1078 Basic Algebra questions: solving an equation,
equalities and inequalities
The last five people to edit this page are
Edward Morey October 6, 2006
(note that the questions are numbered in groups)
(note that this page is a work in progress)
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Solve for the indicated variable in terms of the remaining variables:
1. A= P+Prt; for r
y  mx  b ; for m
1. Solve for x:
log b (2 x)  12 log b 16  log b 2
1
log b (2 x)  log b 16 2  log b 2
log b (2 x)  log b 4  log b 2
4
 log b 2
2
So 2 x  2  x  1
log b (2 x)  log b
1. Solve this equation for u: 2 
5 3

u u2
5 3

times u 2 on both sides  u 2  2    2 u 2
u u

2
 2u  5u   3
2
5 3

u u2
2u 2  5u  3  0
 u  3 2u  1  0   u  3  0
u  3 or u 
1
2
or
 2u  1  0
2. You need a compressor and there are two choices for you. You can rent a compressor
for the cost of $90/day. Alternatively, you can buy one at $4,000 with maintenance
costs around $10/day. Let X denote the number of days that the compressor is needed.
Find out the value of X that makes these two choices equivalent.
Total cost for buying a compressor after x days: 4,000+10x
Total cost for renting one after x days:
90x
Set two costs function equalized and solve for x:
4000+10x = 90x  80x=4000  x=50 days
1. Solve for x.
(A) 3x  5  2
(B)
2
1
1
x x
3
2
4
(C) x 2  3x  0
1. Supply and demand
Suppose that the supply and demand equations for printed T-shirts in a little town during
Christmas week are
Supply: p = q + 12
Demand: p = - 2q + 36
Where p is the price in dollars and q is the quantity in hundreds.
A) Find the supply and demand quantity if p = $16.
B) Find the equilibrium price and quantity. (when the price of supply equals the price
of demand).
1. Solve.
x 9
x


5 10 2
A x  3
Solve and graph
1.  8  2x  4  12
A 2 x 8
(
]
–2
8
x
2. Solve the inequality for y and graph your answer:
2
1  3  y  9
3
A 9  y  6
(
]
–9
6
y
3. You need a compressor and there are two choices for you. You can rent a compressor
for the cost of $80/day. Alternatively, you can buy one at $7,000 with maintenance
costs around $10/day. Let X denote the number of days that the compressor is needed.
Find out the value of X that makes these two choices equivalent.
A Solving 80x  10x  7000 , x = 100 days.
Check the following equalities.
2
1
x4 x4
3
2
y y 1
2.  
4 3 2
1.
3. 3(x-2 ) = 2(x-3) + x
4.
2
1
x 1  x  3
3
2
5.  1 
2
y  5  11
3
6. 10  8  3u  6
Demonstrate the equalities:
1.
a, a 2  2ab  b 2  (a  b) 2
b, a 2  2ab  b 2  (a  b) 2
c, a 2  b 2  (a  b)(a  b)
d , a 3  b3  (a  b)(a 2  ab  b 2 )
e, a 3  b3  (a  b)(a 2  ab  b 2 )
1. Solve the inequality for y and graph your answer:
3
5  4  y  10
2
3
5  4   y  10  4
2
3
9   y  6
2
 2   2  3 
 2
9          y   6   
 3   3  2 
 3
6  y  4
or
3
3
y y9 y6
2
2
3
3
62
4  y  10   y  6  y  
 y  4
2
2
3
Combine : 4  y  6
5  4 
Solve for x:
1. x 2  20  5
2. 2 x 2  7 x  3  0
3. 2 x 2  3 x  x 2  2 x  12
4. 2 x 2  20 x  6  0
1. The supply for broccoli is described by the equation Q = P/6 where P is the price of broccoli
and Q is the amount that will be supplied at price P. The demand equation is described by Q
= 330-9P where P is the price of broccoli and Q is the amount that will be demanded. What is
the equilibrium PRICE of broccoli where demand equals supply?
2. Solve for y in terms of x the expression ( y  3x) 2  9 x 2
3. Solve by factoring:
x 2  2 x  35  0
A (Since ( x  7)( x  5)  0 ,) x  7,5
4. Solve using the quadratic formula
x 2  7 x  11  0
7 5
A
2
5. Suppose the demand equation (inverse demand function) is written as:
D  150 / p
And the supply equation is:
S  10 p  20
What is the equilibrium price p when D = S. (Choose only the positive price.)
A (By ( p  5)( p  3)  0 ,) p  5 .
1. Solve this equation for u: 2 
A 2
 2u
2
5 3

u u2
 5u   3
5 3

u u2
5 3

times u 2 on both sides  u 2  2    2 u 2
u u

2u 2  5u  3  0
 u  3 2u  1  0   u  3  0
u  3 or u 
1
2
2. Solve for x:
x 2  7 x  12
A x 2  7 x  12  0
or
 2u  1  0