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Math 084.03 - Practice Final Solutions
(1) Consider the line with equation
2x + 6y = 12.
(a) Complete the table below, so that each row in the table is a point on the line.
To fill in each row in the table, we can take the given value of x or y, put it into the
equation, and find the value of the remaining variable.
x=0
⇒
2(0) + 6y = 12
⇒
y=2
y=0
⇒
2x + 6(0) = 12
⇒
x=6
x=2
⇒
2(2) + 6x = 12
⇒
y = 43 .
So the table looks like this:
x
y
0
2
6
0
2
4
3
(b) Graph the line below.
We can just plot the points in the table and draw the line through them.
2
(2) Write the equation (in slope-intercept form) of the line through the point (−2, −4) that is
parallel to the line − 21 x − 14 y = −2.
We have a point on the line we want, so we just need its slope. Since the line we want
is parallel to the line − 21 x − 14 y = −2, it will have the same slope. To find the slope of
− 21 x − 41 y = −2, we can put it in slope-intercept form:
− 12 x − 41 y = −2
− 14 y = 12 x − 2
y = −2x + 8.
So the slope of this line is −2. So the slope of the line we want is also −2. Now that we
have a point on our line (−2, −4) and the slope of our line, −2, we can write the point-slope
equation of our line, then put it in slope-intercept form:
y − (−4) = −2(x − (−2))
y + 4 = −2x − 4
y = −2x − 8.
Math 084.03
3
(3) Draw a diagram of which points in the plane satisfy the inequality
2(x + y) > 3x − y + 6.
First, let’s put the inequality into slope-intercept form:
2(x + y) > 3x − y + 6
2x + 2y > 3x − y + 6
2x + 3y > 3x + 6
3y > x + 6
y > 13 x + 2.
So the boundary line is y = 13 x + 2. The boundary line does not satisfy the inequality,
because it is “>” and not “≥”. Finally, we want the half-plane above the line, since y is
greater than the line. This give us the following picture:
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(4) You have one vat of paint that is 20% red paint, and another that is 80% red paint. (The
rest is white.) If you want to make 15 gallons of 55% red paint, how much paint of each
type should you mix?
Let’s let x be the number of gallons of 20% red paint we use, and let y be the number
of gallons of 80% red paint we use. When we mix these together, we get x + y gallons of
paint, and we want 15 gallons. So we get our first equation: x + y = 15. On the other
hand, the total amount of red paint in our mix is 0.2x + 0.8y, and the total amount of red
paint we want is 0.55 · 15, so we get a second equation: 0.2x + 0.8y = 0.55 · 15. Now we
have a system we can solve:
(
x + y = 15
0.2x + 0.8y = 8 14 .
This looks like it’s set up for elimination, so I’ll multiply the second equation by −5 to set
up the x’s to cancel. Now my system is
(
x + y = 15
−x − 4y = −41 41 .
When I add equations, the x’s are gone, and I can solve for y:
−3y = −26 14
⇒
y = 8 34 .
Now I can put this value back in to the first equation to find x:
x + (8 43 ) = 15
⇒
x = 6 14 .
So I need to mix 6 14 gallons of the 20% red paint and 8 34 gallons of the 80% red paint. (I
put these back into the original system to check that they work. They do!)
Math 084.03
5
(5) Simply the following expressions.
(a) 2x23 · 4x11
Using commutativity of multiplication and the the product rule for exponents, we have
2x23 · 4x11 = 2 · 4 · x23 · x11 = 8 · x23+11 = 8x34 .
(b) 3(x2 y 3 )4
Using the many bases rule and then the power rule for exponents, we have
3(x2 y 3 )4 = 3(x2 )4 (y 3 )4 = 3x2·4 y 3·4 = 3x8 y 12 .
(c)
3k 4 0
·k
9k 7
Remember that k 0 is just 1. Using negative powers and the product rule for exponents,
we have
3k 4 0
· k = 3 · k 4 · 3−2 · k −7 · 1 = 3−1 k −3 = 3k13 .
9k 7
(d)
(5a)2
5(a2 )3
Here we’ll use the power rule, the many bases rule, negative exponents, and the product
rule:
(5a)2
52 a2
5
=
= 52 a2 5−1 a−6 = 51 a−4 = 4 .
5(a2 )3
5a6
a
6
(6) (a) Multiply the following polynomials. (You may use any shortcuts you remember.)
(x3 − 3x + 1)(2x2 − 7)
Remember that our shortcut for multiplying polynomials is to multiply every combination of a term from the first polynomial and a term from the second polynomial,
then add it all up. We get
(x3 − 3x + 1)(2x2 − 7)
= 2x5 − 7x3 − 6x3 + 21x + 2x2 − 7
= 2x5 − 13x3 + 2x2 + 21x − 7.
(b) Indicate which expressions below are polynomials. For those that are not, briefly
explain why not.
−5x3 + x − 1
polynomial
0.0574z + 4.21
polynomial
y+1
y−1
x2 + x1 + x0
x2 + x1 + x0 + x−1
not a polynomial, since there is a y in the denominator
polynomial (x0 is just 1
not a polynomial, since we have a negative power on x.