Solving Equations and Inequallities - College Algebra

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Question

Simplify the radical:

$\frac{2\sqrt{20}}{\sqrt{4}}$

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Answer

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.

$\frac{2\sqrt{20}}{\sqrt{4}}=\frac{2\sqrt{4}\sqrt{5}}{\sqrt{4}}=2{\sqrt{5}}$

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Question

Solve the following

$\left | 3x-5\right |=20$

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Answer

We have to set up two equations, which are.

Now lets solve for x in each equation.

$3x=25\rightarrow x=\frac{25}{3}$

$3x=-15\rightarrow x=-5$

So the solutions are and $x=\frac{25}{3}$

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Question

Solve:

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Answer

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

$\x-10=20\x-10=-20$

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

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Question

Solve:

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Answer

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

$\x-45=2\x-45=-2$

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

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Question

Solve the following:

$\left | x-2 \right |>1$

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Answer

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

and

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

$x-2>1\Rightarrow x>3$

$x-2<-1\Rightarrow x<1$

$(-\infty,1)\cup (3,\infty)$

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Question

Solve the following equation for .

$\left |x^2-8 \right |=1$

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Answer

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

$x=\pm3$

Now for the second equation,

$x=\pm\sqrt{7}$

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Question

Solve the following equation:

$-3-\left | 8x-6\right |=3$

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Answer

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

$-3-\left | 8x-6\right |=3$

$-\left | 8x-6\right |=6$

$\left | 8x-6\right |=-6$

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

$x=\frac{12}{8}=\frac{3}{2}: :or: :x=0$

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Question

Solve the equation:

$-5\left | 4+2x\right |=-100$

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Answer

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

$-5\left | 4+2x\right |=-100$

$\left | 4+2x\right |=20$

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

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Question

Solve for x:

$\left | x-7 \right | = 3$

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Answer

To solve $\left | x-7\right | = 3$ 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, $\left | x-7\right | = \left | (-1) (x-7)\right |$ . So actually solving the original equation comes down to solving the following two equations:

So we get the two solutions as:

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Question

Solve the inequality:

$8+\left | 4x-7\right |\geq17$

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Answer

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

$8+\left | 4x-7\right |\geq17$

$\left | 4x-7\right |\geq9$

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

$4x-7\geq9;or;4x-7\leq-9$

$4x\geq16;or;4x-7\leq-2$

$x\geq4:or: x\leq-\frac{1}{2}$

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Question

Larry has a handful of dimes and quarters. In total, he has 14 coins with a value of $2.60. How many of each coin does he have?

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Answer

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.

$Q=\frac{120}{15}=8$

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

Larry has 6 dimes and 8 quarters

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Question

Solve:

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Answer

Add on both sides.

Add 8 on both sides.

Divide by two on both sides.

$\frac{2x}{2}<\frac{28}{2}$

The answer is:

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Question

Quadratic equations appear often in physics. The basic kinematic equations for the position of a particle as a function of time , with an initial velocity (a constant) and constant acceleration can be written as,

$x(t)=x_o +v_ot+\frac{1}{2}at^2$

This is a quadratic function in . The function therefore gives the position as a quadratic function of time . If we are dealing with a free-falling object under Earth's gravitational field, we might write this function in the form,

$h(t)=h_o +v_ot +\frac{1}{2}gt^2$

to express the "height" of the object at a given time falling with a constant acceleration . Here the initial height (a constant). The units for acceleration are meters-per-square second . The negative acceleration is a convention to signify that the direction of the acceleration is downward.

Find the time required for a ball dropped from a height of 100 m from rest to reach the ground using the quadratic function for height written below,

$h(t)=h_o +v_ot +\frac{1}{2}gt^2$

(Hint, what is the value of the height when the ball strikes the ground?).

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Answer

We're given the function,

$h(t)=h_o +v_ot +\frac{1}{2}gt^2$

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:

$h(t)=100 + (0)t-\frac{1}{2}(9.8)t^2=0$ (units omitted in the equation).

$t =\sqrt{ \frac{100}{4.9}}\approx 4.5 s$

(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).

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Question

A sidewalk on a street corner reaches from a library to a bookstore making an L shape around that corner. The L shaped path has one length that is twice as long as the other. The diagonal path being built between the two buildings will be 102 feet long. How many feet shorter is the diagonal path than the traditional L shaped path around the corner?

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Answer

Use Pythagorean Theorem to solve.

$5x^2=10,404\rightarrow x^2=2080.8 \rightarrow x=\pm \sqrt{2080.8}$

$x=\sqrt{2080.8}$

Since the path around the corner is 3x, 3 times the $\sqrt{2080.8}$ 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.

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Question

Solve

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Answer

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.

$x^2=3\rightarrow x=\pm\sqrt{3}$

$x^2=4\rightarrow x=\pm\sqrt{4}=\pm2$

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Question

Solve for x and y in the following pair of equations

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Answer

There are two ways to solve for x and y in this pair of equations. First, you can add the two equations together and cancel out one of the variables. In this case, you can see that there is no need to multiply one of the equations by a positive or negative number, as 2y and -2y already cancel each other out. The second way is to solve for one of the variables, then substitute that value into the other equation to solve for the other variable. In this case, I will show how to add both equations together to solve for x and y since the equations as they are cancel the y variable out.

You can see how adding the two equations cancel the y variable.

now just divide both sides by four to solve for x:

${\color{Red} x=1}$

now just plug the value of x into either equation to solve for y.

Let's pick the top equation:

now subtract 1 from each side to isolate y:

now divide both sides by two to solve for y:

${\color{Red} y=2}$

You can check your answers by plugging them into either equation, and since they are equations, both sides will be equal if the answers are correct.

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Question

Solve for X:

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Answer

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

$\frac{Y-3}{8}=X$

Re-written the answer becomes

$X=\frac{Y-3}{8}$

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Question

Solve:

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Answer

Step 1: Subtract 6 from both sides...

Step 2: Divide by 2.

$\frac {2x}{2}=\frac {6}{2}$

Simplify:

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Question

The town of Marysville is hosting its annual festival. For admission tickets, adults are charged $8, and children are charged $4. The town sold 2300 tickets and took in $13,688 in total ticket revenue. How many more children were admitted to the festival than adults?

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Answer

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.

$4c+8a=13\textup{,}688$

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

$-4(c+a=2300) \rightarrow -4c-4a=-9200$

Now add the two equations together to eliminate a variable.

$4c+8a=13\textup{,}688$

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

Solve for .

$a=\frac{4488}{4}=1122$

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.

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Question

Solve for X and Y for the following pair of equations

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Answer

There are two ways to solve for x and y in a pair of equations. One way is to add the two equations together and eliminate one of the variables. It may be necessary to multiply one equation by a positive or negative number in order to cancel out one of the variables. The second way is to pick one of the equations and solve for one of the variables. Let's pick the top equation and solve for x:

Now substitute x on the other (bottom) equation:

${\color{Red} y=5}$

Now, since we know the value of y, use either equation and "plug in" the value of y:

$2x-4({\color{Red} 5})=24$

${\color{Red} x=2}$

To check your answers, you can plug in both answers to either equation, and since they are equations, if both sides are equal, your answers are correct.

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