Basic Engineering Mathematics Problems And Solutions - Tech Projects/Documentations
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Basic Engineering Mathematics Problems And Solutions

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Problem 1: Solve the following equations for x and y, (a) by substitution and (b) by elimination.

x+2y= -1                  (1)

4x-3y=18                 (2)

Solution:

(a) By substitution

From equation (1):

x=-1 -2y

Substituting this expression for x into equation (2) gives

4\left(-1-2y\right)-3y=18

This is now a simple equation in y. removing the bracket gives

−4 − 8y − 3y = 18

−11y = 18 + 4 = 22

y=\frac{22}{-11}=-2

Substituting y = −2 into equation (1) gives

x+2\left(-2\right)=-1

x − 4 = −1

x = −1 + 4 = 3

Thus, x = 3 and y = −2 is the solution to the simultaneous equations. Check: in equation (2), since x = 3 and y = −2

Check: in equation (2), since x = 3 and y = −2

LHS = 4(3) − 3(−2) = 12 + 6 = 18 = RHS

(b) By elimination

+ 2y = −1          (1)

4x − 3y = 18      (2)

If equation (1) is multiplied throughout by 4, the coefficient of x will be the same as in equation (2), giving

4x + 8y = −4        (3)

Subtracting equation (3) from equation (2) gives

4x − 3y = 18    (2)

\underline{4x + 8y = −4} (3)

0 − 11y = 22

Hence, y=\frac{22}{-11}=-2

Substituting y = −2 into either equation (1) or equation (2) will give x = 3 as in method (a). The solution x = 3, y = −2 is the only pair of values that satisfies both of the original equations.

Problem 2: Use an elimination method to solve the following simultaneous equations:

3x + 4y = 5          (1)

2x − 5y = −12      (2)

Solution: If equation (1) is multiplied throughout by 2 and equation (2) by 3, the coefficient of x will be the same in the newly formed equations. Thus

2 × equation (1) gives 6x + 8y = 10         (3)

3 × equation (2) gives 6x − 15y = −36     (4)

Equation (3) – equation (4) gives

0 + 23y = 46

y=\frac{46}{23}=2

(Note+8y − −15y = 8y + 15y = 23y and 10− −36 = 10 + 36 = 46.)

Substituting y = 2 in equation (1) gives

3x + 4(2) = 5

from which

3x = 5 − 8 = −3

and

x = −1

Checking, by substituting x = −1 and y = 2 in equation (2), gives

LHS = 2(−1) − 5(2) = −2 − 10 = −12 = RHS

Hence, x = −1 and y = 2 is the solution of the simultaneous equations.

The elimination method is the most common method of solving simultaneous equations.

Problem 3: solve

\frac{x}{8}+\frac{5}{2}=y (1)

13 -\frac{y}{3}=3x (2)

Solution: Whenever fractions are involved in simultaneous equations it is often easier to firstly remove them. Thus, multiplying equation (1) by 8 gives.

8\left(\begin{array}{c}\frac{x}{8}\end{array}\right)+8\left(\begin{array}{c}\frac{5}{2}\end{array}\right)=8y

x + 20 = 8y      (3)

Multiplying equation (2) by 3 gives

39 − y = 9x      (4)

Rearranging equations (3) and (4) gives

x − 8y = −20     (5)

9x + y = 39     (6)

Multiplying equation (6) by 8 gives

72x + 8y = 312    (7)

Adding equations (5) and (7) gives

73x + 0 = 292

Substituting x = 4 into equation (5) gives

x=\frac{292}{73}=4

4 − 8y = −20

4 + 20 = 8y

24 = 8y

y=\frac{24}{8}=3

Checking, substituting x = 4 and y = 3 in the original equations, gives

(1) LHS=\frac{4}{8}+\frac{5}{2}+\frac{1}{2}+2\frac{1}{2}=3=y=RHS

(2) LHS=13\frac{3}{3}=13-1=12

RHS = 3x = 3(4) = 12

Hence, the solution is x = 4,y = 3.

Problem 4: Determine the value of 3e−1, correct to 4 decimal places, using the power series for ex

Solution: Substituting x = −1 in the power series

e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}….

e^{-1}=1+(-1)+\frac{(-1)^{2}}{2!}+\frac{(-1)^{3}}{3!}+\frac{(-1)^{4}}{4!}….

= 1 − 1 + 0.5 − 0.166667 + 0.041667− 0.008333 + 0.001389 − 0.000198 +……

= 0.367858

0.367858 correct to 6 decimal places

Hence, 3e−1 = (3)(0.367858) = 1.1036, correct to 4 decimal places

Problem 5: Expand ex (x2 − 1) as far as the term in x5

Solution: The power series for ex is

e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}….

Hence

ex (x2 − 1)

=\left(\begin{array}{c}1+ x+ \frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}….\end{array}\right)(x^{2}-1)

=\left(\begin{array}{c}x^{2}+ x^{3}+ \frac{x^{4}}{2!}+\frac{x^{5}}{3!}+….\end{array}\right)-\left(\begin{array}{c}1+ x+ \frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}….\end{array}\right)

Grouping like terms gives

ex (x2 − 1)

=-1-x+\left(\begin{array}{c}x^{2}- \frac{x^{2}}{2!}\end{array}\right)+\left(\begin{array}{c}x^{3}- \frac{x^{3}}{3!}\end{array}\right)+\left(\begin{array}{c}x^{4}- \frac{x^{4}}{4!}\end{array}\right)+\left(\begin{array}{c}x^{5}- \frac{x^{5}}{5!}\end{array}\right)…

=-1-x+\frac{1}{2}x^{2}+\frac{5}{6}x^{3}+\frac{11}{24}x^{4}+\frac{19}{120}x^{5}

when expanded as far as the term in x5.

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