Intermediate Algebra Problems With Answers 
Sample 1
A set of intermediate algebra problems, with answers, are presented. The solutions are at the bottom of the page.

For what value of the constant K does the equation 2x + K x = 3 have one solution?.

For what value of the constant K does the equation K x^{2} + 2x = 1 have two real solutions?

For what value of the constant K does the system of equations 2x  y = 4 and 6x  3y = 3K have an infinite number of solutions?

Find Y so that the points A(0 , 0), B(2 , 2) and C(4 , Y) are the vertices of a right triangle with hypotenuse AC.

Find M so that the lines with equations 2x + My = 5 and 4y + x = 9 are perpendicular.

The length of a rectangular field is 7/5 its width. If the perimeter of the field is 240 meters, what are the length and width of the field?

Peter wants to get $200,000 for his house. An agent charges 20% of the selling price for selling the house for Peter.
a) What should be the selling price?
b) What will be the agent's commission?

John's annual salary after a raise of 15% is $45,000. What was his salary before the raise?

It took Malcom 3.5 hours to drive from city A to city B. On his way back to city A, he increased his speed by 20 km per hour and it took him 3 hours.
a) Find the average speed for the whole journey.

Let R be a relation defined by R = {(4,3),(x^{ 2},2),(1,6),(4,0)}.
a) Find all values of x so that R is not a function.

Solve for x to obtain: x = 3/(2 + k)
The given equation has one solution for all real values of k not equal to 2.

Rewrite the given quadratic equation in standard form: Kx^{ 2} + 2x  1 = 0
Discriminant = 4  4(K)(1) = 4 + 4K
For the equation to have two real solutions, the discriminant has to be positive. Hence we need to solve the inequality 4 + 4K > 0.
The solution set to the above inequality is given by: K > 1 for which the given equation has two real solutions.

For the system of equations 2x  y = 4 and 6x  3y = 3 K to have an infinite number of solutions, the two equations must be equivalent.
If we multiply all terms of the first equation by 3, we obtain
6x  3y = 12
For the two equations to be equivalent we must have 12 = 3K which when solved gives K = 4.

Let us first find the square of the lengths of the sides of triangle A, B and C.
Hypotenuse: AC^{ 2} = 16 + y^{ 2}
Side AB: AB^{ 2} = 4 + 4 = 8
Side BC: BC^{ 2} = (y  2)^{ 2} + 36
We now apply Pythagora's theorem: 16 + y^{ 2} = 8 + (y  2)^{ 2} + 36
Solve the equation to find y = 8.

We first find the slopes of the two lines: 2/M and 1/4
For two lines to be perpendicular, the product of their slopes must be equal to 1. Hence the equation (2/M)*(1/4) = 1. Solve for M to find M = 1/2.

Let L be the length and W be the width. L = (7/5)W
Perimeter: 2L + 2W = 240, 2(7/5)W + 2W = 240
Solve the above equation to find: W = 50 m and L = 70 m.

Let x be the selling price: x  20%x = 200,000
a) Solve for x to find x = $250,000
b) 20% *250,000 = $50,000

Let x be the salary before the increase. Hence x + 15% x = $45,000
Solve for x to find x = $39,130

Let x and x + 20 be the speeds of the car from A to B and then from B to A. Hence the distance from A to B may expressed as 3.5 x and the distance from B to A as 3(x + 20)
The average speed = total distance / total time = (3.5 x + 3 (x + 20)) / (3.5 + 3)
The distance from A to B is equal to the distance from B to A, hence: 3.5 x = 3(x + 20). Solve for x to obtain x = 120 km/hr.
We now substitute x by 120 in the formula for the average speed to obtain.
average speed = 129.2 km/hr

Relation R is not a function if x^{ 2} = 4 or x^{ 2} = 1
Hence, R is not a function fo the following values of x: 2, 1, 1 and 2.
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