Part I: 1. !
!!!− !
!!!
For this problem, we first need to find the least common multiple of the denominators in order to combine fractions through adding or subtracting. The least common multiple of 𝑎!𝑏 and 𝑎𝑏! is 𝑎!𝑏!. So, we need to get the
denominators of the original problem to both be 𝑎!𝑏!. Don’t forget whatever you multiply the bottom by, you must multiply the top by as well.
2∙𝑏! 𝑎!𝑏∙𝑏! −
4∙𝑎 𝑎𝑏!∙𝑎 =
2𝑏! −4𝑎 𝑎!𝑏!
2. 4𝑥! !
You just need to remember that exponents distribute across multiplication (order 3 to order 2). So we, get.
4!∙ 𝑥! ! 64𝑥!"
It’s important to remember that when you have an expression with an exponent to an exponent you multiply the two exponents. For instance, in this problem we’re not doing 5!, but 𝑥! !. These are very different problems.
3. 𝑥−5 𝑥+5
This is a special pattern that you need to have memorized. This is called the product of the sum and difference which is the difference of two squares. We can also figure it out by double-switch and distribute, but you need to recognize this pattern when you see it.
𝑥! −5!
4. (−4)!!
Negative exponents make reciprocals and become positive. So, we get:
(−4)!!
1 −4 !
1 −64
5. !"!"∙!"!"
This is just a problem where you have to cross cancel. You should see that the 10 and the 15 becomes 2 over 3 and the 14 over 21 becomes 2 over 3.
10 2 21 3∙
14 2 15 3=
4 9
6. ! 36𝑥!"
This problem is similar to problem 4. Instead of distributing an exponent, we are distributing a root. So, we take the square root of 36, and then the square root of 𝑥!"
36𝑥!"
!
36 !
∙ ! 𝑥!" 6∙𝑥!
7. 3𝑥!−5𝑥!+𝑥−3 − 𝑥!−2𝑥!+𝑥−8
The most important thing here is to watch out for the subtraction sign in the middle. You must double-switch and distribute the negative. If you do that correctly, you will almost certainly get the problem right.
3𝑥! +−5𝑥!+𝑥+−3 +− 𝑥!+−2𝑥!+𝑥+−8
3𝑥!+−5𝑥! +𝑥+−3+−𝑥! +2𝑥!−𝑥+8
Part II:
1. 𝑥− 3𝑥−2 =0
First, we double-switch, distribute, and combine in order to get 𝑥 written once.
𝑥+ − 3𝑥+ −2 =0
𝑥+ −3𝑥+2=0
−2𝑥+2=0
Then, we get 𝑥 by itself using the undressing principle.
−2𝑥+2=0 −2 −2
−2𝑥= −2
−2𝑥
−2 =
−2 −2
2. 2𝑥!−𝑥−21=0
The first step when solving an equation is getting the 𝑥 written once. In this case, we can’t combine the 𝑥’s.
When we can’t combine the 𝑥’s, you must:
1. Get 0 on one side 2. Factor
3. Set each factor equal to 0 4. Solve each new equation.
We already have 0 on one side, so now we have to factor. The first step in factoring is taking out the greatest common factor. There is none other than 1. Then, we have to use our special factoring trick for trinomials. We have to find what multiplies to −42 and adds to −1. The two numbers that work are −7 and 6. So, we split up the middle term into two terms with those numbers as coefficients.
2𝑥!−7𝑥+6𝑥−21=0
Then we split it up into two parts and factor out the greatest common factor:
2𝑥! −7𝑥 | +6𝑥−21=0
𝑥 2𝑥−7 +3 2−7 = 0
We check that we haven’t made a mistake since the two factors in parentheses are the same. Then, you combine:
𝑥+3 2𝑥−7 =0
After you factor, you must set each factor equal to 0.
𝑥+3= 0 𝑜𝑟 2𝑥−7=0
Then, we solve each:
3. 3𝑥!+4𝑥= 0
The first step when solving an equation is getting the 𝑥 written once. In this case, we can’t combine the 𝑥’s.
When we can’t combine the 𝑥’s, you must:
1. Get 0 on one side 2. Factor
3. Set each factor equal to 0 4. Solve each new equation.
We already have 0 on one side, so now we have to factor. The first step in factoring is taking out the greatest common factor. The greatest common factor here is 𝑥. So, we first factor out the 𝑥 and get:
3𝑥!+4𝑥= 0
𝑥 3𝑥+4 =0
Then, we set each factor equal to 0.
𝑥=0 𝑜𝑟 3𝑥+4=0
Then, we solve each equation.
4. 𝑥!+4𝑥= 12
The first step when solving an equation is getting the 𝑥 written once. In this case, we can’t combine the 𝑥’s.
When we can’t combine the 𝑥’s, you must:
1. Get 0 on one side 2. Factor
3. Set each factor equal to 0 4. Solve each new equation.
In order to get 0 on one side we subtract 12 from each side.
𝑥!+4𝑥= 12 −12 −12
𝑥!+4𝑥−12= 0
Now that we have 0 on one side, we factor. The first step in factoring is to take out the greatest common factor. The greatest common factor here is 1. So, nothing to factor out. Then, we have to use our special factoring trick for trinomials. We need to find what two numbers multiply to −12 and add to 4. The two numbers that work are 6 and −2. So, we get:
𝑥+6 𝑥−2 =0
After you factor, you set each factor equal to 0. So, we get:
𝑥+6= 0 𝑜𝑟 𝑥−2=0
Then, we solve each new equation.
5. 𝑥 𝑥−4 −3 𝑥−3 = −1
The first step when solving an equation is getting the 𝑥 written once. So, we will first try to combine the 𝑥’s by following double-switch, distribute, combine.
𝑥 𝑥+ −4 + −3 𝑥+ −3 =−1
𝑥!+ −4𝑥+ −3𝑥+9=−1
𝑥!+ −7𝑥+9= −1
Now, we can’t combine anymore. When we can’t combine the 𝑥’s, you must:
1. Get 0 on one side 2. Factor
3. Set each factor equal to 0 4. Solve each new equation.
So, first we’ll get 0 on one side by adding 1 to both sides.
𝑥!+ −7𝑥+9= −1 +1 +1
𝑥!+ −7𝑥+10=0
Then, we factor. The first step in factoring is taking out the greatest common factor. The greatest common factor here is 1. So, nothing to take out. Then, we need to use our special factoring trick for trinomials. We must see which two numbers multiply to 10 and add to −7. The two numbers that work are −5 and −2. So, we get:
𝑥−5 𝑥−2 =0
Then, we set each factor equal to 0:
𝑥−5= 0 𝑜𝑟 𝑥−2=0
Then, we solve each of the new equations:
6. 2𝑥 𝑥−3 = 14−3 𝑥−2
The first step when solving an equation is getting the 𝑥 written once. So, we will first try to combine the 𝑥’s by following double-switch, distribute, combine.
2𝑥 𝑥+ −3 =14+ −3 𝑥+ −2 2𝑥!+ −6𝑥=14+ −3𝑥+6
2𝑥!+ −6𝑥=20+ −3𝑥
Now, we can’t combine the 𝑥’s. When we can’t combine the 𝑥’s, you must: 1. Get 0 on one side
2. Factor
3. Set each factor equal to 0 4. Solve each new equation. So, we’ll first get 0 on one side.
2𝑥!+ −6𝑥 = 20+ −3𝑥 +3𝑥−20 −20 +3𝑥
2𝑥!+ −3𝑥−20=0
Now that we have 0 on one side, we factor. The first step in factoring is to take out the greatest common factor. The greatest common factor here is 1. So, nothing to factor out. Then, we have to use our special factoring trick for trinomials. We have to find what multiplies to −40 and adds to −3. The two numbers that work are −8 and 5. So, we split up the middle term into two terms with those numbers as coefficients.
2𝑥!+ −8𝑥+5𝑥−20=0
Then we split it up into two parts and factor out the greatest common factor:
2𝑥!+−8𝑥 | +5𝑥−20= 0
2𝑥 𝑥−4 +5 𝑥−4 =0
We check that we haven’t made a mistake since the two factors in parentheses are the same. Then, you combine:
2𝑥+5 𝑥−4 =0
Then, you set each factor equal to 0.
2𝑥+5= 0 𝑜𝑟 𝑥−4=0 Then, solve each new equation:
𝑥=−5
Part III:
1. 𝑦=𝑥!−1
The easiest way to do this it to set up a table of values and then graph the points.
2. 𝑦=2𝑥−4
Likewise, the easiest thing to do is to set up a table of values and graph the points. 𝑥 𝑦 Ordered
Pair −3 8 (−3,8) −2 3 (−2,3)
−1 0 (−1,0)
0 −1 (0,−1) 1 0 (1,0) 2 3 (2,3)
𝑥 𝑦 Ordered Pair −3 −10 (−3,−10) −2 −8 (−2,−8)
−1 −6 (−1,−6)
0 −4 (0,−4)
1 −2 (1,−2)
3. 𝑦=3𝑥−5 𝑦= 𝑥−7
We need to solve using substitution. So, we find what either y or x equals and circle the other side. Then, plug it into the y or x in the other equation. In this case, we circle 3𝑥−5 and then plug it in for the y in the second equation, which gets us.
3𝑥−5= 𝑥−7 Then, we solve this.
3𝑥−5= 𝑥−7
−𝑥 −𝑥
2𝑥−5= −7 +5 +5
2𝑥 =−2
𝑥=−1
Then, we plug this in for 𝑥 in one of the original equations and solve for y. Plugging it back into the first equation we get:
𝑦 =3 −1 −5 𝑦=−3−5
𝑦 =−8
So, our solution is −1,−8
4. This is solved in the exact same manner as number 3. The solution is (3,−1).
5. To write a linear equation you first find the slope, then the y-intercept. The slope is the rate of change. So, in this case the slope is 4, because that’s how much money he gets every hour. The y-intercept is how much he starts with, which the problem tells us is 6. So, since we know the slope and y-intercept, we can just plug them into the form 𝑦 =𝑚𝑥+𝑏 where m is the slope and b is the y-intercept.
So, we get 𝑦=4𝑥+6
6. Likewise, we need to find the slope and the y-intercept. However, this one is a bit trickier because we need to figure those out since they’re not given. The slope is how much his number of tokens changes by ever hour. After 5 hours, he had 25 tokens and after 8 hours, he had 19. So, that means he lost 6 tokens in 3 hours, which means he loses 2 tokens per hour. So, his slope is −2. Then, we need to figure out how many he started with to get the y-intercept. He had 25 tokens after 5 hours. Since he uses 2 per hour. He must have used 10 tokens in those 5 hours, which means he must have started with 35, which is our y-intercept. So, our equation is:
