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

What number is named by 3 × 10² ? Problem #1 © 2007 Herbert I. Gross Answer: 300 next

Answer: 300 Solution: 3 × 10² The PEMDAS agreement tells us that we do exponents before we multiply. Since 10² = 100, the given expression is equivalent to 3 × 100 or 300. next © 2007 Herbert I. Gross

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. next Note 1 © 2007 Herbert I. Gross

next What number is named by (3 × 10)² ? Problem #2 © 2007 Herbert I. Gross Answer: 900 next

Answer: 900 Solution: (3 × 10)² Every thing within parentheses is one number. In this case, since 3 × 10 = 30, the given expression is equivalent to 30² or 30 × 30, which is 900 next © 2007 Herbert I. Gross

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. next Note 2 © 2007 Herbert I. Gross

next What number is named by × 10² ? Problem #3 © 2007 Herbert I. Gross Answer: 703 next

Answer: 703 Solution: × 10² Using PEMDAS we square before we multiply, and we multiply before we add. 10² = 100, so 7 × 10² = 700 and therefore 3 + (7 × 10²) is equivalent to or 703. next © 2007 Herbert I. Gross

In looking at × 10², it is tempting to want to read the expression from left to right and thus add 3 and 7 before multiplying by 10². If this is what we had intended and we were using the PEMDAS agreement we would have had to write (3 + 7) × 10² next Note 3 © 2007 Herbert I. Gross

next Evaluate 5[ 2 (x – 3) + 4] – 6 when x = 7 Problem #4 © 2007 Herbert I. Gross 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 or 12. next © 2007 Herbert I. Gross

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 © 2007 Herbert I. Gross

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. next Note 4 © 2007 Herbert I. Gross

next Note 4 © 2007 Herbert I. Gross a sequence of keystrokes for the algebraic expression would translate as… x- 3×2+4×5×5-6y 7- 3×2+4×5× and if we now replace x by 7 we obtain… next

In recipe format we have … next Note 4 © 2007 Herbert I. Gross Input x7 Subtract 34 Multiply by 28 Add 412 Multiply by 560 Subtract 654 answer next

Evaluate 5[ 2 (x – 3) + 4] ÷ 6 when x is = 7 Problem #5 © 2007 Herbert I. Gross 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. next © 2007 Herbert I. Gross

next For what value of x is it true that… 5[ 2 (x – 3) + 4] – 6 = 84 Problem #6 © 2007 Herbert I. Gross 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 © 2007 Herbert I. Gross

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 © 2007 Herbert I. Gross

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 © 2007 Herbert I. Gross

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. next © 2007 Herbert I. Gross

next Note 6 © 2007 Herbert I. Gross 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 ×2+4×5× next

Note 6 © 2007 Herbert I. Gross Once we decide to use the key stroke model we can solve the problem directly by writing the “undoing” key stroke program. Namely… 10+3÷2- 4÷5÷ next 10- 3×2+4×5×

In recipe format we have … next Note 6 © 2007 Herbert I. Gross Input? Subtract 3? Multiply by 2? Add 4? Multiply by 5? Subtract 6? next Recipe Undoing Recipe ? ? ? ? ? ? Output is Add 3 Divide by 2 Subtract 4 Divide by 5 Add 6 Output is84 ?Input Output is Add 3 Divide by 2 Subtract 4 Divide by 5 Add 6 Input Subtract 3 Multiply by 2 Add 4 Multiply by 5 Subtract 6 Output is

next For what value of x is it true that… 5[ 2 (x – 3) + 4] ÷ 6 = 84 Problem #7 © 2007 Herbert I. Gross 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… next © 2007 Herbert I. Gross

5[ 2 (x – 3) + 4] ÷ 6 = 84 For example, using the calculator key stroke model we see that… next © 2007 Herbert I. Gross ÷2- 4÷5÷5× next

Note 7 © 2007 Herbert I. Gross 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. next

Note 7 © 2007 Herbert I. Gross 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

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 Problem #8 © 2007 Herbert I. Gross Answer: {5[2(x+4)+3]- 23} ÷ 2

Solution: We can begin by rewriting the given recipe step by step in algebraic notation. next © 2007 Herbert I. Gross Step 1:Input xx next Step 2:Add 4x + 4 next Step 3:Multiply by 22(x + 4) Step 4:Add 32(x + 4) + 3 Step 5:Multiply by 55[2(x + 4) + 3] Step 6:Subtract 235[2(x + 4) + 3]- 23 Step 7:Divide by 2 y = {5[2(x + 4) + 3]- 23} ÷ 2Step 8:Output is y {5[2(x + 4) + 3]- 23} ÷ 2

next Note 8 © 2007 Herbert I. Gross It is just as logical to write = as it is to write = 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. next

Note 8 © 2007 Herbert I. Gross 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. next

Note 8 © 2007 Herbert I. Gross 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 next

For what value of x is it true that… {5[ 2 (x +4) +3] – 23} ÷ 2 = 51 Problem #9 © 2007 Herbert I. Gross Answer: 7 next

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… next © 2007 Herbert I. Gross x+4×2+3×5×5-23÷251 next

Solution: {5[ 2 (x +4) +3] – 23} ÷ 2 = 51 and the “undoing” program is then… next © 2007 Herbert I. Gross x+4×2+3×5×5-23÷251 next 7-4÷2-3÷5÷5+23×

next Note 9 © 2007 Herbert I. Gross 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. next

Note 9 © 2007 Herbert I. Gross In this case we start with x and because is inside x the parentheses we next add 4. next 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 next

Note 9 © 2007 Herbert I. Gross 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 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 next

Note 9 © 2007 Herbert I. Gross Notice that if we omitted the braces in the previous step, it would read… 5[2( x + 4) + 3] – 23 ÷ 2 next 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

next Note 9 © 2007 Herbert I. Gross next 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].