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# Solving Linear Equations

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Thus, so long as there is possibility to determine the two points which a line passes through, then it would be possible to generate the equation of the line (Rattan and Klingbeil 24).

For instance, the following two points have been provided:

(-3, 3) and (2, -7)

X values = 2 and -3

Y values = -7 and 3

In that case: Change in y versus Change in x =

Therefore, the calculation is as shown below:

Then, using the point slope formula further calculations would be as follows:

First pick -3,3: Y – Y1 = m (X – X1)

Y – 3 = -2 (X – (-3)

Y – 3 = -2x – (6)

Y = -2x – (6) + (3)

Y = -2X -3

Question 2

Substitution method can be used to solve linear systems; thus requires substituting the one y-value with the other (Rattan and Klingbeil 34). In this case, where it is determinable that y = y. then substitution method would be applicable. System B, therefore, would be suitable for the substitution method since y = y. scenario is available.

Question 3

By substitution method for System B.

y =  + 8

2x – 3y = 12

There, solving for y in equation 2 is as shown below.

2x – 3(-2/3x + 8) = 12

2x + 2x – 24 = 12

4x -24 = 12

4x = 12 + 24

4x = 36

x = 9

Hence;

y = -2/3(9) + 8

y = -6 +8

y = 2

By elimination method for System A:

3x – 2y = 7

-4x – 6y = -5

The first step is to “cancel out” the y term by adding the two equations as shown below:

3x – 2y = 7

-4x – 6y = -5

The first step is to multiply one of the equation with a value to enable it cancel out: for instance, multiple the first equation by -3 to get the following results:

-9x + 6y = -21

-4x – 6y = -5

When two equations have been added shall give the result:

-13x = -26

x = 2

Thus, knowing the value of x the choice is to pick equation 2 and substitute the value of x to get that of y as shown below.

-4(2) – 6y = -5

-8 – 6y = -5

-6y = 3

y = -0.5

The solution is x, y = (2, -0.5)

Question 4

The provided systems have 3 variables and can be solved as shown next.

x – 2y –z = 8

2x – 3y + z = 23

4x – 5y + 5z = 53

Consideration is to undertake elimination by addition by taking any two of the questions as shown next.

x – 2y –z = 8

2x – 3y + z = 23

= 3x – 6y = 31

Then take the third equation and subtract from the second equation to get extra equation with two variables:

2x – 3y + z = 23 …multiply by 5 to eliminate z

4x – 5y + 5z = 53

5x – 15y + 5z = 115

4x – 5y +5z = 53

= x – 10y = 62

The system of equations with two variables would look as follows:

3x – 6y = 31

x – 10y = 62 … Multiply by -3

3x – 6y = 31

-3x – 30y = -186

- 36y = -155

y = 155/36

Then, taking 3x – 6y = 31 the value of y as 155/36 has been plugged into the equation to solve for x as shown below.

3x – 6(155/36) = 31

3x – 28.83 = 31

3x = 31 + 28.83

3x = 56.83

x = 57/3

Going back to the first original system the value of z could be solved as follows:

x – 2y –z = 8

57/3 – (2) * (155/36) – z = 8

19 – 8.6 = 10.4

z = 8 – 10.4

z = -2.4

The solution is x, y, z = (19, 155/36, 2.4)

Question 5

A system of equation can have three solutions namely: (a) no solution (b) a unique solution and (c) infinitely many solutions (Gel’fand & Shilov 39). If a system has inconsistency, then it means there can be no solutions but consistency means there is unique solution. Infinitely many solutions assert a level of consistency and the number of variables exceed the number of nonzero outcomes (Gel’fand & Shilov 43).

Works Cited

Gel’fand, Ian & and Shilov, Gabriel. Theory of Differential Equations. Elsevier Science

Publishers, 2014. Print.

Rattan, S. Kuldip and Klingbeil, W. Nathan. Introductory Mathematics for Engineering

Applications. John Wiley & Sons, 2014. Print

September 18, 2023
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