Part I: Simplifying and Combining 1. 2 30∙3 45

When multiplying square roots, we first rearrage the expression so that the radicals are next to each other. We’re allowed to do this because of the commutativity of multiplication

2∙3 ∙ 30∙ 45

Then, since we are multiplying two square roots, we look for the greatest common factor of the inside part, which is 15 and factor it out as a root:

6 ∙ 15 2∙ 15 3

Then, again, because of the commutativity of multiplication we can rearrange the expression to get:

6 ∙ 15 15 2 3

Then, we simplify:

6 ∙15 6

90 6 2. !! !"− ! !"+ ! !

When combining fractions that are adding and/or subtracting, we need to first find a common denominator. The least common multiple of 21, 14, and 3 is 42. So, we make all the denominators 42 by multiplying by the same number on the top and bottom of the fraction:

−7∙2 21∙2 −

1∙3 14∙3+

2∙14 3∙14

−14−3+28 42 11 42 3. !"# !"#× !" !""

When multiplying fractions, you always look to simplify before you multiply. We first check that each fraction on its own is reduced. In this case, they are both reduced. Then, we check to see if we can cross cancel. In this case, we can because 13 goes into both 169 and 39, and 20 goes into both 140 and 200. So, by cross cancelling we get:

20∙7 13∙13 ×

13∙3 20∙10

4. −1− −9 − −3

Absolute value turns negative values into positive values. So, we get:

−1−9−3

−13

5. −3!! !!_{−} _{−2}!! !!

In this question you need to watch out for following the correct order of evaluation and correctly understanding negative exponents. Negative exponents make reciprocals of the base. So, working from inside out, we get:

−1 9

!!

− −1 4

!!

−9− −64

55

6. 2 75−3 50

When adding or subtracting radicals, we first write each radical in simplest radical form. We do that by factoring out the greatest perfect square factor. The greatest perfect square that goes into 75 is 25. The greatest perfect square that goes into 50 is also 25. So, we factor that out as a radical to get:

2 25 3 −3 25 2

2∙5 3 − 3∙5 2

10 3−15 2

We can’t combine any further because the insides of the radical are different even after reducing.

7. _{!"}! −_{!}!_{!}

When combining fractions by adding or subtracting, you must first find a common denominator. The
least common multiple of 𝑎𝑏 and 𝑎!_{ is }_{𝑎}!_{𝑏}_{. So, we change the denominators of each fraction to }_{𝑎}!_{𝑏}

by multiplying by the same thing on the top and bottom:

2∙𝑎!

𝑎𝑏∙𝑎!−

3∙𝑏
𝑎!_{∙}_{𝑏}

2𝑎!_{−}_{3𝑏}

8. 3𝑥+5 !

Whenever you square a sum or difference, you get the **square-double-square pattern**. So, although
you can figure this out by doing (3𝑥+5)(3𝑥+5) and then distributing, you really need to be able to
go straight to:

square of the first term - double the product - square of the last term

3𝑥 !_{ }_{+} _{ 2 } _{3𝑥} _{5} _{ }_{+}_{ }_{5}!

9𝑥!_{+}_{30𝑥}_{+}_{25}

9. !!"

! ±

!"!!!! !!

!

This problem is just an order of operations question along with evaluating ±. So, we’ll first simplify:

−3± 36−4 5 4

−3± 36−20

4

−3± 16 4

−3± 4 4

−3±1

−3+1 𝑜𝑟 −3−1

Part II: Solving Equations.

1. −14−5𝑥=4𝑥

The first step when solving an equation is to get the variable written once. Since the variable is written twice here and on opposite sides, we need to get rid of all the variables on one side. The easiest way to do that is to get rid of the 5𝑥 on the left side. So, we add 5𝑥 to each side first:

−14−5𝑥=4𝑥

+5𝑥 +5𝑥

−14 =9𝑥

Now, we divide by 9 on both sides to get the variable by iteslf:

−14 9 =

9𝑥 9

−14 9 =𝑥

2. 5+!3𝑥−2=9

The first step when solving an equation is to get the variable written once. That’s already done. So, we move onto the second step when solving an equation, which is to get the variable by itself using the undressing principle. There are 4 numbers on the side of the variable, which means it will take 4 steps to solve this equation. The 5 is happening last, so we must get rid of it first:

5+!3𝑥−2=9

−5 −5

!3𝑥−2=4

Now, the 2 in the root is happening last. So, we will use an exponent to get rid of the 2:

3𝑥−2

! !_{=} _{4} _{!}

3𝑥−2=16

Now, the 2 that is subtracting is happening last. So, we will add 2 to both sides:

3𝑥−2=16 +2 +2

3𝑥 =18

Now, the 3 that is multiplying is happening last. So, we divide by 3 on each side:

3𝑥 3 =

18 3

3. 𝑥−1 3𝑥−5 =(4𝑥+5)(𝑥−5)

The first step when solving an equation is to get the variable written once. In this equation the variable is written 4 times. So, we will:

1. Double-switch

𝑥 + −1 3𝑥 + −5 = 4𝑥+5 𝑥 + −5

2. Distribute

3𝑥!_{+ }_{−5𝑥}_{+} _{−3𝑥}_{+}_{5}_{=}_{4𝑥}!_{+} _{−20𝑥}_{+}_{5𝑥}_{+} _{−25}

3. Combine

3𝑥!_{−}_{8𝑥}_{+}_{5}_{=}_{4𝑥}!_{−}_{15𝑥}_{−}_{25}

Now, we will get rid of the 𝑥’s on one side of the equation:

3𝑥!_{−}_{8𝑥}_{+}_{5} _{=}_{ 4𝑥}!_{−}_{15𝑥}_{−}_{25}

−3𝑥!_{+}_{8𝑥}_{−}_{5 }_{−}_{3𝑥}!_{+}_{ 8𝑥 }_{−}_{5 }

0 = 𝑥! _{−}_{ 7𝑥 }_{−}_{ 30}

Since we can’t combine anymore and the 𝑥 is stil not written once, we need to: 1. Get 0 on one side, which is already done:

0 = 𝑥! _{−}_{ 7𝑥 }_{−}_{ 30}

2. Factor. First we would take out the greatest common factor, but there is none other than 1. So, then we use our factoring trinomials trick, where we find two numbers that multiply to −30 and add to

−7. So, −10 and 3.

0= 𝑥−10 (𝑥+3)

3. Set each factor equal to 0

𝑥−10=0 or 𝑥+3=0

4. Solve each new equation.

𝑥=10 or 𝑥=−3

4. 2𝑥!_{−}_{20𝑥}!_{+}_{42𝑥}_{=}_{0}

The first step when solving an equation is to get the variable written once. The variable is written three times in this equation, so first we try combining. We can’t combine because there are no like terms. So, we must first

1. Get 0 on one side. This is already the case.

2𝑥!_{−}_{20𝑥}!_{+}_{42𝑥}_{=}_{0}

2. Factor. The first step when factoring is to take out the greatest common factor, which in thise case is 2x. So, we get:

2𝑥 𝑥!_{−}_{10𝑥}_{+}_{21} _{=}_{0}

Then, we must factor the trinomial by finding two numbers that multiply to 21 and add to −10. Those two number are −7 and −3. So, we get:

2𝑥 𝑥−7 𝑥−3 =0

3. Set each factor equal to 0:

2𝑥=0 or 𝑥−7=0 or 𝑥−3=0

4. Solve each new equation:

5. _{!!!}! =_{!"}!

The standard way of solving an equation thast has fraction = fraction is to cross multiply.

5 24 = 𝑥+7 𝑥

The first step when solving an equation is to get the variable written once. Since it is written twice, we will

1. Double-switch (nothing to double-switch)

2. Distribute:

120=𝑥!_{+}_{7𝑥}

3. Combine (nothing to combine)

Since, the variable is still written more than once, we must: 1. Get 0 on one side.

120=𝑥!_{+}_{7𝑥}
−120 −120

0=𝑥!_{+}_{7}_{𝑥}_{−}_{120}

2. Factor. First we would take out the greatest common factor, but there is none other than 1. So, then we factor the trinomial by finding two numbers that multiply to −120 and add to 7.

These are −8 and 15. So we get:

0=(𝑥−8)(𝑥+15)

3. Set each factor equal to 0:

𝑥−8=0 or x+15=0

4. Solve each new equation:

6. !!!!!!!

! =5

The first step when solving an equation is to get the variable written once. In this case, the variable is already written once. So, we will move to step two, which is to get the variable by itself by using the undressing principle. In this equation, there are 5 numbers on the same side as the variable. So, it will take 5 steps to solve.

First, since the 9 is happening last, we must get rid of it first.

9∙3 𝑥−4

!_{−}_{6}

9 =5∙9

3 𝑥−4 !_{−}_{6}_{=}_{45}

Then, the 6, which is subtracting, is happening last. So, we will add 6 to both sides:

3 𝑥−4 !_{−}_{6}_{=}_{45}

+6 +6

3 𝑥−4 !_{ }_{=}_{51 }

Then, the 3, which is multiplying, is happening last. So, we will divide by 3 on both sides:

3 𝑥−4 !

3 =

51 3

𝑥−4 !_{=}_{17}

Now, the 3, which is an exponent, is happening last. So, we will take the 3rd_{ root of both sides: }

! 𝑥−4 !_{=}! _{17}

𝑥−4= !17

Lastly, the 4, which is subtracting is happening last. So, we will add 4 to both side:

𝑥−4= !17

+4 +4

𝑥 = 4+! 17

It is not possible to reduce the radical or to combine. So, this is the simplest way to write the answer.

Part III: Functions
1. 𝑦= 𝑥+5 !_{−}_{4}

The vertex on a quadratic occurs whenever it is squaring 0. That means that 𝑥+5 equals 0. So, the 𝑥-coordinate of the vertex will be −5.

Then, plugging that in, we see that the y-coordinate is −4.

So, the vertex is (−5,−4).

The y-intercept occurs when 𝑥=0. So, we plug in 0 for x and get:

𝑦=25−4=21

So, the 𝑦-intercept is (0,21)

The x-intercepts occur whenever 𝑦=0. So, we plug in 0 for 𝑦 and get:

0= 𝑥+5 !_{−}_{4}

We solve this and get

𝑥=−5±!4

𝑥=−5+2 or 𝑥=−5−2

𝑥=−3 or x=−7

So, the 𝑥-intercepts are −3,0 and −7,0 .

2. 𝑦=𝑥!_{+}_{6𝑥}_{+}_{5}

This quadratic is not yet written in vertex form. So, we will first complete the square. To complete the
square when there is no number in front of the 𝑥!_{ term, we do the square of the sum of }_{𝑥}_{ and half the }

middle number, and then subtract the square of that number from the constant term.

𝑦=𝑥!_{+}_{6𝑥}_{+}_{5}
−𝟗
𝑦= 𝑥+𝟑 −4

We got 3 from half of 6. Then, we subtract the square of 3, which is 9, from the 5 to get −4.

Now, we just solve it the same way we solved the previous problem.

The vertex is −3,−4

The 𝑦-intercept is 0,5

The 𝑥-intercepts are −5,0 and −1,0

3. 9𝑥+𝑦=5

−3𝑥−2𝑦=−10

We must solve the system of equations. The easiest way to do that here is through elimination. If we multiply the bottom equation by 3, then the 𝑥 terms will be opposites and we can eliminate them.

9𝑥+𝑦=5 leave alone 9𝑥+𝑦=5

−3𝑥−2𝑦=−10 multiply each term by 3 −9𝑥−6𝑦=−30

Then, we add the two equations:

9𝑥 + 𝑦 = 5 −9𝑥−6𝑦 =−30

−5𝑦 = −25

Then, we solve:

𝑦=5

Then, we plug this back into one of the equations to solve for 𝑥. Let’s plug it into the top one:

9𝑥+5=5

Solving for 𝑥 we get:

𝑥=0

So, the solution is:

0,5

4. −2𝑥+5𝑦=−1

5𝑥+3𝑦=18

Like with problem 3, it will be easiest to solve this by elimination. If we multiply the top equation by 5 and the bottom by 2 then the 𝑥 terms will have coeefficients of −10 and 10 and will eliminate when we add the two equations.

−2𝑥+5𝑦=−1 multiply each term by 5 −10𝑥+25𝑦=−5

5𝑥+3𝑦=18 multiply each term by 2 10𝑥+ 6𝑦 = 36

Then, we add the two new equations:

−10𝑥+25𝑦=−5

10𝑥 + 6𝑦 = 36

31𝑦 = 31

Solving for 𝑦, we get

𝑦=1

We plug this back in to one of the equations to solve for 𝑥. Let’s plug it into the bottom equation:

5𝑥+ 3=18

Solving for 𝑥, we get:

𝑥=3

So, the solution is:

3,1

5. Find the equation of the line through the points 2,−3 with a slope of ! ! .

The first step when finding the equation of a line is to find the slope. Luckily, the problem gives us the slope. So, we have:

𝑦=!_{!}𝑥+𝑏

We just need to figure out what 𝑏 is and then we’ll be done. Since it gives us a point, we can plug the point in to solve for 𝑏. Pluggin in (2,−3) we get:

−3=!

! 2 +𝑏

−3=3+𝑏 −3 −3

−6=𝑏

Now, we put −6 in for b in the original equaion we had and we get:

𝑦=!

!𝑥−6

6. Find the equation of the line through the points −4,3 and 1,−7 .

The first step when finding the equation of a line is to find the slope. The slope is how much the 𝑦

changes every time the 𝑥 goes up by 1. So, we calculate it by finding the change in 𝑦 and divide by the change in 𝑥. In this problem, the y goes from 3 to −7 when the 𝑥 goes from −4 to 1, which means the

𝑦 decreases by 10 when the 𝑥 increases by 5. So, the slope is −10 divided by 5, which is −2.

So, we have:

𝑦=−2𝑥+𝑏

In order to find b, we must plug in one of the points. It doesn’t matter which one. Let’s plug in the first one, (−4,3)

3=−2 −4 +𝑏

3=8+𝑏

−5=𝑏

Plugging 11 back in for b, we get:

𝑦=2𝑥−5