next Algebra Problems… Solutions Set 2 © 2007 Herbert I. Gross By Herbert I. Gross and Richard A.

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Algebra Problems … Solutions

Set 2

Problem #1

What number is named by 3 × 10

?

Answer: 300

next Answer: 300 Solution:

3 × 10²

The P E MDAS agreement tells us that we do exponents before we multiply. Since 10 ² = 100, the given expression is equivalent to 3

100 or 300.

next Note 1 • If we were using a left-to-right agreement we would first multiply 3 by 10 and then square the result. There is nothing illogical about this. However to avoid misinterpretation, we must all accept the same agreement; and PEMDAS , by and large, is the one that is usually chosen.

Problem #2

What number is named by (3 × 10)

?

Answer: 900

next Answer: 900 Solution:

(3 × 10)²

Every thing within parentheses is 3 one number. In this case, since

30, the given expression is equivalent to 30 ² or 30 × 30, which is 900

next Note 2 • If we are using PEMDAS but wish to multiply 3 by 10 before we square the result, we must use grouping symbols as we did in this problem. In other words for anyone using PEMDAS , 3 × 10² means 3 × 100 or 300.

Problem #3

What number is named by 3 + 7 × 10

?

Answer: 703

next Answer: 703 Solution:

3 + 7 × 10²

Using PEMDAS we square before we multiply, and we multiply before we add. 10 ² = 100, so 7 × 10² = 700 and therefore 3 + (7 × 10²) is equivalent to 3 + 700 or 703.

next Note 3 • In looking at 3 + 7 × 10², it is tempting to want to read the expression from left to right and thus add 3 and 7 before multiplying by 10 ². If this is what we had intended and we were using the PEMDAS agreement we would have had to write (3 + 7) × 10²

Problem #4

Evaluate 5[ 2 (

– 3) + 4] – 6 when

= 7 Answer: 54

next Answer: 54 Solution: 5[ 2 ( x – 3) + 4] – 6 when x = 7 x is inside the parentheses. Replacing x by 7, the number within the parentheses is 7 – 3 or 4.

Thus the bracketed expression becomes 2(4)

4; and since we multiply before we add, the expression within the brackets becomes 8 + 4 or 12.

next 5[ 2 ( x – 3) + 4] – 6 when x = 7 Replacing the bracketed expression by 12, we obtain 5[12] – 6.

And since we multiply before we subtract, we obtain 60 – 6 or 54 as our answer.

next Note 4 • If you are uncomfortable working with the algebraic expression, translate it first into either a verbal recipe or as a step by step key stroke sequence using a calculator.

Note 4 • a sequence of keystrokes for the algebraic expression would translate as… x - 3 ×2 +4 ×5 -6 y and if we now replace x by 7 we obtain… 7 - 3 4 ×2 8 +4 12 ×5 60 -6 54 54

Note 4 • In recipe format we have …

answer

Problem #5

Evaluate 5[ 2 (

– 3) + 4] ÷ 6 when x is = 7 Answer: 10

next Answer: 10 Solution: 5[ 2 ( x – 3) + 4] ÷ 6 when x = 7 Up to the last step the solution here is identical to the solution we had in problem 4. The only difference is that in the last step we divide 60 by 6 rather than subtract 6 from 60.

Problem #6

For what value of

that… is it true 5[ 2 (

– 3) + 4] – 6 = 84 Answer: 10

next Answer: 10 Solution: 5[ 2 ( x – 3) + 4] – 6 = 84 Reading the left side of the equation we see that we obtained 84 after we subtracted 6. So to undo this, we add 6 to both sides of the equation to obtain… 5[ 2 ( x – 3) + 4] = 90 ( 1 )

next 5[ 2 ( x – 3) + 4] = 90 ( 1 ) Referring to equation ( 1 ) we obtained 90 after we multiplied by 5. So to undo this we divide both sides of equation ( 1 ) by 5 to obtain… 2 ( x – 3) + 4 = 18 ( 2 )

next 2 ( x – 3) + 4 = 18 ( 2 ) Referring to equation ( 2 ) we obtained 18 after we added 4 so to undo this we subtract 4 from both sides of equation ( 2 ) to obtain… 2 ( x – 3) = 14 ( 3 )

next 2 ( x – 3) = 14 ( 3 ) We obtained 14 after we multiplied by 2 so to undo this we divide both sides of equation (3) by 2 to obtain… x – 3 = 7 ( 4 ) And finally we add 3 to both sides of equation ( 4 ) to obtain… x = 10.

Note 6 • Remember that you can always check whether your answer is correct. Namely let x = 10 as the input and verify that the output will be 84.

10 - 3 7 ×2 14 +4 18 ×5 90 -6 84 84

Note 6 10 - 3 ×2 +4 ×5 -6 84 • Once we decide to use the key stroke model we can solve the problem directly by writing the “undoing” key stroke program. Namely… 10

10 +3 7 ÷2 14 - 4 18 ÷5 90 +6 84

Note 6 • In recipe format we have …

Recipe Undoing Recipe

Problem #7

For what value of

that… is it true 5[ 2 (

– 3) + 4] ÷ 6 = 84 Answer: 51.4

next Answer: 51.4

Solution: 5[ 2 ( x – 3) + 4] ÷ 6 = 84 Except for the first step in which we begin the “undoing” by multiplying by 6 rather than by adding 6, the procedure in this part is exactly the same as it was in problem 6. For example , using the calculator key stroke model we see that…

5[ 2 ( x – 3) + 4] ÷ 6 = 84

For example , using the calculator key stroke model we see that… 51.4

+3 51.4 ÷2 - 4 ÷5 ×6 84

Note 7 • There is a tendency for students to be suspicious if an answer is anything but a whole number; especially when all of the numbers in the computation are whole numbers. However in the “real world” answers, more often than not, are not whole numbers.

Note 7 • One way of seeing this is to think of the real numbers as being points on the number line. If a point was chosen at random it most likely would fall between two consecutive whole numbers rather than on a whole number.

next Problem #8 Rewrite the following recipe using the PEMDAS agreement Step 1:Input x Step 2:Add 4 Step 3: Multiply by 2 Step 4: Add 3 Step 5: Multiply by 5 Step 6: Subtract 23 Step 7: Divide by 2 Step 8: The output is y

Answer: {5[2(

+4)+3]- 23} ÷ 2

Solution: Answer: {5[2( x +4)+3]- 23} ÷ 2 We can begin by rewriting the given recipe step by step in algebraic notation.

Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: Step 7: Step 8:

Input x Add 4 x x + 4 Multiply by 2 Add 3 2( x + 4) 2( x + 4) + 3 Multiply by 5 Subtract 23 Divide by 2 Output is y 5[2( x + 4) + 3] {5[2( x + 4) + 3]- 23} ÷ 2 y 5[2( x + 4) + 3]- 23 = {5[2( x + 4) + 3]- 23} ÷ 2

Note 8 • It is just as logical to write 4 + 1 = 3 + 2 as it is to write 3 + 2 = 4 + 1

That is the equality of two expressions does not depend on the order in which the expressions are written. Hence it would be just as logical to write the answer in the form. y = {5[2( x + 4) + 3]- 23} ÷ 2 In this respect, it is often customary to write the output by itself on the left side of the equal sign.

Note 8 • We could have used brackets in Step 4 and written [2( x + 4)] + 3 rather than 2( x + 4) + 3 to indicate that we were adding 3 after we multiplied by 2. However this is unnecessary because the PEMDAS agreement tells us that we multiply (and/or divide) before we add (and/or subtract). However, there is no harm in using more grouping symbols than necessary when in doubt.

Note 8 • Again because of the PEMDAS agreement in Step 6, we wrote 5[2( x + 4) + 3]- 23 rather than, say, { 5[2( x + 4) + 3] } - 23 • The grouping symbols can be whatever you wish to use. For example , we could have written the answer as… 5{2[ x + 4] + 3} - 23 ÷ 2

Problem #9

For what value of

that… is it true {5[ 2 (

+4) +3] – 23} ÷ 2 = 51 Answer: 7

Answer: 7 Solution: {5[ 2 ( x +4) +3] – 23} ÷ 2 = 51 The left side of the equation is the same expression that we dealt with in Question 4. In key stroke format the problem becomes… x

+4 ×2 +3 ×5 -23 ÷2 51

Solution: x +4 {5[ 2 ( x +4) +3] – 23} ÷ 2 = 51 ×2 +3 ×5 -23 ÷2 and the “undoing” program is then… 51 7 7 -4 11 ÷2 22 -3 25 ÷5 +23 125 ×2 102

Note 9 • In problem 4 we started with the verbal recipe and converted it into an algebraic equation. Problem 9 was in essence the inverse problem. That is we started with the algebraic equation and converted to the verbal recipe. If we had been given problem 9 without having been given problem 4 first, we might have wanted to convert it first into the less threatening verbal recipe format.

Note 9 • In this case we start with x and because is inside x the parentheses we next add 4.

• We are then inside the brackets where we are multiplying by 2 and then adding 3. Because of PEMDAS we first multiply by 2 and then add 3 to obtain the expression… 2( x + 4) + 3

Note 9 • The entire expression 2( x + 4) + 3 is being multiplied by 5 so we enclose it in brackets and write… 5[2( x + 4) + 3] • Next we subtract 23 from 5[2( x + 4) + 3] and since we multiply before we subtract there is no need to enclose 5[2( x + 4) + 3] in grouping symbols. In this way we obtain the expression… 5[2( x + 4) + 3] – 23

Note 9 • Finally we divide the entire expression 5 [ 2( x + 4) + 3 ] – 23 by 2. So we enclose the expression in braces and write… { 5 [ 2( x + 4) + 3 ] – 23 } ÷ 2 • Notice that if we omitted the braces in the previous step, it would read… 5 [ 2( x + 4) + 3 ] – 23 ÷ 2

Note 9 5[2( x + 4) + 3] – 23 ÷ 2 • In this case the PEMDAS agreement tells us that we divide before we subtract. So we would first divide 23 by 2 and then subtract the result from the expression 5[2( x + 4) + 3].

next Algebra Problems… Solutions Set 2 © 2007 Herbert I. Gross By Herbert I. Gross and Richard A.

Transcript next Algebra Problems… Solutions Set 2 © 2007 Herbert I. Gross By Herbert I. Gross and Richard A.

Algebra Problems … Solutions

Set 2

What number is named by 3 × 10

?

Answer: 300

What number is named by (3 × 10)

?

Answer: 900

What number is named by 3 + 7 × 10

?

Answer: 703

Evaluate 5[ 2 (

– 3) + 4] – 6 when

= 7 Answer: 54

Evaluate 5[ 2 (

– 3) + 4] ÷ 6 when x is = 7 Answer: 10

For what value of

that… is it true 5[ 2 (

– 3) + 4] – 6 = 84 Answer: 10

For what value of

that… is it true 5[ 2 (

– 3) + 4] ÷ 6 = 84 Answer: 51.4

Answer: {5[2(

+4)+3]- 23} ÷ 2

For what value of

that… is it true {5[ 2 (

+4) +3] – 23} ÷ 2 = 51 Answer: 7

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