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Solve the system of equations: {3x - 6y + z = 0 -x + y - z = -1 x - 2y = -1 solv

ID: 3031911 • Letter: S

Question

Solve the system of equations: {3x - 6y + z = 0 -x + y - z = -1 x - 2y = -1 solve the system of equation {2x - 5y + z = -8 -x + y - z = 1 x - 3y = -6 Maricopa's Success scholarship fund receives a gift of $ 130000. The money is invested in stocks, bonds. and CDs. CDs pay 4.75 % interest, bonds pay 4.4 % interest, and stocks pay 6.8 % interest. Maricopa Success invests S 50000 more in bonds than in CDs. If the annual income from the investments is $ 6750, how much was invested in each account? Maricopa Success invested Maricopa Success invested Maricopa Success invested

Explanation / Answer

8)

3x - 6y + z = 0                              (1)

-x + y - z = -1                                (2)

x - 2y = -1                                     (3)

we have 3 variables "x" , "y" and "z".

we can see that equation (3) has only 2 variables

so we try to remove the third variable from equation (1) and (2)

Adding equation (1) and (2) we get                     // by adding these two eq we can remove "z"

3x - 6y + z = 0

-x + y -z = -1

=>  

2x - 5y = -1                (4)

now from (3) and (4)

multiplying eq (3) with "2"

we get

2x - 4y = -2                          (5)

subtracting eq (5) and eq (4)

2x - 5y = -1

2x - 4y = -2

-     +       +                         // on subtracting we change the sign

=>

-y = 1

we get

y = -1

now from equation (3) we can find the value of "x"

x - 2y = -1

x - 2 * (-1) = -1

x + 2 = -1

x = -3

now from equation (1)

3x - 6y + z = 0

3 * (-3) - 6 * (-1) + z = 0

-9 + 6 + z = 0

z = 3

.

9)

2x - 5y + z = -8                      (1)

-x + y - z = 1                           (2)

x - 3y = -6                               (3)

adding eq (1) and (2)                                   // we can eliminate "z"

2x - 5y + z = -8

-x    + y   - z = 1

we get

x - 4y = -7                        (4)

subtracting eq (3) and (4)

x - 3y = -6

x - 4y = -7

-     +      +

y = 1

now from equation (3)

x - 3y = -6   

x - 3 * (1) = - 6                      // since y = 1

x - 3 = -6

we get

x = -3

now from equation (2)

-x + y - z = 1   

- (-3) + (1) - z = 1

3 + 1 - z = 1

z = 3

.

10)

Total fund recieved = $ 130,000

This recieved money is invested in stocks , bonds and CDs

Let the amount spent in CDs be "x".

Investment in Bonds = "50000 + x"                  // the investment in Bonds is 50000 more then CDs

Investment in Stocks = Total Investment - (Investment in Bonds + Investment in Stocks)

                                  = 130000 - (50000 + x + x)

                                  = 130000 - 50000 - 2x

                                  = 80000 - 2x

Annual Income from the Investment = $ 6750

Annual Income from the Investment = Income from CDs + Income from Bonds + Income from Stocks                 (1)

CDs pays 4.75 % interest

Bonds pays 4.4% interest

Stocks pays 6.8 % interest

Income from CDs = amount invested * Interest payed / 100

                             = x * 4.75 / 100 = > 4.75x/100

Income from Bonds = amount invested * Interest payed / 100

                                =  ( 50000 + x) * 4.4 / 100

                                = 2200 + 4.4x/100       

Income from Stocks = amount invested * Interest payed / 100

                                 = ( 80000 - 2x ) * 6.8 / 100

                                 = 5440 - 13.6x/100

now from eq (1)

6750 = 4.75x/100 + 2200 + 4.4x/100 + 5440 - 13.6x / 100

6750 = 4.75x/100 + 4.4x/100 -13.6x/100 + 7640

-890 = -4.45x/100

890 = 4.45x/100

x = 20000

So the amount invested in CDs = $ 20,000

Amount invested in Bonds = 50000 + x   => 50000 + 20000 = $ 70,000

Amount Invested in Stocks = 80000 - 2x => 80000 - 2*(20000) => 80000 - 40000 = $ 40,000