### All Algebra 1 Resources

## Example Questions

### Example Question #1 : How To Find Out When An Equation Has No Solution

Solve:

**Possible Answers:**

**Correct answer:**

First factorize the numerator.

Rewrite the equation.

The terms can be eliminated.

Subtract one on both sides.

However, let's substitute this answer back to the original equation to check whether if we will get as an answer.

Simplify the left side.

The left side does not satisfy the equation because the fraction cannot be divided by zero.

Therefore, is not valid.

The answer is:

### Example Question #51 : Systems Of Equations

Solve for :

**Possible Answers:**

No solution

**Correct answer:**

No solution

Combine like terms on each side of the equation:

Next, subtract from both sides.

Then subtract from both sides.

This is nonsensical; therefore, there is no solution to the equation.

### Example Question #1 : How To Find Out When An Equation Has No Solution

Solve the equation:

**Possible Answers:**

No solution

**Correct answer:**

No solution

Notice that the end value is a negative. Any negative or positive value that is inside an absolute value sign must result to a positive value.

If we split the equation to its positive and negative solutions, we have:

Solve the first equation.

The answer to is:

Solve the second equation.

The answer to is:

If we substitute these two solutions back to the original equation, the results are positive answers and can never be equal to negative one.

The answer is no solution.

### Example Question #52 : Systems Of Equations

Solve for x.

**Possible Answers:**

x = –9, –2

x = 6, 3

Cannot be factored by grouping

x = –2

x = –6, –3

**Correct answer:**

x = –6, –3

1) This is a relatively standard quadratic equation. List and add factors of 18.

1 + 18 = 19

2 + 9 = 11

3 + 6 = 9

2) Pull out common factors of each pair, "x" from the first and "6" from the second.

3) Factor again, pulling out "(x+3)" from both terms.

4) Set each term equal to zero and solve.

x + 3 = 0, x = –3

x + 6 = 0, x = –6

### Example Question #53 : Systems Of Equations

Solve for x.

**Possible Answers:**

x = –1

x = –4

x = 4

x = 2, 4

x = 1

**Correct answer:**

x = –1

1) After adding like terms and setting the equation equal to zero, the immediate next step in solving any quadratic is to simplify. If the coefficients of all three terms have a common factor, pull it out. So go ahead and divide both sides (and therefore ALL terms on BOTH sides) by 4.

Since zero divided by four is still zero, only the left side of the equation changes.

2) Either factor by grouping or use the square trick.

Grouping:

1 + 1 = 2

(The "1" was pulled out only to make the next factoring step clear.)

x + 1 = 0, x = –1

OR

Perfect Square:

x = –1

### Example Question #54 : Systems Of Equations

Solve for x.

**Possible Answers:**

x = –1/4

Cannot be factored by grouping

x = –4, 4

x = 4, –1/4

x = –1, 1

**Correct answer:**

x = 4, –1/4

1) Quadratics tend to be easier to follow when stated in order of descending degree. In other words, we need to rearrange the euqation.

2) No other simplification is possible, as there are no common factors between 15 and 4. Multiply the first coefficient by the final term and list off factors.

4 * –4 = –16

Factors of –16 include:

–1 + 16 = 15

1 + –16 = –15

3) Split up the middle term so that factoring by grouping is possible.

4) Factor by pulling the greatest common factor out of each pair of terms, "x" from the first and "-4" from the second.

5) Factor out the "4x+1" from both terms.

6) Set both parts equal to zero and solve.

x – 4 = 0, x = 4

4x + 1 = 0, x = –1/4

### Example Question #55 : Systems Of Equations

Solve for x.

**Possible Answers:**

No solution

**Correct answer:**

There are two ways to do this. One way involves using the quadratic formula. The quadratic formula is written below.

By looking at , a = 7, b = –4, and c = 13. Plug these values into the quadratic equation to find x.

Note that .

Factor out the two, then cancel out that two and separate terms.

This is our answer by the first merthod.

The other method to solve involves completing the square.

Subtract 13 to both sides.

Divide 7 to both sides.

Take the –4/7 from the x-term, cut it in half to get –2/7. Square that –2/7 to get 4/49. Finally, add 4/49 to both sides

Factor the left hand side and simplify the right hand side.

Square root and add 2/7 to both sides.

Don't forget to write it in terms of 'i'.

Note that we should find the same answer by either method.

### Example Question #56 : Systems Of Equations

Billy is several years older than Johnny. Billy is one less than twice as old as Johnny, and their ages multiplied together make ninety-one. When will Billy be 1.5 times Johnny's age?

**Possible Answers:**

When Johnny is 4 and Billy is 6

When Johnny is 14 and Billy is 21

When Johnny is 2 and Billy is 3

When Johnny is 7 and Billy is 13

When Johnny is 12 and Billy is 18

**Correct answer:**

When Johnny is 12 and Billy is 18

1) Before we can figure out when Billy will be 1.5 times Johnny's age, we have to figure out their current ages. So let's define our variables in terms of the first part of the question.

B = Billy's age and J = Johnny's age

It's easier to solve if we put one variable in terms of the other. If Billy were just twice as old as Johnny, we could write his age as B = 2J.

But Billy is *one less* than twice as old as Johnny, so B = 2J – 1

2) We know that the two boys' ages multiply together to make ninety-one.

B * J = J(2J – 1) = 91

3) Now we have our factored quadratic. We just need to multiply it out and set everything equal to zero to begin.

4) Now we need to factor back out. We start by multiplying the first coefficient by the final term and listing off the factors.

2 * –91 = –182

1 + –182 = –181

2 + –91 = –89

7 + –26 = –19

13 + –14 = –1

5) Split up the middle term in so that factoring by grouping is possible.

6) Factor by grouping, pulling out "2J" from the first set of terms and "13" from the second.

7) Factor out the "(J-7)" from both terms.

8) Set both parentheses equal to zero and solve.

2J + 13 = 0, J = –13/2

J – 7 = 0, J = 7

Clearly, only of the two solutions works, since Johnny's age has to be positive. Johnny is 7, therefore Billy is 2(7) – 1=13. But we're not done yet!

9) We need to figure out at what point Billy will 1.5 times Johnny's age. Guess and check would be a fairly efficient way to do this problem, but setting up an equation would be even faster. First, though, we need to figure out what our variable is. We know Billy's and Johnny's current ages; we just need to figure out their future ages. One variable is always better than two, so instead of using two different variables to represent their respective future ages, we'll use one variable to represent the number of years we have to add to each of their current ages in order to make Billy 1.5 times older than Johnny. Let's call that variable "x."

1.5(J + x) = B + x

We know the values of J and B, so we can go ahead and fill those in.

1.5(7 + x) = 13 + x

10) Then we solve for x algebraically, with inverse order of operations.

10.5 + 1.5x = 13 + x

0.5x = 2.5

x = 5

J = 7 + 5 = 12

B = 13 + 5 = 18

### Example Question #1 : Quadratic Equations

Find all of the solutions to the following quadratic equation:

**Possible Answers:**

None of the above

**Correct answer:**

This requires the use of the quadratic formula. Recall that:

for .

For this problem, .

So,

.

.

Therefore, the two solutions are:

### Example Question #57 : Systems Of Equations

Solve for .

**Possible Answers:**

No solution

**Correct answer:**

Write the equation in standard form by first eliminating parentheses, then moving all terms to the left of the equal sign.

First:

Inside:

Outside:

Last:

Now factor, set each binomial to zero, and solve individually. We are lookig for two numbers with sum and product ; these numbers are .

and

or

The solution set is .