Systems of Linear Equations, Statistics Problem Example

Suppose Bob owns 2,000 shares of Company X and 10,000 shares of Company Y.  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.  The total value of Frank’s holdings of these two companies is $400,000.

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

Bob’s holding 2 000 in X and 10 000 in Y.

2 000X + 10 000Y = 372 000 (Bob’s holding)

Frank’s holding will be 8 000X + 6 000Y = 400 000

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

Let Bob’s holding be 1^st equation and Frank’s be 2^nd.

2 000X + 10 000Y = 372 000………………………1^st

8 000X + 6 000Y = 400 000…………………………..2^nd

Dividing 2^nd by 4 we get 2 000X + 1 500Y = 100 000

Therefore to eliminate X we subtract equation two from the 1^st.

(2 000X + 10 000Y) – (2 000X + 1 500Y) = 372 000 – 100 000

8 500Y = 272 000

Y = 32

Is obtained when we substitute the value of Y in either 1^st or 2^nd equations. Substituting in the 2^nd we get;

8 000X + 6 000(32) = 400 000

8 000X + 192 000= 400 000

8 000X = 400 000 – 192 000

8 000X = 208 000

X = 26

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

X + 2Y +  Z = 22

X + Y  = 15

3X + Y + Z =  37

Let X + 2Y + Z = 22 be the first equation, X + Y = 15 the second and 3X + Y + Z = 37 the third equation.

Eliminating Z using by subtracting 3^rd from first we get

(3X + Y + Z )-(X + 2Y + Z) = 37-22

3X-X + Y-2Y + Z-Z = 15

2X –Y = 15 let this be equation xx

Taking 2^nd equation and xx to get the value of y and x

(2X –Y) +( X + Y) = 15+15

3X = 30

X = 10

When X is 10, Y becomes 10+ Y = 15; Y = 5.

From the 1^st equation, we can compute fore Z

X + 2Y + Z = 22 but X= 10 Y = 5; therefore

10 + 2(5) + Z = 22

10 + 10 + Z = 22

Z = 22 – 20

Z = 12

X + Y + Z = 603………………….k

8X + 2Y + Z  = 603………………….kk

20X – 10Y – 2Z = -6………………………kkk

We use k and kk to eliminate Z. This is done through subtraction of  kk from k

(10X + Y + Z) – (8X + 2Y + Z) = 603-603

2X –Y = 0 and let this be equation d.

We again use kkk and k to eliminate Z. we add the two equations. To have the coefficient of Z rhyming in the two, we multiply k by two.

2(10X + Y + Z) = 20X + 2Y + 2Z = 1206

(20X + 2Y + 2Z) + (20X – 10Y – 2Z) = -6 +1206

40X – 8Y = 1200 and let this be dd.

2X –Y = 0……………………..d

40X – 8Y = 1200…………… dd

Multiplying the d by 8 we get, 16X – 8Y = 0. We then use elimination method(by subtraction d from dd) to get X

(40X – 8Y) – (16X –8Y) = 1200 – 0

24X = 1200

X = 50

40(50) – 8Y = 1200 ( from dd)

200 – 8Y = 1200

Y = (1200 – 200)/8

Y = 125

Obtaining Z

We use equation k 10X + Y + Z = 603; where X = 50 Y = 125

10×50 + 125 + Z = 603

500 + 125 – 603 = Z

Z = 22

 22X + 5Y + 7Z = 12…………………………..p

10X + 3Y + 2Z = 5…………………………….pp

9X  +  2Y  + 12Z = 14…………………………ppp

 Eliminating Y, from p and pp, we multiply p by 3 and pp by 5. This will bring

66X + 15Y + 21Z = 36……………………..p1

50X + 15Y + 10Z = 25………………………..pp2

Subtracting the two (66X + 15Y + 21Z) – (50X + 15Y + 10Z) = 36 – 25.

16X + 11Y = 11 and let this be equation f

Eliminating Y using pp and ppp, we multiply pp by 2 and ppp by 3

20X + 6Y + 4Z = 10…………………pp2

27X + 6Y + 36Z = 42……………….ppp3

Subtracting pp2 from ppp3, (27X + 6Y + 36Z) – (20X + 6Y + 4Z) = 42 – 10

7X + 32Z = 32 and let this be ff.

We use f and ff to find the values of X and Z

16X + 11Y = 11 ………………f

7X + 32Z = 32 …………………ff

Making X the subject of the formula from f, we get X = (11 – 11Y)/16 and we substitute in ff

7[(11 – 11Y)/16] + 32Z = 32

7[ 11/16 – 11Y/16] + 32Z = 32

77/16 – 77Y/16 + 32Z = 32

27.1875Z = 27.1875

Z = 1

We use ff to get X; 7X + 32×1 = 32

7X = 32 – 32

7X = 0; X = 0

We use pp to get Y

10X + 3Y + 2Z = 5

10 x 0 + 3Y + 2×1 = 5

3Y + 2 = 5

3Y = 3

Y = 1