1. Suppose we are given real numbers w, x, y, and z such that x < y and w <
z.

Determine if each of the following inequalities is true or false or inconclusive
(i.e., it

could be either true or false).

Solution:

(a) Let us look at some examples: If we take numbers greater than 0 then the

inequality seems to hold. For example,

x = 3 < 4 = y

w = 2 < 7 = z

xw = 6 < 28 = yz

However, if we take numbers less than 0, then the inequality does not hold.
For

example take

x = −3 < −1 = y

w = −4 < −2 = z.

But

xw = 12 > 2 = yz.

So this inequality one may or may not be true, depending on the real numbers

chosen.

(b) This inequality will again depend on the numbers chosen. To see this, let
us

again choose numbers that are greater than 0 and then numbers that are less

than 0. It is clearly true that x = 3 < 4 = y and x^2 = 9 < 16 = y^2, but if we

take x = −4 < −3 = y then x^2 = 16 > 9 = y^2.

(c) Again, let us choose specific numbers, say, −3, −2, 0, and 5. We have
that

−3 < 5 and −2 < 0. It is clear that −3 + (−2) < 0 + 5. The fact is that the

inequality is always true, given the conditions.

How shall we prove it? From x < y we can say y = x + p with p > 0. From

w < z we can say that z = w + q with q > 0. So

Thus y + z is x + w plus (p + q). Note that (p + q) is positive since each of
p, q

is positive. So we get x +w < y + z.

(d) Summary. How do we decide to prove or disprove the inequality? In
general,

remember that multiplication by negative numbers reverses an inequality, so

whenever products are involved, we suspect failure.

When only additions or multiplications by known positives are involved, we
expect

the inequality to be true, provided the algebraic manipulations can show its

truth.

In general, proving the inequality true involves general calculations, whereas

proving it to be false or inconclusive is done by a simple example!

Even though our examples are fairly simple, in real life, inequalities can be
lot

more complicated than equations.

2. Let f(x) = x^3 + x + 2 and g(x) = 3x^2 + x. We will find deg_{x}(fg).

**Solution:**

One approach would be to multiply the polynomials term by term and note the
degree.

But this is not necessary. Let us recall the the product rule for degrees. The
degree

of the product of two non-zero polynomials is the sum of their degrees.
Therefore

We note that deg_{x}(f) = 3 and deg_{x}(g) = 2. And so deg_{x}(fg)
= 3+2 = 5.

From experience you could also say

Thus we even know the highest degree term!

3. Let us define s to be the monomial
Find the coefficient of s in general as well

as the coefficient of s when it is regarded as a monomial in x or y.

**Solution:**

• To begin, it is good to simplify s to the expression

• Now simply note that for s =
the coefficient is 6 and the degree is the

sum of the powers of the variables, i.e., 6 + 4 = 10.

• If we regard s as a monomial in x then let we rewrite it as
As a monomial

in x its coefficient is
and the degree is simply 6.

• Finally, let us regard s as a monomial in y and rewrite it as
So, as a

monomial in y, the coefficient is
and the degree is 4.

4. Let
and
Find the coefficient of x

in

Solution:

One approach would be to findexplicitly.

A simpler approach is to recognize that since we are adding polynomials, the
only

terms that contribute to the coefficient of x in
are those coefficients of x

in f and g. In f we have 12x and in g we have −11x. Then the term with x in

is

So the term with x in
is

The desired coefficient is then 6.

A better strategy would be to do a more general problem and avoid numerical
errors.

Consider the more advanced problem:

Find the coefficient of x in af + bg where a, b are constants. Clearly the
x-term in

af + bg will be

Thus the desired coefficient is 12a−11b. For the given problem we have a = 1,
b =

so the answer is
as before.

5. Let and Find

Solution:

We can, of course add the two polynomials, collect the coefficients and then
find the

degree.

But that might be more work than needed.

Recall the facts:

• Also the coefficients of x^{7} in f, g are respectively 0, 1. So the
coefficient of x^{7} in

f + g is exactly 0 + 1 = 1.

• The above two facts tell us that we must have:

lower degree terms .

Hence the desired degree of f + g is exactly 7.

• To remember: We may remember once for all that the degree of a sum f + g

is equal to the maximum of the degrees of f and g as long as the two degrees

are distinct! When the degrees of f, g are equal, then the answer depends on

the cancellation of terms and any degree less than or equal to the maximum is

possible, including the chance that f + g may be zero!

6. Use Cramer’s rule to find the solution to a system of two equations in two variables:

Solution:

Calculate

So

Recommendation:

It is an excellent idea to double check your answer. Plug the answers in each of
the

equations and verify that they are true:

7. Solve a system of two equations with two variables using substitution.

Solve:

Solution:

Solving Eq2 for x gives

Solution1

x = 2+3y.

Plugging this into Eq1 gives

Now we use this value for y in Solution 1,

So the final solution is

As before, we recommend that you double check the answers. Either use
Cramer’s

Rule or substitute the answer and verify each equation.

Check:

Eq1:

Eq2:

8. Let f(x) = 2x^2 −3x+4. Find the equation for the line tangent to y = f(x) at x = 2.

Solution:

To write the equation of any line, it is enough to know a point on it and its slope.

To find the point we simply note that x = 2, y = f(2) is on the line.

For the slope of the line, we know that the slope of a tangent line is given by
the

derivative of f(x) at x = 2, i.e. f'(2).

• Let us evaluate f(2).

So the point is (2, 6).

• Next, let us find f'(x) in general. By the power rule,

So f'(2) = 4(2) − 3 = 5 and the slope of the tangent line at (2, 6) is 5.

• Using the point slope formula, (y − y_{1}) = m(x − x_{1}) where m = 5, x_{1} = 2, and

y_{1} = 6, we see that the equation for the tangent line is

(y − 6) = 5(x − 2), or y = 5x − 4.

There is an alternate technique based on theory. We
consider the equation y =

2x^2 −3x+4 and to find the tangent at x = 2, y = f(2) = 6 we rewrite x = 2+u and

y = 6+u.Substitute this in the given equation and keep only the linear terms.

Thus y = f(x) becomes

Thus the linear part becomes

u = 5u

or

(y − 6) = 5(x − 2)

This is the desired tangent line!

Let A(2, 7) and B(4, 5) be points in the plane. Find an
equation of the line passing

through A and B and put it in slope-intercept form.

Solution:

(a) First, we find the slope of the line by using the
slope formula. Recall that the

slope of the line joining P_{1}(a_{1}, b_{1}), P_{2}(a_{2},
b_{2}) is given by the formula

Using P_{1} = (2, 7) and P_{2} = (4, 5) we
get:

So the slope of the line is −1.

(b) Now, we choose a point, say, P_{1}(2, 7) and use the point-slope
formula.

Rearranging:

y = −x + 9

(c) Thus the slope is −1 and y-intercept is 9.

(d) Short cut! Recall the two point formula from the book (for same points P_{1},
P_{2}

as above):

Using P_{1} = (2, 7) and P_{2 }= (4, 5),
we get:

(y − 7)(4 − 2) = (x − 2)(5 − 7)

or when simplified:

2y − 14 = −2x + 4 | which further becomes |

2y = −2x + 18 | or |

y = −x + 9. |

10. Given the points P(2, 3) and Q(5,−1) find the equation
of the line passing through

these points in both standard and parametric form.

Solution:

(a) To find the standard form we find the slope, m, and
use the point-slope form of

a line. Recall that and the point slope form
is

(b) To find the parametric form, recall that possible
equations for parametric form

are So we have

x = 2+t(5 − 2) = 2 + 3t , and

y = 3+t(−1 − 3) = 3 − 4t.

Note that if we eliminate t from these we get

4x + 3y = 4(2+3t) + 3(3 − 6t) = 8+9 = 17.

Which also leads to

3y = −6x + 17

or

11. Given a line 2x+4y−7 = 0 and a line given
parametrically by x = 3t+3 and y = t−2,

determine where they intersect.

Solution:

First, plug in the parametric values for x and y from the
second line into the equation

of the first to determine the t value at the point of intersection.

Now find the x and y values for

Checking:

12. Convert the parametrically given line, x = −t+2 and y
= 2t−1, into standard from,

y = mx + c.

Solution :

We wish to eliminate t from the parametric equations

x = −t + 2

y = 2t − 1.

If we multiply the first equation by 2, the second by 1 and add them together we get:

Thus 2x + y = 3 is an equation of the line. In
slope-intercept form it becomes

y = −2x + 3.

13. Find the point where the line y = −2x and the circle x^2 + y^2 = 20 intersect.

Solution:

Since the equation for the line is already solved for y we
simply plug the value for y

into the equation for the circle:

Thus the line intersects the circle at the x-coordinates x
= 2 and x = −2. The

corresponding y-coordinates come by using y = −2x. Thus y = −4, 4 respectively.

The two points, respectively, are (2,−4), (−2, 4).