### All Algebra 1 Resources

## Example Questions

### Example Question #11 : Equations / Solution Sets

Find the solutions to the following equation.

**Possible Answers:**

**Correct answer:**

The first step in solving the equation is to add to both sides so that we can set the new equation equal to so that we can factor it.

We now need to factor it. We need to find two numbers, and , such that and .

Those two numbers are

Our new factored expression becomes

Now we can easily identify that the values that satisfy this equation are .

### Example Question #11 : Equations / Solution Sets

Solve by completing the square:

**Possible Answers:**

no real solution

**Correct answer:**

In order to set this up for completing the square, we need to move the 135 to the other side:

Now the equation is in the form:

To complete the square we need to add to both sides the folowing value:

So we need to add 9 to both sides of the equation:

Now we can factor the left side and simplify the left side:

Now we need to take the square root of both sides:

**NOTE: Don't foget to add the plus or minus symbol. We add this becase there are two values we can square to get 144:

and

**End note

Now we can split into two equations and solve for x:

and

So our solution is:

### Example Question #13 : Equations / Solution Sets

Factor:

**Possible Answers:**

**Correct answer:**

There is a common variable in each term. Pull this out as a common factor.

Factor .

The common factors that will achieve the middle term and will have a product of 36 is four and nine. Write the binomials.

The cubics here cannot be simplified any further.

The answer is:

### Example Question #1 : How To Find A Solution Set

Solve for :

**Possible Answers:**

The equation has no solution.

**Correct answer:**

Move one radical to the other side, then square, thereby yielding an equation with only one radical.

Isolate the radical on one side, then square.

Substitution confirms this to be the only solution.

### Example Question #2 : How To Find A Solution Set

If the area of a rectangle is 100 square feet and the width is 20 feet, then what is the perimeter?

**Possible Answers:**

20 feet

30 feet

80 feet

50 feet

**Correct answer:**

50 feet

The area of a rectangle is , where A is the area, L is the length, and W is the width. The perimeter is given by . We know that and . We can solve for L using The perimeter is then feet.

### Example Question #12 : Equations / Solution Sets

Solve for :

**Possible Answers:**

The equation has no solution.

**Correct answer:**

Cube both sides of the equation to form a linear equation, then solve:

### Example Question #1 : Solving Equations

Solve this system of equations.

**Possible Answers:**

, ,

, ,

, ,

, ,

, ,

**Correct answer:**

, ,

Equation 1:

Equation 2:

Equation 3:

Adding the terms of the first and second equations together will yield .

Then, add that to the third equation so that the y and z terms are eliminated. You will get .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

We can use this z value to find y

So the solution set is x = 1, y = 2, and z = –5/3.

### Example Question #2 : Solving Equations

Solve for :

**Possible Answers:**

**Correct answer:**

To solve this problem we can first add to each side of the equation yielding

Then we take the square root of both sides to get

Then we calculate the square root of which is .

### Example Question #1 : Graphing Linear Functions

Determine where the graphs of the following equations will intersect.

**Possible Answers:**

**Correct answer:**

We can solve the system of equations using the substitution method.

Solve for in the second equation.

Substitute this value of into the first equation.

Now we can solve for .

Solve for using the first equation with this new value of .

The solution is the ordered pair .

### Example Question #1 : Solving Non Quadratic Polynomials

Give all real solutions of the following equation:

**Possible Answers:**

**Correct answer:**

By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

Set each factor to zero and solve:

Since no real number squared is equal to a negative number, no real solution presents itself here.

The solution set is .

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