Download Midterm 1 Solutions

Math 097 - Midterm 1 Solutions
1. Compute
−32 + 27
5
( 12 − 3)2 +
−32 + 27
5
( 21 − 3)2 +
1
2
1
2
=
−9 + 27
5
(− 52 )2 + 21
=
− 18
5
25
1
+
4
2
=
=
=
−18
5
27
4
4
− 18
5 · 27
8
− 15
.
2. Explain when two terms in an expression can be combined and why. (Saying that they are
“like terms” is not suffiecient.)
Two terms can be combined when you factor out a common factor and only arithmetic is left
in the parentheses. For example, we can factor an x2 out of 2x2 + 4x2 and get just arithmetic:
2x2 + 4x2 = (2 + 4)x2 = 6x2 .
If we factor the common y out of 3y 2 + 7y, we get
3y 2 + 7y = (3y + 7)y.
Since the parentheses don’t just contain arithmetic with numbers, we can do the computation
in the parentheses. So we’re stuck with uncombined terms.
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3. Simplify the following expressions.
(a) −(3 + 2y) + x(y − x + 2)
We’ll distribute everything we can distribute:
−(3 + 2y) + x(y − x + 2) = −3 − 2y + xy − x2 + 2x.
There aren’t any terms we can combine, so we’re done.
(b) (a − b) + 12 (a + b) − 7(2b + 4a)
(a − b) + 12 (a + b) − 7(2b + 4a) = a − b + 12 a + 12 b − 14b − 28a
= −26 12 a − 14 12 b
= − 53
2 a−
29
2 b.
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4. A pizza shop has two pricing options for pizza delivery. For option A, the price per pizza is
$15 with a $5 delivery fee for the order, and for option B, the price per pizza is $13.50 with
a $20 delivery fee for the order.
(a) Write an equation that expresses the question “For what number of pizzas will the two
options cost the same?”
If we let x be the number of pizzas, we get
cost for option A = cost for option B
$
$
$5 + 15
· x pizzas = $20 + 13.50 pizza
· x pizzas
pizza
5 + 15x = 20 + 13.50x.
(b) For what number of pizzas will the two options cost the same?
We just need to solve the equation from part a.
5 + 15x = 20 + 13.50x
−13.5x − 5
− 13.5x − 5
1.5x = 15
x = 10.
So if we order 10 pizzas, the two pricing options will cost the same.
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5. Solve the equations below. Describe the collection of solutions in words.
(a) 2x − 3 = 12
2x − 3 = 12
+3
+3
2x = 15
/2
/2
x=
We can put
15
2
15
2 .
back in the original equation to check that it works:
2( 15
2 ) − 3 = 15 − 3 = 12.
(b) −(2 + a) − 2a = 12 − 3(a + 1)
−(2 + a) − 2a = 12 − 3(a + 1)
−2 − a − 2a = 12 − 3a − 3
−2 − 3a = 9 − 3a
+3a
+ 3a
−2 = 9
This equation is always false, so there is no solution.
(c)
1
100 y
+
1500
1000
=
3
500
−
17
1000 y
We can clear the denominators first by multiplying both sides by 1000:
1
100 y
+
1500
1000
=
3
500
−
17
1000 y
10y + 1500 = 6 − 17y
+17y − 1500
+ 17y − 1500
27y = −1494
/27
/27
y = − 1494
27
y = − 166
3
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6. (a) Solve the inequality and draw a picture of the collection of solutions:
7z − 3 ≤ 4(3z − 4) + 18
7z − 3 ≤ 4(3z − 4) + 18
7z − 3 ≤ 12z − 16 + 18
7z − 3 ≤ 12z + 2
−12z + 3
− 12z + 3
−5z ≤ 5
/(−5)
/(−5)
z ≥ −1
(b) Write an inequality whose collection of solutions is pictured below:
14
We want w-values to the right of 14
3 , that is, bigger than 3 . Also, the solid dot indicates
that we want to include 14
3 itself as a solution to our inequality. So we want
w≥
14
3 .