Download practice test 2

Math 097 - Midterm 1 practice #2 Solutions
(1) For each equation below, solve for x.
(a) 2(x + 1) + 1 = 2x + 3
2(x + 1) + 1 = 2x + 3
2x + 2 + 1 = 2x + 3
2x + 3 = 2x + 3
0 = 0.
All values of x work in this equation, so every number is a solution.
(b) 4(x − 1) = 2x + (2x + 2)
4(x − 1) = 2x + (2x + 2)
4x − 4 = 4x + 2
−4 = 2.
There is no value of x that makes this equation true, so this equation has no solutions.
(c) 2x − 3y = 5x + 6
2x − 3y = 5x + 6
2x − 3y − 5x = 6
−3x − 3y = 6
−3x = 3y + 6
3y + 6
x=
−3
x = −y − 2.
2
(2) Simplify the expression below. For each step, state what property of numbers justifies that
step.
−2x(x + 1) − 4(−3x − 2) + 3x − 4
−2x(x + 1) − 4(−3x − 2) + 3x − 4
−2x2 − 2x + 12x + 8 + 3x − 4
distribution
−2x2 − 2x + 12x + 3x + 8 − 4
commutativity of addition
2
−2x + 13 + 4
factoring and arithmetic
3
(3) Solve the inequality below, and draw a picture of the collection of solutions.
2(z − 3) − 7 ≥ 5z + 12
2(z − 3) − 7 ≥ 5z + 12
2z − 6 − 7 ≥ 5z + 12
2z − 13 ≥ 5z + 12
2z ≥ 5z + 25
−3z ≥ 25
z ≤ − 25
3 .
The solution set for this inequality is below:
(4)
1
2t
+ 7 < 13 t − 8
We can clear the denominators first so we don’t have to deal with so many fractions. To
clear the denominators, we’ll multiply by 6:
6 · ( 12 t + 7) < ( 31 t − 8) · 6
3t + 42 < 2t − 48
t + 42 < −48
t < −90.
The solution set is
4
(5) Two plumbers are bidding on a job. The first estimates that the job will cost $600 in
materials plus $30 per hour of labor. The second plumber estimates that the job will cost
$700 in materials plus $28 per hour of labor. How many hours of labor would make the
two plumbers charge the same for the job?
If the plumbers work for h hours, the first plumber charges
$30
$600 +
h,
1hour
and the second plumber charges
$28
$700 +
h.
1hour
We want these two to be the same, so we declare them to be equal:
600 + 30h = 700 + 28h.
Now we can solve this equation for h.
600 + 30h = 700 + 28h
600 + 2h = 700
2h = 100
h = 50.
So the two plumbers charge the same amount for a job that takes 50 hours.
5
(6) (a) Simplify:
2
3
3 (− 4 x
− 9) + 51 (20y − 3)
2
3
1
3 (− 4 x − 9) + 5 (20y
− 12 x − 6 + 4y − 53
3
− 12 x + 4y − 30
5 − 5
− 12 x + 4y − 33
5
− 3)
(b) Calculate
2
3
5 − 10
1
10 + 1
There’s more than one way to do this, but here’s one possibility:
3
2
5 − 10
1
10 + 1
=
=
2
3
5 − 10 10
·
1
10
10 + 1
2
3
5 · 10 − 10 · 10
1
10 · 10 + 1 · 10
4−3
1 + 10
1
= .
11
=
(c) Calculate
3 · 42 − 52 + (−22 )
3 · 42 − 52 + (−22 ) = 3 · 16 − 25 + (−4)
= 48 − 25 − 4
= 19