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Algebra I · Lesson 2.2

Evaluating in Practice

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Evaluating was settled in 1.6, so nothing about the method changes here, the expressions just get heavier. Substitute in parentheses, compute by the order of operations, read off one number. This lesson is six problems, and each one adds a step the last did not have.

Problem
Evaluate 5−4y at y=−7. There are two minus signs in the second term, one printed in the expression and one that comes with the value, so write the substitution with parentheses to keep them separate.
Show a hint
  • Read 4(−7) as one finished product before subtracting anything.
Show the full solution
Substituting in parentheses gives 5−4(−7). The product is 4(−7)=−28, and subtracting −28 is adding 28, so the value is 5+28=33. Losing the value's sign turns the line into 5−28=−23, which is exactly what the parentheses prevent.
Problem
Evaluate 3k4+20k at k=8. The variable sits under a fraction bar this time, and it appears in two places at once. Enter the value as a fraction in lowest terms.
Show a hint
  • Both ks become 8 at the same moment, one above a bar and one below, and each bar finishes as its own division.
Show the full solution
Substituting gives 3(8)4+208. Each bar finishes as its own division, 244=6 and 208=52, so the value is 6+52=172. Combining across the bars into 24+204+8 merges two separate divisions into one and is wrong.
Problem
Evaluate −12w at w=−3. The minus printed under the bar looks like the even-root trouble from 1.5. Substitute first, then decide whether the root exists.
Show a hint
  • Compute the number actually under the bar once w is replaced, then judge.
Show the full solution
Substituting puts −12(−3)=36 under the bar, and 36=6. The minus printed in −12w is not the sign of the value, so rejecting the root on sight would throw away a perfectly good number.
Judge the value after substituting, not the printed sign√(-4q)at q = -25looks negativeWrong√100 = 10CorrectWork inside out(2u + 3)²at u = 9(2·9 + 3)²21²441
On top, rejecting −4q at sight from the printed minus judges the expression instead of the value. Substituting q=−25 puts −4(−25)=100 under the bar, so the root exists and equals 10. Below, (2u+3)2 at u=9 runs inside out, the parentheses finish as 21, and squaring last gives 441.
Problem
Evaluate n2−5n+3 at n=−2. Three terms, three sign decisions, one value, and every appearance of n gets its own parentheses.
Show a hint
  • Write the parentheses at both appearances of n before doing any arithmetic.
  • Subtracting 5(−2) is subtracting a negative.
Show the full solution
Substituting at both appearances gives (−2)2−5(−2)+3. That is 4+10+3=17. Two slips are common here, reading the square as −22=−4, and dropping the sign in the middle term when −5(−2) is really +10.
Problem
Evaluate (2h−3)2+h at h=6. The parentheses hold a whole expression this time. Build the inside value first, square the result, then finish the sum.
Show a hint
  • The exponent applies to the whole parenthesized value, so finish 2h−3 before squaring anything.
Show the full solution
Inside first, 2(6)−3=9. Squaring gives 92=81, and the total is 81+6=87. The exponent takes the finished inside value, so squaring term by term into 144−9+6=141 is wrong.
Problem
This one uses every skill from the lesson. Evaluate 1−12vv at v=−4. The bar covers all of 1−12v, the input is negative, and a division finishes the job. Enter the value as a fraction in lowest terms.
Show a hint
  • Substitute before touching the root, so compute 1−12(−4) under the bar first.
  • After taking the root, divide by v itself. Its value is −4, sign included.
Show the full solution
Under the bar, 1−12(−4)=1+48=49, so the top is 49=7. The bottom is v=−4, and 7÷(−4)=−74. Two slips lurk here, stopping at the radical because of the printed minus, and dropping the denominator's sign to land on 74.

The expressions got heavier and the method never changed. Parentheses on every substitution, signs judged after substituting, computation run from the inside out. Next, 2.3 turns the distributive property around, reading ab+ac as a(b+c), a rewrite called factoring.

Practice these ideas

Practice
Evaluate 9−2t at t=−3.
Show the solution
Substituting gives 9−2(−3). The product is −6, and subtracting −6 is adding 6, so the value is 9+6=15. Without the parentheses the term reads 2⋅3 and the answer comes out 3.
Practice
Evaluate m3+m at m=12.
Show the solution
Only the first term sits over the bar, so it is 12÷3=4, and the value is 4+12=16. The bar groups only its own top, so pulling the lone m underneath into 12+123 is wrong.
Practice
Evaluate 2−7d at d=−2. Substitute before judging the root.
Show the solution
Substituting gives −7(−2)=14, so the bar holds 2+14=16 and 16=4. What decides whether an even root exists is the sign of the value after substituting, not the sign printed in the expression.
Practice
Evaluate w2+3w at w=−18.
Show the solution
Substituting gives (−18)2+3(−18), which is 324−54=270. The square takes the whole value with its sign, so (−18)2=324. Reading it as −182=−324 drops the total to −378.
Practice
Evaluate (3g−4)2 at g=5.
Show the solution
Inside first, 3(5)−4=11, and 112=121. The exponent applies to the finished parenthesized value, so squaring g alone into 3(25)−4=71 is wrong.
Practice
Evaluate 40p−p at p=−5.
Show the solution
The first term is 40÷(−5)=−8, and subtracting p means subtracting −5, so the value is −8+5=−3. The divisor and the subtracted p both keep their signs, and dropping either one lands you on 13.
Practice
Evaluate h2+63 at h=−1.
Show the solution
(−1)2=1, so the bar holds 1+63=64, and 64=8. Reading the square as −12=−1 leaves 62 under the bar, which has no clean root, a good sign you took the sign wrong.
Practice
Evaluate a2+ba+b at a=−3 and b=6.
Show the solution
The top is (−3)2+6=15 and the bottom is −3+6=3, so the value is 15÷3=5. Finish the whole top and the whole bottom before dividing, and remember a2 means (−3)2=9, not −9.
Practice
Evaluate 2c2−c3 at c=−3.
Show the solution
The even power clears the sign, so 2(−3)2=18, and the odd power keeps it, so (−3)3=−27. Then 18−(−27)=18+27=45. Treating the cube as +27 gives −9 instead.
Practice
Evaluate 50−t2t+12 at t=−5. Enter the value as a fraction in lowest terms.
Show the solution
The bar holds 50−(−5)2=50−25=25, so the top is 5. The bottom is −5+12=7, and the value is 57. Both halves turn on signs, (−5)2 is 25 rather than −25, and the bottom is −5+12 rather than −5−12.