- April 3, 2012
- Posted by: essay
- Category: Sample essay papers

In general, a system of equations is a set of equations containing multiple variables; the solution of such set of equations is the specification of values of all variables that simultaneously satisfies every equation in the set (Larson & Hostetler & Kelly 2007). A particular form of systems of equations is the system of linear equations. The equation which can be expressed using only linear operations with numbers and variables is called linear equation (Larson & Hostetler & Kelly 2007). A set of two or more such equations is called system of linear equations.

The simplest method of solving systems of equations is elimination, when one variable is expressed from one equation, and substituted in the other equation by appropriate expression. It is valid to divide or multiply any equation of the system to a given number, thus obtaining equivalent equation. Also, it is possible to add or subtract equations in the system obtaining equivalent equations, or replace an equation from the set by a linear combination of current equations. The latter idea is the basis for Gaussian elimination used to solve more complex system of linear equations (Dahlquist & Bjorck 2004). The methods of solving will be used according to the structure of the system.

*a.Β X + Y= 10 ,Β 3X + Y = 12;*

Let us subtract equation 1 from equation 2, and replace equation 2.

X+Y=10; 2X = 2;

X+Y=10; X=1;

Y = 10 X = 9;

X=1, Y=9 – solution

*b.Β 2X + 5Y = 19 ,Β 3X + 3Y = 15*

Let us multiply equation 2 by 2/3.

2X + 5Y = 19, 2X + 2Y = 10

Let us subtract equation 2 from equation 1, and replace equation 1; divide equation 2 by 2.

3Y = 9, X + Y = 5

Y = 3; X = 5 Y = 5-3 = 2

X=2, Y=3 – solution

*c.Β 4X +Β Y = 22 ,Β Β Β 2X + 3Y = 16*

Let us express Y from equation 1, and replace it by this expression in equation 2.

Y = 22 4X, 2X+3(22-4X)=16;

Transforming equation 2, we obtain:

2X + 66 12X = 16;

-10X = -50; X = 5;

Y = 22 4X = 22 20 = 2

X = 5, Y = 2 – solution

*d.Β 12X + Y = 174 ,Β 8X – 2Y = 36*

Let us express Y from equation 1, and replace it by this expression in equation 2.

Y = 174 12X, 8X 2 (174 12X) = 36

Transforming equation 2, we obtain:

8X 348 + 24X = 36;

32X = 384; X = 12;

Y = 174 144 = 30.

X = 12, Y = 30 solution

2. Suppose Bob owns 2,000 shares of Company X and 10,000 shares of Company Y.Β The total value of Bob’s holdings of these two companies is $372,000.

Suppose Frank owns 8,000 shares of Company X and 6,000 shares of Company Y.Β The total value of Frank’s holdings of these two companies is $400,000.

*a. Write equations for Bob and Frank’s holdings.Β Use the variables X and Y to represent the values of shares of Company X and Company Y.*

Let us consider X value of share of company X, and Y value of share for company Y. Then, the total value of Bob’s holding may be expressed by the following equation:

2000X + 10000Y = 372000. Total value of Frank’s holdings is expressed as 8000X + 6000Y = 400000. Thus, we obtained a system of linear equations.

*b. Solve for the value of a share of Company X and Company Y.Β Show your work so you can get partial credit even if your final answer is wrong.*

2000X + 10000Y = 372000, 8000X + 6000Y = 400000.

Let us divide each equation by 2000, thus obtaining equivalent system of equations.

X + 5Y = 186, 4X + 3Y = 200;

Let us express X from equation 1, and replace X by appropriate expression in equation 2:

X = 186 5Y, 4 (186 5Y) + 3Y = 200.

Let us transform equation 2:

744 20Y + 3Y = 200;

-17 Y = -544;

Y = 32; X = 186 160 = 26.

Thus, value of share of company X is $32, and value of share of company Y is $26.

*3. Solve for X, Y, and Z in the following systems of three equations:*

*a. X + 2Y + Z = 22, X + YΒ = 15, 3X + Y + Z =Β 37.*

Let us express Y from equation 2: Y = 15 X; using this expression, let us express Z from equation 1:

X + 2 (15 X) + Z = 22;

X + 30 2X + Z = 22;

-X + Z = -8; Z = X-8;

Using expressions for Y and Z in equation 3, it is possible to get the solution.

3X + (15 X) + (X-8) = 37;

3X = 30; X=10;

Y = 15 X = 5; Z = X -8 = 2;

X = 10, Y = 5, Z = 2 solution.

*b. 10X + Y + Z = 603, 8X + 2Y + ZΒ = 603, 20X – 10Y – 2Z = -6*

Let us divide equation 3 by 2.

10X + Y + Z = 603, 8X + 2Y + ZΒ = 603, 10X – 5Y – Z = -3;

Then, it is possible to subtractΒ equation 2 from equation 1, and replace equation 1:

2X Y = 0; 8X + 2Y + ZΒ = 603, 10X – 5Y – Z = -3;

Let us express Y from equation 1.

Y = 2X.

Substituting Y in both equation 2 and equation 3, we obtain:

Y = 2X, 8X + 4X + Z = 603; 10X 10X Z = -3;

From equation 3, we get that Z = -3; Z = 3;

From equation 2: 12XΒ = 603 Z = 600; X = 50;

From equation 1: Y = 2X = 100;

X = 50, Y = 100, Z = 3 solution.

*c. 22X + 5Y + 7Z = 12, 10X + 3Y + 2Z = 5, 9XΒ +Β 2YΒ + 12Z = 14*

Let us add equation 2 and equation 3, and replace equation 2 by the result:

22X + 5Y + 7Z = 12, 19X + 5Y + 14Z = 19; 9XΒ +Β 2YΒ + 12Z = 14

Then, let us subtract equation 2 from equation 1, and replace equation 1.

3X 7ZΒ = -7; 10X + 3Y + 2Z = 5, 9XΒ +Β 2YΒ + 12Z = 14

Let us express 3X from equation 1, and substitute 9X in equation 3.

3X = 7Z 7; 10X + 3Y + 2Z = 5; 3 (7Z 7) + 2Y + 12Z = 14;

3X = 7Z 7; 10X + 3Y + 2Z = 5; 33Z + 2Y = 35;

Let us multiply equation 2 by 6.

3X = 7Z 7; 60X + 18Y + 12Z = 30; 33Z + 2Y = 35;

Using equation 1, let us express 60X: 3X = 7Z -7; 60X = 140Z 140;

Using equation 3, let us express 18Y: 2Y = 35 33Z; 18Y = 315 297Z;

Let us place appropriate substitutes to equation 2:

140Z 140 + 315 297Z + 12Z = 30;

-145Z = -145; Z = 1;

3X = 7Z 7 = 0; X = 0;

2Y = 35 33Z = 2; Y = 1;

X = 0, Y = 1, Z = 1 = solution