Practice Problems MAT 102, Test #1

(1) Tell whether the following pairs are solutions of the system

(a) (12,0) (b) (5,2)

(2) Graph the two lines in the system and ‘guess’ the solution as the intersection:

(a)

(3) Solve the linear system by substitution:

(a) (b)

(4) Solve the linear system by elimination (addition).

(a) (b)

(5) (a) K deposits $9000 in two CD’s, one paying 5% interest and the other paying 6% interest. If he earns $505 in cumulative interest from the two accounts, how much did he invest in each?

(b) Vito issues two loans for a total of $40,000 – one to a guy who’s going to pay him 20% on the original loan, and another to someone who’s going to pay him 25%. If he stands to collect $8500 in interest from the two loans, how much was each original loan?

(6) (a) A soccer league awards 3 points for a win and 1 for a tie (none for a loss). During the season, Columbia FC wins 4 more games than they tie, and collects 36 points. How many games did they win and how many did they tie?

(b) A restaurant owner takes money to the bank to make a deposit, all in 5-dollar and 10-dollar bills. He has 100 bills on him for a total of $765. How many of each type of bill did he deposit?

(7) (a) Josef K intends to mix two solutions, one of which is 35%-zolon, and the other which is 50%-zolon. The final mix will be 32 liters of a solution which is 41%-zolon. Find the amount of the original solutions he must start with to get this final mix.

(b) A goldsmith has two gold alloys. The first alloy is 40% gold; the second alloy is 60% gold. How many grams of each should be mixed to produce 20 grams of an alloy that is 52% gold?

(8) Graph the system of linear equations:

(a) (b)

Solutions:

(1) Just plug the pairs into the system and see if they work. Remember, they have to satisfy both equations:

(a) (12,0): 3(12)-2(0)=11?

36=11? no, not a solution (even though it does satisfy the 2nd equation:)

2(12) + 7(0) = 24

(b) (5,2): 3(5) – 2(2) = 11? yes:

15 – 4 = 11

2(5) + 7(2) = 24? yes:

10 + 14 = 24. So (5,2) is a solution.

(2)

The intercepts of the first line are (9,0) and .

The intercepts of the second line are (0,2), and (-1,0). Graph the intercepts and then draw the lines between them:

The solution is (1,4)

(3)

(a) (The x in the first equation has coefficient 1 – solve for it)

x = 2y + 7 (substitute this into the other equation)

2(2y + 7) + 5y = -13,

4y + 14 + 5y = -13,

-14 -13

9y = -27,

(back-solve for x: )

x – 2(-3) = 7,

-6 -6

x = 1. The solution is the ordered pair (1,-3).

(b) None of the variables have coefficient 1 – solve for y in the first:

2x – y = 11,

-2x -2x

-y = 11 – 2x (switch signs)

y = -11 + 2x (substitute into the other equation:)

-8x + 4(-11+2x) = -44

-8x – 44 + 8x = -44

+44 +44

0x = 0, the system is dependent, it has an infinite number of solutions.

(4) (a)

(eliminate x: 2 and 4 both ‘go into 4’ - * first equation by -2):

Back-solve for x (i’m going to use the first equation – you could use either)

2x + 3(-2)=2

+6 +6

2x = 8, The solution is (4,-2).

(b) You can eliminate either variable here. Either way, you’re going to have to multiply both equations. I picked y to eliminate, as the coefficients have opposite signs already. The coefficients of y both ‘go into’ 20, so make one coefficient -20, and the other 20:

The solution is (3,3).

This equation has no solutions. We call the system inconsistent.

(d) Here, we want to begin by clearing the fractions – multiply through by the LCD in both equations:

Now, eliminate a variable – I’m going to eliminate the y since the coefficient have opposite signs – note that 5 and 2 both ‘go into’ 10. So,

The solution is .

(5) (a) In interest problems, you need to keep in mind the fact that

Interest = Principal * rate

Begin by naming the unknowns:

x = amount invested at 6%,

y = amount invested at 5%

Now, translate the other information into equations for these variables:

x + y = 9000 (total invested)

.06x + .05y = 505 (total interest).

Multiply the second equation by 100 to clear the decimals, and you get:

y = 3500.

(b) Let x = amount of loan 1, y = amount of loan 2. Translate the information into equations:

Multiply the second equation through by 100 and solve the system:

(6) (a) Start with the question. What are they asking for?

x = number of ties,

y = number of wins. Now translate the information into equations:

y = x + 4 (four more wins than ties)

1x + 3y = 36 (number points from wins + number points from ties = total points)

This one is good to go for substitution, since the first equation is ‘solved’:

x + 3(x + 4) = 36,

x + 3x + 12 = 36,

-12 -12

4x = 24,

and y = (6) + 4 = 10.

So they won 10 games and tied 6.

(b) Start with the question, and name the unknowns:

x = number of five-dollar bills, y = number of ten-dollar bills.

Translate:

x + y = 100 (a hunderd bills total)

5x +10y = 765 ($ in fives + $ in tens equals $765)

Solve. I’m going to use elimination, eliminating x:

(7) (a) I’m going to construct a table:

% gold

solution 1 | x | .35 | .35x

solution 2 | y | .50 | .5y

mix | 32 | .41 | 32(.41)

x + y = 32,

(.35x + .5y = 13.12)100

-35( x + y = 32),

35x + 50y = 1312

-35x – 35y = -1120

35x + 50y = 1312

15y = 192

(b) I’m going to construct a table:

alloy 1 | x | .4 | .4x

alloy 2 | y | .6 | .6y

mix | 20 | .52 | 20(.52)

x + y = 20

(.4x + .6y = 10.4)*10

-4(x + y = 20) (eliminate x)

4x + 6y = 104

-4x – 4y = -80

4x + 6y = 104

2y = 24,

(8) (a) graph the border lines and find the overlap:

(b)