒ 2 + 2y = 3
∴ y= 1/2
The solution of the system of equations is x = 1/2 and y = 1/2
11. √2x – √3y = 0
√3x − √8y = 0
Solution:
The given pair of equations are
√2x – √3y = 0……………………….. (i)
√3x − √8y = 0……………………….. (ii)
From equation (i),
x = √(3/2)y ……………..(iii)
Substituting this value in equation (ii), we obtain
√3x − √8y = 0
⇒ √3(√(3/2)y) − √8y = 0
⇒ (3/√2)y – √8y = 0
⇒ 3y – 4y = 0
⇒ y = 0
Now, substituting y in equation (iii), we obtain
⇒ x=0
Thus, the value of x and y obtained are 0 and 0, respectively.
12. 3x – (y + 7)/11 + 2 = 10
2y + (x + 11)/7 = 10
Solution:
The given pair of equations are
3x – (y + 7)/11 + 2 = 10……………….. (i)
2y + (x + 11)/7 = 10…………………….. (ii)
From equation (i),
33x – y – 7 + 22 = (10 x 11) [After taking LCM]
⇒ 33x – y + 15 = 110
⇒ 33x + 15 – 110 = y
⇒ y = 33x – 95………. (iv)
From equation (ii),
14 + x + 11 = (10 x 7) [After taking LCM]
⇒ 14y + x + 11 = 70
⇒ 14y + x = 70 – 11
⇒ 14y + x = 59 …………………….. (iii)
Substituting (iv) in (iii), we get
14 (33x – 95) + x = 59
⇒ 462x – 1330 + x = 59
⇒ 463x = 1389
⇒ x = 3
Putting x = 3 in (iii), we get
⇒ y = 33(3) – 95
∴ y= 4
The solution for the given pair of equations is x = 3 and y = 4, respectively.
13. 2x – (3/y) = 9
3x + (7/y) = 2 ,y ≠ 0
Solution:
The given pair of equations are
2x – (3/y) = 9……………………………. (i)
3x + (7/y) = 2…………………………… (ii)
Substituting 1/y = u, the above equations become
2x – 3u = 9 ………………………..(iii)
3x + 7u = 2………………………..(iv)
From (iii)
2x = 9 + 3u
⇒ x = (9+3u)/2
Substituting the value of x from above in equation (iv), we get
3[(9+3u)/2] + 7u = 2
⇒ 27 + 9u + 14u = (2 x 2)
⇒ 27 + 23u = 4
⇒ 23u = -23
⇒ u = -1
So, y = 1/u = -1
And putting u = -1 in x = (9 + 3u)/2, we get
⇒ x = [9 + 3(−1)]/2 = 6/2
∴ x = 3
The solutions of the pair of equations given are y = 3 and x = -1, respectively.
14. 0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
Solution:
The given pair of equations are
0.5x + 0.7y = 0.74……………………… (i)
0.3x – 0.5y = 0.5 ………………………….. (ii)
Now, let’s multiply LHS and RHS by 100 for both (i) and (ii) to make integral coefficients and constants.
(i) x100 ⇒
50x +70y = 74 ……………………….. (iii)
(ii) x100 ⇒
30x + 50y = 50 …………………………… (iv)
From (iii),
50x = 74 – 70y
x = (74−70y)/ 50 ……………………………… (v)
Now, substituting x in equation (iv), we get
30[(74−70y)/ 50] + 50y = 50
⇒ 222 – 210y + 250y = 250 [After taking LCM]
⇒ 40y = 28
⇒ y = 0.7
Now, by putting the value of y in the equation (v), we get
⇒ x = [74 − 70(0.7)]/ 50=0
⇒ x =25/ 50 = 1/2
∴ x = 0.5
Thus, the values of x and y so obtained are 0.5 and 0.7, respectively.
15. 1/(7x) + 1/(6y) = 3
1/(2x) – 1/(3y) = 5
Solution:
The given pair of equations are
1/(7x) + 1/(6y) = 3………………………….. (i)
1/(2x) – 1/(3y) = 5……………………………. (ii)
Multiplying (ii) by 1/2, we get
1/(4x) – 1/(6y) = 52……………………………. (iii)
Now, solving equations (i) and (iii),
1/(7x) + 1/(6y) = 3………………………….. (i)
1/(4x) – 1/(6y) = 5/2……………………………. (iii)
Adding (i) + (iii), we get
1/x(1/7 + 1/4 ) = 3 + 5/2
⇒ 1/x(11/28) = 11/2
⇒ x = 1/14
Using x =1/ 14, we get, in (i)
1/[7(1/14)] + 1/(6y) = 3
⇒ 2 + 1/(6y)=3
⇒ 1/(6y) = 1
⇒ y = 1/6
The solution for the given pair of equations is x=1/14 and y=1/6, respectively.
16. 1/(2x) + 1/(3y) = 2
1/(3x) + 1/(2y) = 13/6
Solution:
Let 1/x = u and 1/y = v,
So, the given equations become
u/2 + v/3 = 2 ………………(i)
u/3 + v/2 = 13/6 ……………(ii)
From (i), we get
u/2 + v/3 = 2
⇒ 3u + 2v = 12
⇒ u = (12 – 2v)/3 ………….(iii)
Using (iii) in (ii),
[(12 – 2v)/3]/3 + v/2 = 13/6
⇒ (12 – 2v)/9 + v/2 = 13/6
⇒ 24 – 4v + 9v = (13/6) x 18 [after taking LCM]
⇒ 24 + 5v = 39
⇒ 5v = 15
⇒ v = 3
Substituting v in (iii),
u = (12 – 2(3))/3
⇒ u = 2
Thus, x = 1/u ⇒ x = 1/2 and
y = 1/v ⇒ y = 1/3
The solution for the given pair of equations is x = 1/2 and y = 1/3, respectively.
17. 15/u + 2/v = 17
1/u + 1/v = 36/5
Solution:
Let 1/x = u and 1/y = v
So, the given equations become
15x + 2y = 17 ………………………….. (i)
x + y = 36/5………………………. (ii)
From equation (i), we get
2y = 17 – 15x
=y = (17 − 15x)/ 2 …………………. (iii)
Substituting (iii) in equation (ii), we get
= x + (17 − 15x)/ 2 = 36/5
2x + 17 – 15x = (36 x 2)/ 5 [after taking LCM]
-13x = 72/5 – 17
= -13x = -13/5
⇒ x = 1/5
⇒ u = 1/x = 5
Putting x = 1/5in equation (ii), we get
1/5 + y = 36/5
⇒ y = 7
⇒ v = 1/y = 1/7
The solutions of the pair of equations given are u = 5 and v = 1/7, respectively.
18. 3/x – 1/y = −9
2/x + 3/y = 5
Solution:
Let 1/x = u and 1/y = v
So, the given equations become
3u – v = -9…………………..(i)
2u + 3v = 5 ……………………….(ii)
Multiplying equation (i) x 3 and (ii) x 1, we get
9u – 3v = -27 ………………………….. (iii)
2u + 3v = 5 ……………………………… (iv)
Adding equations (iii) and (iv), we get
9u + 2u – 3v + 3v = -27 + 5
⇒ 11u = -22
⇒ u = -2
Now putting u =-2 in equation (iv), we get
2(-2) + 3v = 5
⇒ 3v = 9
⇒ v = 3
Hence, x = 1/u = −1/2 and,
y = 1/v = 1/3
19. 2/x + 5/y = 1
60/x + 40/y = 19
Solution:
Let 1/x = u and 1/y = v
So, the given equations become
2u + 5v = 1…………………..(i)
60u + 40v = 19 ……………………….(ii)
Multiplying equation (i) x 8 and (ii) x 1, we get
16u + 40v = 8 ………………………….. (iii)
60u + 40v = 19 ……………………………… (iv)
Subtracting equation (iii) from (iv), we get
60u – 16u + 40v – 40v = 19 – 8
⇒ 44u = 11
⇒ u = 1/4
Now putting u = 1/4 in equation (iv), we get
60(1/4) + 40v = 19
⇒ 15 + 40v = 19
⇒ v = 4/ 40 = 1/10
Hence, x = 1/u = 4 and
y = 1/v = 10
20. 1/(5x) + 1/(6y) = 12
1/(3x) – 3/(7y) = 8
Solution:
Let 1/x = u and 1/y = v
So, the given equations become
u/5 + v/6 = 12…………………..(i)
u/3 – 3v/7 = 8……………………….(ii)
Taking LCM for both equations, we get
6u + 5v = 360………. (iii)
7u – 9v = 168……….. (iv)
Subtracting (iii) from (iv),
7u – 9v – (6u + 5v) = 168 – 360
⇒ u – 14v = -192
⇒ u = (14v – 192)………. (v)
Using (v) in equation (iii), we get
6(14v – 192) + 5v = 360
⇒ 84v -1152 + 5v = 360
⇒ 89v = 1512
⇒ v = 1512/89
⇒ y = 1/v = 89/1512
Now, substituting v in equation (v), we find u.
u = 14 x (1512/89) – 192
⇒ u = 4080/89
⇒ x = 1/u = 89/ 4080
Hence, the solution for the given system of equations is x = 89/4080 and y = 89/ 1512.
21. 4/x + 3y = 14
3/x – 4y = 23
Solution:
Taking 1/x = u, the given equations become
4u + 3y = 14…………………….. (i)
3u – 4y = 23…………………….. (ii)
Adding (i) and (ii), we get
4u + 3y + 3u – 4y = 14 + 23
⇒ 7u – y = 37
⇒ y = 7u – 37……………………… (iii)
Using (iii) in (i),
4u + 3(7u – 37) = 14
⇒ 4u + 21u – 111 = 14
⇒ 25u = 125
⇒ u = 5
⇒ x = 1/u = 1/5
Putting u= 5 in (iii), we find y
y = 7(5) – 37
⇒ y = -2
Hence, the solution for the given system of equations is x = 1/5 and y = -2.
22.
4/x + 5y = 7
3/x + 4y = 5
Solution:
Taking 1/x = u, the given equations become
4u + 5y = 7…………………….. (i)
3u + 4y = 5…………………….. (ii)
Subtracting (ii) from (i), we get
4u + 5y – (3u + 4y) = 7 – 5
⇒ u + y = 2
⇒ u = 2 – y……………………… (iii)
Using (iii) in (i),
4(2 – y) + 5y = 7
⇒ 8 – 4y + 5y = 7
⇒ y = -1
Putting y = -1 in (iii), we find u.
u = 2 – (-1)
⇒ u = 3
⇒ x = 1/u = 1/3
Hence, the solution for the given system of equations is x = 1/3 and y = -1.
23.
2/x + 3/y = 13
5/x – 4/y = -2
Solution:
Let 1/x = u and 1/y = v
So, the given equations become
2u + 3v = 13………………….. (i)
5u – 4v = -2 ………………………. (ii)
Adding equations (i) and (ii), we get
2u + 3v + 5u – 4v = 13 – 2
⇒ 7u – v = 11
⇒ v = 7u – 11…….. (iii)
Using (iii) in (i), we get
2u + 3(7u – 11) = 13
⇒ 2u + 21u – 33 = 13
⇒ 23u = 46
⇒ u = 2
Substituting u = 2 in (iii), we find v.
v = 7(2) – 11
⇒ v = 3
Hence, x = 1/u = 1/2 and,
y = 1/v = 1/3
24. 2/x + 3/y = 2
4/x – 9/y = -1
Solution:
Let 1/√x = u and 1/√y = v,
So, the given equations become
2u + 3v = 2………………….. (i)
4u – 9v = -1 ………………………. (ii)
Multiplying (ii) by 3 and
Adding equations (i) and (ii)x3, we get
6u + 9v + 4u – 9v = 6 – 1
⇒ 10u = 5
⇒ u = 1/2
Substituting u = 1/2 in (i), we find v
2(1/2) + 3v = 2
⇒ 3v = 2 – 1
⇒ v = 1/3
Since, 1/√x = u we get x = 1/u
2
⇒ x = 1/(1/2)
2
= 4
And,
1/√y = v y = 1/v
2
⇒ y = 1/(1/3)
2
= 9
Hence, the solution is x = 4 and y = 9.
25. (x + y)/xy = 2
(x – y)/xy = 6
Solution:
The given pair of equations are
(x + y)/xy = 2 ⇒ 1/y + 1/x = 2……. (i)
(x – y)/xy = 6 ⇒ 1/y – 1/x = 6………(ii)
Let 1/x = u and 1/y = v, so the equations (i) and (ii) become
v + u = 2……. (iii)
v – u = 6……..(iv)
Adding (iii) and (iv), we get
2v = 8
⇒ v = 4
⇒ y = 1/v = 1/4
Substituting v = 4 in (iii) to find x,
4 + u = 2
⇒ u = -2
⇒ x = 1/u = -1/2
Hence, the solution is x = -1/2 and y = 1/4.
26. 2/x + 3/y = 9/xy
4/x + 9/y = 21/xy
Solution:
Taking LCM for both the given equations, we have
(2y + 3x)/ xy = 9/xy ⇒ 3x + 2y = 9………. (i)
(4y + 9x)/ xy = 21/xy ⇒ 9x + 4y = 21………(ii)
Performing (ii) – (i)x2⇒
9x + 4y – 2(3x + 2y) = 21 – (9×2)
⇒ 3x = 3
⇒ x = 1
Using x = 1 in (i), we find y
3(1) + 2y = 9
⇒ y = 6/2
⇒ y = 3
Thus, the solution for the given set of equations is x = 1 and y = 3.
Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3
Solving RD Sharma’s Class 10 Maths Chapter 3, Exercise 3.3 on Pair of Linear Equations in Two Variables offers multiple benefits:
Foundation in Algebra
: Builds a strong base for understanding algebraic methods, essential for higher mathematics.
Problem-Solving Skills
: Develops critical thinking by using substitution, elimination, and cross-multiplication techniques.
Real-World Application
: Helps students model and solve real-life problems involving relationships between two variables, such as financial calculations, distances, and rates.
Competitive Exam Preparation
: Enhances accuracy and speed, useful for exams.
Logical Thinking
: Strengthens analytical abilities, essential for advanced studies in science and engineering.
