# Intermediate Algebra Problems With Answers - Sample 3

A set of intermediate algebra problems related to systems of equations, quadratic equations, ...with answers, are presented. The solutions are at the bottom of the page.

1. Solve the following system of equations
- x / 2 + y / 3 = 0
x + 6 y = 16

2. How many real solutions does each quadratic equation shown below have?
a) x 2 + (4 / 5) x = - 1/4
b) x 2 - 7 x + 10 = 0
c) x 2 - (2/3) x + 1/9= 0

3. Does the data in the table below represents y as a function of x? Explain.
x y
1 - 10 4
11 - 20 6
21 - 30 8
30 - 40 12
40 - 50 16
51 - 60 34

4. Solve the following quadratic equation.
0.01 x 2 - 0.1 x - 0.3 = 0

5. In a cafeteria, 3 coffees and 4 donuts cost $10.05. In the same cafeteria, 5 coffees and 7 donuts cost$17.15. How much do you have to pay for 4 coffees and 6 donuts?

6. Find the slope of the lines through the given points and state whether each line is vertical, horizontal or neither.
Line L1 : (-2 , 3) and (8 , 3)
Line L2 : (4 , 3) and (4 , -3)
Line L3 : (-1 , 7) and (3 , -3)

7. Find four consecutive even integer numbers whose sum is 388.

8. Going for a long trip, Thomas drove for 2 hours and had lunch. After lunch he drove for 3 more hours at a speed that is 20 km/hour more than before lunch. The total trip was 460 km.
a) What was his speed after lunch?

9. Compare the following expressions: a) 2 -4 , b) (-2) -4 , c) (-1/2) 4

10. Evaluate and convert to scientific notation.
(3.4 � 10 11)(5.4 � 10 -3)

1. We first multiply all terms of the first equation by the LCM of 2 and 3 which is 6.
6(-x/2 + y/3) = 6(0)
x + 6y = 16
We then solve the following equivalent system of equations.
-3x + 2y = 0
x + 6y = 16
which gives the solution
x = 8/5 and y = 12/5

2. Find the discriminant of each equation.
a) (4/5) 2 - 4(1)(1/4) = 16/25 - 1 < 0 , no real solutions.
b) (-7) 2 - 4(1)(10) = 49 - 40 = 9 > 0 , 2 real solutions.
c) (-2/3) 2 - 4(1)(1/9) = 4/9 - 4/9 = 0 , 1 real solutions.

3. No y is not a function of x. According to the table, for x = 30 or x = 40 there are two possible values for the output.

4. Multiply all terms of the equation by 100 to obtain an equivalent equation with integer coefficients.
x 2 - 10 x - 30 = 0
Discrminant = (-10) 2 - 4(1)(-30) = 220
Solutions: x = (10 ~+mn~ 2√55) / 2 = 5 ~+mn~ √55

5. Let x be the price of 1 coffee and y be the price of 1 donut.
We now use "3 coffees and 4 donuts cost $10.05" to write the equation 3x + 4y = 10.05 and use "5 coffees and 7 donuts costs$17.15 " to write the equation
5x + 7y = 17.15
Subtract the terms of the first equation from the terms of the second equation to obtain
2x + 3y = 7.10
Mutliply all terms of the last equation to obtain
4x + 6y = 14.2
4 coffees and 6 donuts cost \$14.2.

6. Slope of line L1 = (3 - 3)/(8 + 2) = 0 , horizontal line.
Slope of line L2 = (-3 - 3)/(4 - 4) = -6/0 undefined, vertical line.
Slope of line L3 = (-3 - 7) / (3 + 1) = -10/3, L3 is neither horizontal nor vertical.

7. Let x, x + 2, x + 4 and x + 6 be the 4 consecutive integer.
Their sum is 388, hence the equation: x + (x + 2) + (x + 4) + (x + 6) = 388
Solve the above equation to find x = 94.
The four numbers are: 94, 96, 98 and 100. Add them to check that their sum is 388.

8. Let x be the speed before lunch, hence the distance driven before lunch is equal to 2 x. After lunch his speed is 20 km/hr more than before lunch and is therefore x + 20. The distance after lunch is 3 (x + 20).
The total distance is 460 , hence the equation: 2 x + 3 (x + 20) = 460
Solve the above equation to find speed before lunch x = 80 km/hr
The speed after lunch is 20 km/hr more than before lunch and is therefore equal to 80 km/hr + 20 km/hr = 100 km/hr.

9. We first simplify each expression.
2 -4 = 1 / 2 4 = 1/16
(-2) -4 = 1 / (-2) 4 = 1/16
(-1/2) 4 = (-1) 4 / 2 4 = 1 / 16
The 3 expression simplify to the same value.

10. (3.4 � 10 11)(5.4 � 10 -3) = (3.4 � 5.4) 10 11 x 10 -3 = 18.36 � 10 8 = 1.836 � 10 9

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