Solving Equations and Inequallities - College Algebra

We need to set up two equations since we are dealing with absolute value

We solve for , in each equation to get the solutions.

We need to set up two equations since we are dealing with absolute value

We solve for , in each equation to get the solutions.

To solve, you must split the absolute value into the two following equations.

and

Now, solve for x and right it in interval form.

We first need to get rid of the absolute value symbol to solve the equation. TO break this absolute value, we assign two values to the right hand side, as shown below.

We now proceed to solve each equation independently.

Starting with the first equation, we get

Now for the second equation,

To simplify the radical, break the radical in the numerator down into its factors. When doing so, the radical in the bottom will call with one from the numerator.

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

Recall that the absolute value can come from either a negative or positive value therefore two possible equations are set up.

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

Since absolute values can come from either negative or positive values, two equations need to be set up and solved for.

To solve we need to find a way to reverse the operation of taking the absolute value. What we need to do is think about what the absolute value operation does to an expression. Since it makes everything positive, . So actually solving the original equation comes down to solving the following two equations:

So we get the two solutions as:

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

Since absolute values can come from either negative or positive values, two equations need to be set up and solved for.

Add on both sides.

Add 8 on both sides.

Divide by two on both sides.

The answer is:

We're given the function,

We know that the gravitational acceleration on earth is:

(meters-per-square second)

Because the ball starts at rest, the initial velocity is zero,

(meters-per-second)

We are given the initial height,

(meters)

When the ball reaches the ground, the height is "zero," so the value of is zero at this time, and so we have:

(units omitted in the equation).

(Note that although taking the square root of both sides of an equation will produce positive and negative solutions, we ignore the negative solution since "negative time" makes no sense in the context of this problem).

Use Pythagorean Theorem to solve.

Since the path around the corner is 3x, 3 times the is approximately 137. The distance saved is 137-102= 35 feet. Note that we only needed the positive square root since we cannot have a negative distance.

To make this problem easier, lets start off by doing a u-substitution.

Let .

Now we can factor the left hand side.

We have two solutions for , now we can plug those into , to get all the solutions.

Find the roots of the polynomial,

Set equal to

Factor out ,

Notice that the the factor is a quadratic even though it might not seem so at first glance. One way to think of this is as follows:

Let

Then we have , substitute into to get,

Notice that the change in variable from to has resulted in a quadratic equation that can be easily factored due to the fact that it is a square of a simple binomial:

The solution for is,

Because we go back to the variable ,

Therefore, the roots of the factor are,

The other root of is since the function clearly equals when .

The solution set is therefore,

Below is a plot of . You can see where the function intersects the -axis at points corresponding to our solutions.

Further Discussion

The change of variable was a tool we used to write the quadratic factor in a more familiar form, but we could have just factored the original function in terms of as follows,

Setting this to zero gives the same solution set,

can be rewritten in quadratic form by setting , and, consequently, ; the resulting equation is as follows:

By the reverse-FOIL method we can factor the trinomial at left. We are looking for two integers with sum 8 and product 12; they are 2 and 6, so the equation becomes

Setting both binomials equal to 0, it follows that

or .

Substituting for , we get

,

in which case

,

or

in which case

.

The solution set is .

can be rewritten in quadratic form by setting , and, consequently, ; the resulting equation is as follows:

By the reverse-FOIL method we can factor the trinomial at left. We are looking for two integers with sum 8 and product 12; they are and , so the equation becomes

Setting both binomials equal to 0, it follows that

or .

Substituting for , we get

in which case

,

and

,

in which case

The set of real solutions is therefore .

In , if we're solving for x, we first need to get the "x" term isolated. We do this by subtracting 3 from both sides so:

becomes

Now we divide both sides by 8

Re-written the answer becomes

Since this problems has 2 variables (D-dimes and Q-quarters) we need 2 equations. Because Larry has 14 coins, the first equation can be written as:

The value of those coins equals $2.60 or 260 cents. If Dimes are worth 10C and quarters are 25C, the next equation can be written as

To solve this write both equations on top of each other

Now we eliminate 1 variable by multiplying 1 equation by the lowest common denominator (as a negative) and adding the equations together.

becomes

adding the equations

-----------------------------------

now we solve for Q.

Since we know Q, now we plug it back in to an equation and find D

Larry has 6 dimes and 8 quarters

This question requires setting up two equations and then solving the system of two equations. One equation represents the total number of tickets sold, and the other equation represents the total revenue. Both equations will use the same two variables, and , for the number of children and adults who were admitted, respectively.

The first equation is the easier of the two. The sum of the number of children and the number of adults admitted to the fair will equal 2300.

The second equation will incorporate the price charged to children and adults

Children's tickets cost $4 and adults' tickets cost $8. The total amount of money raised was $13,688.

To solve the system of equations, you can eliminate one of the variables. One way to do this is via the elimination method. To eliminate

Now add the two equations together to eliminate a variable.

--------------------------------

Solve for .

Substitute into either of the original equations and solve for .

There were 1178 children and 1122 adults admitted to the festival. Finally, subtract the results.