# Week 2: Equations – Answers

Week 2: Equations – Day 5

Deriving Equations

When deriving equations, there are three steps:
– assign variables to quantities
– figure out what principle, law or rule is involved
– use the principle, law or rule to set up an equation involving the variables

1. Set up the equation for solving for the intersection between the line joining (0, 0) and (1, 4) and the line x + y = 1.

(In this problem the variables are already set as x and y. The principle involved is that any two lines that are not parallel intersect at one point. The two equations for the lines will agree at that point.)

The line joining (0, 0) and (1, 4) has slope (4-0)/(1-0) = 4 and y-intercept 0. So its equation is y = 4x. To find the intersection between the two lines we can either substitute y = 4x into x + y = 1 to get:

x + 4x = 1

Or we can solve both equations for y and set them equal to one another:

x + y = 1 corresponds to y = – x + 1

so the equation for the intersection is:

– x + 1 = 4x

2. Set up the equation for solving for the zeros of the function f(x) = 2x3 + 4x2 – 17x – 3.

2x3 + 4x2 – 17x – 3 = 0

3. Set up the equation for solving for the time it takes to drive 432 kilometers at a speed of 65 kilometers per hour.

Let T = time in hours.
Let D = distance = 432 km.
Let S = speed = 65 km/h.

When moving at constant speed, Distance = Speed x Time.
Therefore, T = D/S

Let T = time in hours. Then:
T = 432/65.

4. An isosceles triangle has two angles that are the same: 15 degrees. Set up the equation for solving for the other angle of the triangle.

Let the unknown angle be X.
Principle: the sum of angles of a triangle is 180 degrees.

Equation: X + 15 + 15 = 180

5. Set up the equation for solving for the principal amount of a loan, if the amount of interest owed in one year is \$700 and the interest rate is 5% per year.

Let the principal amount = P.
Let the interest = I = 700.
Let the rate = R = 0.05
Let the time period = T = 1 year.

The principle or rule involved in this problem is that: Interest = Principal x Rate x Time

Therefore I = PRT

And P = I/RT = 700/0.05/1.

Week 2: Equations – Day 4

Without solving these equations, predict the maximum and minimum number of solutions:

1. 3x + 54 = 8x – 6

This equation describes the intersection of two lines, y = 3x + 54 and y = 8x – 6. The lines have different slopes (3 and 8), are therefore not parallel, and will intersect at one point. There is exactly one solution to this equation.

2. 4x – 7 = 4x + 20

This equation describes the intersection of y = 4x – 7 and y = 4x + 20. These two lines have the same slope, 4, and are therefore parallel. They are not the same line, because they have different y-intercepts. Therefore there are exactly zero solutions (no solutions) to this equation.

3. 5 + 4x = 3x2 + 6x

This equation can be rearranged to be the zero of a polynomial of degree 2:

0 = 3x2 + 6x – 4x – 5 = 3x2 + 2x – 5

A polynomial of degree 2 has at most 2 real zeros. There might be 0, 1, or 2 solutions.

4. 10x5 – 47x4 + 3x3 – x2 + 8 = 0

This equation is the zero of a polynomial of degree 5. Because this polynomial has odd degree, there must be at least one solution. There might be 1, 2, 3, 4, or 5 solutions.

5. 1 = x2 + 3

This equation is a zero of a polynomial of degree 2:

0 = x2 + 2

The above equation has no real solutions.

Week 2: Equations – Day 3

Equations Involving Exponential Expressions

Exponential expressions have two basic parts: exponent and base. The simplest equations involve one or the other.

1. If 2300 x 2n = 2501 what is n?

The left hand side is equivalent to:
2300+n
Now the bases are the same on both sides, so we can equate exponents according to the rules of exponents:
300 + n = 501

n = 201.

2. If 2100 x 3100 = a100 what is a?

The left hand side is the product of two bases with the same exponent. So it combines to:
(2 x 3)100
Therefore a = 6.

3. If 100n x 53 = 20n x 563 what is n?

100 = 22 x 52
Therefore the left hand side becomes:
22n x 52n + 3
20 = 22 x 5
Therefore the right hand side becomes:
22n x 5n + 63
The factors involving 2 are the same on both sides, and cancel out.
We are left with the 5s. We can equate the exponents of 5:
2n + 3 = n + 63

Therefore, n = 60.
There are many ways to do this problem.

4. If 249 x 2n = 1/4, what is n?

To solve the equation, we have to group terms with the same base.
The left hand side becomes 249+n and the right hand becomes to 2-2.
Equating exponents:
49 + n = -2

n = -51.

5. If (2n)10 = 326 what is n?

The left hand side is 210n.
The right hand side is (25)6 = 230

Equating exponents, we have:
10n = 30

n = 3.

Week 2: Equations – Day 2

Review of easy equation-solving techniques.

1. How many solutions are there to:

1/(x2 + 1) = 0

There are no solutions. The only way there is a solution is if the numerator is zero. The numerator is 1 which is never zero, so there are no solutions. The denominator is never zero, so you don’t have to worry about undefined values of x.

2. What are the solutions of:

(x – 2)/(x2 + 1) = 0

The only solution is if (x – 2) = 0, or x = 2. The denominator causes no problems because it is always positive.

3. What are the solutions of:

x2 – 28 = 3x

There are two standard ways to solve a problem like this.

One: turn it into an equation equal to zero and factor: x2 – 3x – 28 =0
which factors into (x – 7)(x + 4) = 0 and there are two solutions, x = 7 and x = -4.

Two: complete the square. This method works even if you can’t factor. If there are no solutions, this method will expose that fact.

Bring the x-terms to one side and the constant to the other:
x2 – 3x = 28
Complete the square on the left hand side:
x2 – 3x + 9/4 – 9/4 = 28
(x – 3/2)2 = 28 + 9/4 = (112 + 9)/4 = 121/4
Taking square roots of both sides:
x – 3/2 = + or – 11/2, or x = 3/2 + or – 11/2 = 14/2 or -8/2 = 7 or -4.

4. What are the solutions of:

(x + 3)2 = 25

Like method Two above, take square roots of both sides:
x + 3 = + or – 5 or x = – 3 + or – 5 which is x = 2 or x = -8.
It’s important to check both values of x in the original equation. They both work.

5. Solve: Cube both sides of the equation to get:

x + 4 = 27
x = 23
Plugging into the original equation works, and so this is a valid solution.

Week 2: Equations – Day 1

1.   3x – 3 = 5 – x

Adding x to both sides: 4x – 3 = 5
Adding 3 to both sides: 4x = 8
x = 2

2.   1/2 – 1/x = 1/x
Note: x=0 cannot be a solution.
Multiplying both sides by 2x: x – 2 = 2
x = 4

3. x4 – 1 = 0

Factoring: (x2 + 1)(x2 – 1)
And again: (x2 + 1)(x + 1)(x – 1) = 0
Solutions are x = -1 and x = 1

4. 5x2 – 3 = 2x

Subtracting 3x from both sides: 5x2 – 2x – 3 = 0

Multiply coefficients of x2 and 1 gives 5*(-3) = -15. Factors of -15 that add up to -2 are -5 and 3.
5x2 -5x + 3x – 3 = 0
5x(x – 1) + 3(x – 1) = (5x + 3)(x – 1) = 0
Solutions are x = -3/5 and x = 1.

5. (x2 – 2x – 3)/(x + 1) = 0

Note: x = -1 cannot be a solution.
Multiply both sides by (x + 1): x2 – 2x -3 = 0
Factoring: Factors of -3 that add up to -2 are -3 and 1: (x – 3)(x + 1) = 0
Solution is only x = 3.
x = -1 cannot be a solution.