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VOCABULARY

LINEAR EQUATIONS

exponents

100

axis of symmetry

axis of symmetry is:

the line that divides a figure into two parts that are mirror images

100

4x−7(2−x)=3x+24𝑥−7(2−𝑥)=3𝑥+2

First, we need to clear out the parenthesis on the left side and then simplify the left side.

4x−7(2−x)=3x+24x−14+7x=3x+211x−14=3x+24𝑥−7(2−𝑥)=3𝑥+24𝑥−14+7𝑥=3𝑥+211𝑥−14=3𝑥+2Hide Step 2

Now we can subtract 3x𝑥 and add 14 to both sides to get all the x𝑥’s on one side and the terms without an x𝑥 on the other side.

11x−14=3x+28x=1611𝑥−14=3𝑥+28𝑥=16Hide

Finally, all we need to do is divide both sides by the coefficient of the x𝑥 (i.e. the 8) to get the solution of x=2𝑥=2.

Now all we need to do is check our answer from and verify that it is a solution to the equation. It is important when doing this step to verify by plugging the solution from Step 3 into the equation given in the problem statement.

Here is the verification work.

4(2)−7(2−2)?=3(2)+28=8

100

2⁹

2^{9 =}512

200

binomial

a polynomial of two terms

200

2(w+3)−10=6(32−3w)

First, we need to clear out the parenthesis on each side and then simplify each side.

2(w+3)−10=6(32−3w)2w+6−10=192−18w2w−4=192−18w2(𝑤+3)−10=6(32−3𝑤)2𝑤+6−10=192−18𝑤2𝑤−4=192−18𝑤

Now we can add 18w𝑤 and 4 to both sides to get all the w𝑤’s on one side and the terms without an w𝑤 on the other side.

2w−4=192−18w20w=1962𝑤−4=192−18𝑤20𝑤=196

Finally, all we need to do is divide both sides by the coefficient of the w𝑤 (i.e. the 20) to get the solution of w=19620=495𝑤=19620=495.

Don’t get excited about solutions that are fractions. They happen more often than people tend to realize.

Now all we need to do is check our answer from Step 3 and verify that it is a solution to the equation. It is important when doing this step to verify by plugging the solution from Step 3 into the equation given in the problem statement.

Here is the verification work.

2(495+3)−10?=6(32−3(495))2(645)−10?=6(135)785=785

200

1²

1²=1

300

coefficient

the numerical factor in a term

300

4−2z3=34−5z6

The first step here is to multiply both sides by the LCD, which happens to be 12 for this problem.

12(4−2z3)=12(34−5z6)12(4−2z3)=12(34)−12(5z6)4(4−2z)=3(3)−2(5z)12(4−2𝑧3)=12(34−5𝑧6)12(4−2𝑧3)=12(34)−12(5𝑧6)4(4−2𝑧)=3(3)−2(5𝑧)

Now we need to find the solution and so all we need to do is go through the same process that we used in the first two practice problems. Here is that work.

4(4−2z)=3(3)−2(5z)16−8z=9−10z2z=−7z=−724(4−2𝑧)=3(3)−2(5𝑧)16−8𝑧=9−10𝑧2𝑧=−7𝑧=−72Hide Step 3

Now all we need to do is check our answer from Step 2 and verify that it is a solution to the equation. It is important when doing this step to verify by plugging the solution from Step 2 into the equation given in the problem statement.

Here is the verification work.

4−2(−72)3?=34−5(−72)64+73?=34−−3526113?=34+3512113=113

300

10¹⁰

10¹⁰=10000000000

400

complex conjugates

number pairs of the form a + bi and a - bi

400

4tt2−25=15−t

Let’s first factor the denominator on the left side so we can identify the LCD. While we are at it we will also factor a minus out of the denominator on the right side.

4tt2−25=15−t4t(t−5)(t+5)=1−(t−5)4t(t−5)(t+5)=−1t−54𝑡𝑡2−25=15−𝑡4𝑡(𝑡−5)(𝑡+5)=1−(𝑡−5)4𝑡(𝑡−5)(𝑡+5)=−1𝑡−5

So, after factoring the left side and factoring the minus sign out of the denominator on the right side we can quickly see that the LCD for this equation is,

(t−5)(t+5)(𝑡−5)(𝑡+5)

From this we can also see that we’ll need to avoid t=5𝑡=5 and t=−5𝑡=−5. Remember that we have to avoid division by zero and we will clearly get division by zero with each of these values of t𝑡.

Next, we need to do find the solution. To get the solution we’ll need to multiply both sides by the LCD and the go through the same process we used in the first couple of practice problems. Here is that work.

(t−5)(t+5)(4t(t−5)(t+5))=−(1t−5)(t−5)(t+5)4t=−(t+5)4t=−t−55t=−5t=−1(𝑡−5)(𝑡+5)(4𝑡(𝑡−5)(𝑡+5))=−(1𝑡−5)(𝑡−5)(𝑡+5)4𝑡=−(𝑡+5)4𝑡=−𝑡−55𝑡=−5𝑡=−1

Finally, we need to verify that our answer from Step 2 is in fact a solution.

The first thing to note is that it is not one of the values of t𝑡 that we need to avoid. Having determined that we know that we do have a potential solution (i.e. it’s not a value of t𝑡 we need to avoid) all we need to do is plug the solution into the equation given in the problem statement.

Here is the verification work.

4(−1)(−1)2−25?=15−(−1)−41−25?=15+116=16

500

complex number

the real numbers AND the imaginary numbers

500

3y+4y−1=2+7y−1

First, we can see that the LCD for this equation is,

y−1𝑦−1

From this we can also see that we’ll need to avoid y=1𝑦=1. Remember that we have to avoid division by zero and we will clearly get division by zero with this value of y𝑦.

(y−1)(3y+4y−1)=(2+7y−1)(y−1)3y+4=2(y−1)+73y+4=2y+5y=1