### All Algebra 1 Resources

## Example Questions

### Example Question #1 : Setting Up Expressions

We have three cats, Chai, Sora, and Newton. Chai is 3 years old. Sora two years older than twice Chai's age. Newton is one year younger than one-fourth of Sora's age. How old are Sora and Newton?

**Possible Answers:**

Sora: 4 years

Newton: one half year

Sora: 1 year

Newton: 8 years

Sora: 5 years

Newton: 6 years

Sora: 8 years

Newton: 1 year

Sora: 3 years

Newton: not born yet

**Correct answer:**

Sora: 8 years

Newton: 1 year

To make this much easier, translate the word problem into a system of three equations.

We have C for Chai, S for Sora, and N for Newton. To find Sora's age, plug in into .

Sora is 8 years old. Use this to find Newton's age.

Newton is one year old. So the answer is:

Sora, 8 years

Newton, 1 year

### Example Question #21 : Finding Roots

Give all real solutions of the following equation:

**Possible Answers:**

The equation has no real solutions.

**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 36 and sum ; these integers are .

Substitute back:

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

The solution set is .

### Example Question #11 : Solving Rational Expressions

Solve for :

**Possible Answers:**

There is no solution.

**Correct answer:**

Subtract 1 from both sides, then multiply all sides by :

A quadratic equation is yielded. We can factor the expression, then set each individual factor to 0.

Both of these solutions can be confirmed by substitution.

### Example Question #11 : How To Find A Solution Set

Solve for :

**Possible Answers:**

**Correct answer:**

Isolate the radical, square both sides, and solve the resulting quadratic equation.

Factor the expression at left by finding two integers whose product is 65 and whose sum is ; they are .

Set each linear binomial to 0 and solve for to find the possible solutions.

or

Substitute each for .

This is a false statement, so 5 is a false "solution".

This is a true statement, so 13 is the only solution of the equation.

### Example Question #21 : Equations / Solution Sets

Solve the equation for .

**Possible Answers:**

**Correct answer:**

The two sides of the equation will be equal if the quantities inside the absolute value signs are equal or equal with opposite signs.

From this, you get two equations.

or

### Example Question #264 : Equations / Inequalities

Find the solution, in the form , to the following system of equations:

**Possible Answers:**

**Correct answer:**

Multiply (1) by 3, multiply (2) by 4:

Add the two resulting solutions:

Substitute into (1) and solve for :

### Example Question #2391 : Algebra 1

**Possible Answers:**

No solution.

None of the other answers.

**Correct answer:**

To find the solution isolate the variable on one side of the equation.

To check to see if this is the correct solution, plug the value of x back into the equation and solve:

### Example Question #11 : How To Find A Solution Set

Solve the following by substitution:

**Possible Answers:**

No solution

**Correct answer:**

To solve this equation you will plug the second equation straight into the first one by substituting what is written there for .

You will then get:

From here you need to simplify by combining like terms:

Bring the over by addition:

Then divide both sides by to get:

You will then take this value of and plug it into either equation.

### Example Question #82 : Number Theory

Set A is composed of all multiples of 4 that are that are less than the square of 7. Set B includes all multiples of 6 that are greater than 0. How many numbers are found in both set A and set B?

**Possible Answers:**

**Correct answer:**

Start by making a list of the multiples of 4 that are smaller than the square of 7. When 7 is squared, it equals 49; thus, we can compose the following list:

Next, make a list of all the multiples of 6 that are greater than 0. Since we are looking for shared multiples, stop after 48 because numbers greater than 48 will not be included in set A. The biggest multiple of 4 smaller that is less than 49 is 48; therefore, do not calculate multiples of 6 greater than 48.

Finally, count the number of multiples found in both sets. Both sets include the following numbers:

The correct answer is 4 numbers.

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