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.
-
Solve the following system of equations
- x / 2 + y / 3 = 0
x + 6 y = 16
-
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
-
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
-
Solve the following quadratic equation.
0.01 x 2 - 0.1 x - 0.3 = 0
-
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?
-
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)
-
Find four consecutive even integer numbers whose sum is 388.
-
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?
-
Compare the following expressions: a) 2 -4 , b) (-2) -4 , c) (-1/2) 4
-
Evaluate and convert to scientific notation.
(3.4 × 10 11)(5.4 × 10 -3)
-
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
-
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.
-
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.
-
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
-
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.
-
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.
-
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.
-
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.
-
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.
-
(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
More ACT, SAT and Compass practice