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May 24th

May 24th

# SOLUTIONS TO ELEMENTARY ALGEBRA TEST 3 SAMPLE PROBLEMS

1. Determine if each of the relations be low is a function or not, giving
a detailed, valid reason in each case.

(a) (5, 8), (3, 8), (6, 2), (1, 9) Function. No number is repeated as a
first component.

2. (a) If f(x) = 3x2 − 5x + 4, compute f(−5), showing correct substitution.
f(−5) = 3(−5)2 − 5(−5) + 4 = 3(25) + 25 + 4 = 75 + 25 + 4 = 104.

(b) State the domain of the relation {(5, 8), (3, 8), (6, 2), (1, 9)}
The domain is {1, 3, 5, 6} .

(c) State the range of the relation {(5, 8), (3, 8), (6, 2), (1, 9)}
The range is {2, 8, 9 }.

3. Use the addition method to solve the system of equations , if possible.
Check the solution, if there is one.

(a)
8x + 5y = 7
2x − 3y = 6
Since one coefficient of x is a multiple of the other,
it is convenient to eliminate x .

Add E1 − 4E2. 17y = −17,

Substitute in E 2, say. 2x − 3(−1) = 6, 2x + 3 = 6, 2x = 3, x =3/2

Check:

3. (b)
− 4x + 6y = 5
2x − 3y = 6
Since one coefficient of x is a multiple of the other,
it is convenient to eliminate x. (We could also eliminate y.)

Add  E1 + 2E2. 0 = 17.No solution.

4. (a) Find a calculator approximation for On the TI-84, press
to get 21.21320344

(b) Simplify
Check by evaluating on your calculator, and comparing the approximations.

(c) A right triangle has sides of lengths 9.60 m and 8.30 m forming the right
angle. Find the length of the hypotenuse, rounded to two decimal places.

c2 = (8.3)2 + (9.6)2 = 68.89 + 92.16 = 161.05.

The hypotenuse has length 12.69 m, approximately.

5. Solve the equation Be sure to check.

6x + 1 = 81 − 18x + x2, 0 = x2 − 24x + 80,
0 = (x − 4)(x − 20), x = 4, 20. Check x = 4.
Check x = 20. Reject 20.

6. Rationalize the denominator and simplify .

7. (a) Compute since 34 = 81.

(b) Give your calculator display for On a TI-84 press
then to get Also

(b) Simplify
(Be careful not to write )

(c) Simplify

8. (a) Solve the system of equations by the substitution method,
and check your answer. The second equation is a formula for y in terms of x.
Substitute the formula for y in the first equation. 8x − 5(3x + 2) = 2,
8x − 15x − 10 = 2, − 7x = 12, x = 12/(−7) = − 12/7. Use the formula to find y.
Since we just did the
computation that checks the second equation, we only have to check the first.
8(−12/7) − 5(−22/7) = − 96/7 + 110/7 = 14/7 = 2.

8. (b) Set up a system of two equations in x and y for the following problem:
“Maria wants to make 5 liters of a 28% alcohol solution. 20% alcohol solution
and 40% acid solution are available . How much of each should she mix?”
(Set up only, do not solve.)

Mix x liters of 20% solution and y liters of 40% solution.
or

9. Solve the equation x2 + 16x − 5 = 0 by completing the square.

x2 + 16x + 64 = 5 + 64, (x + 8)2 = 69,

10. Given the quadratic equation 3x2 − 10x + 5 = 0,
(a) compute the discriminant a = 3, b = −10, c = 5,
b2 − 4ac = (−10)2 − 4(3)(5) = 100 − 60 =

(b) solve the equation by using the quadratic formula.

12. Find the distance between the two points: (−1, 7), (5, − 1)

13. Simplify (Write answer in fraction form .)

14. A rectangle has a length 3 meters longer than its width.
Its area is 154 square meters . Find the dimensions of the rectangle.

Let the width be w meters, and the length
l Meters. l = w + 3, A = lw,
w(w + 3) = 154, w2 + 3w = 154,
w2 + 3w − 154 = 0. If necessary, we could
use the quadratic formula here to solve.
(w − 11)(w + 14) = 0, so w = 11, − 14.
We reject −14 < 0.

Using w = 11, l = w + 3 = 11 + 3 = 14. Check: lw = 11(14) = 154.

The rectangle is 11 meters by 14 meters.

Note that in factoring above, we had to find two integers differing by 3, whose product
is 154. This is actually the original problem. However, we could have used the
quadratic formula if the factoring was difficult, or if the dimensions were irrational
numbers. Also there might be two answers to a problem, and solving the quadratic
equation would find both.

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