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Algebra I · Lesson 1.6

Evaluating Expressions

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Chapter 1 so far has been about reading. 1.2 fixed the order a computation runs in, 1.4 named the parts of an expression, and 1.5 gave exponents their meaning. This lesson runs the calculation. Pick a value for the variable, substitute it, compute, and a single number comes out.

Problem
Take the expression 7+2n2 and set n=3. Replace the letter with its number, then follow the order of operations from 1.2. What single value comes out?
Show a hint
  • The exponent applies only to n, not to 2n.
Show the full solution
Substituting gives 7+2(3)2=7+2⋅9=7+18=25. The exponent binds only to n. Squaring 2n instead gives (6)2=36 and the wrong total 43.
Problem
Now an input with a sign attached. Evaluate 5−x at x=−9. Two minus signs are in play, one written in the expression and one that is part of the value, and the answer depends on keeping them apart.
Show a hint
  • Write the substitution down on paper before computing anything.
  • What exactly replaces x, the number 9 or the number −9?
Show the full solution
The value of x is −9, sign included, so the expression becomes 5−(−9). Subtracting a negative is adding, as covered in 1.3, so 5+9=14. Substituting a bare 9 reads 5−9=−4, which is why the parentheses are worth writing.
Substituting a negative value keeps its sign inside parentheses6 - mat m = -76 - (-7) = 13Correct6 - 7 = -1WrongOne recipe, many inputs2n + 1132537
Substituting m=−7 into 6−m inside parentheses keeps the value's sign, so 6−(−7) comes out to 13. Dropping the parentheses reads 6−7 and gives the wrong −1. Below, one recipe 2n+1 runs on the inputs 1, 2 and 3 and returns 3, 5 and 7, which is the point of writing an expression at all.
Problem
Evaluate 3x2 at x=−5. Decide two things before you compute, what exactly the exponent touches, and what squaring does to the sign of the value.
Show a hint
  • The exponent binds only to x, not to 3x. That reading came from 1.2.
  • The value of x is −5, all of it, so the parentheses habit puts the whole thing under the square.
Show the full solution
Only x is squared, and x is −5 with its sign, so 3(−5)2=3⋅25=75. Squaring the product would give (−15)2=225, and reading the square as −52=−(52) would give −75.
Problem
The letter w appears twice in w(w−5). Evaluate it at w=8. Both appearances take the value at the same moment, and the parentheses still group exactly as they did in 1.2.
Show a hint
  • Substitute 8 at both spots first, then finish inside the parentheses before multiplying.
Show the full solution
Both ws become 8, so the expression reads 8(8−5). The parentheses finish first, and 8⋅3=24. Reading it as 8⋅8−5=59 drops the grouping.
Problem
Two letters, two values, four sign decisions. Evaluate 2x−3y at x=−1 and y=−4, writing each substituted value inside its own parentheses before computing anything.
Show a hint
  • After substituting, the line reads 2(−1)−3(−4), and the middle minus applies to the whole product 3(−4).
Show the full solution
Substituting gives 2(−1)−3(−4). The products are −2 and −12, and subtracting −12 is adding 12, so −2+12=10. Losing the second sign gives −2−12=−14.
Problem
Evaluate m+14m−2 at m=6. Typed on one line as m+14/m−2, the expression loses its grouping, so work it as written, with the whole top as one computation and the whole bottom as another.
Show a hint
  • The fraction bar wraps everything above it and everything below it, the invisible grouping from 1.2.
Show the full solution
The top is 6+14=20, the bottom is 6−2=4, and 20÷4=5. The bar groups the whole top and the whole bottom. Collapsed onto one line without parentheses, that grouping is lost and the pieces combine in the wrong order.
Problem
Nothing about substitution needs whole numbers. Evaluate 8g2+g at g=12. The exponent question comes first, what (12)2 is, and 1.5 already answered it. Enter the value as a fraction in lowest terms.
Show a hint
  • Squaring 12 makes it smaller, not the same and not doubled.
Show the full solution
Squaring gives (12)2=14, so 8⋅14=2, and 2+12=52. Squaring a fraction below 1 makes it smaller. Treating the square as 12 would give 92.
Problem
Formulas are expressions somebody already built for you. A dropped object falls 16t2 feet in its first t seconds. Evaluate the expression at t=3 to find how far, in feet, the object falls in three seconds.
Show a hint
  • The exponent binds only to t, so square first, then multiply by 16.
Show the full solution
Square first, 32=9, then 16⋅9=144 feet. The exponent binds only to t. Applying it to the whole product gives (16⋅3)2=2304, which misreads the formula.
Problem
Everything in this lesson at once. Evaluate x2−5xx+1 at x=−3. Substitute in parentheses at every appearance, finish the top and the bottom separately, and divide last.
Show a hint
  • Three appearances of x means three sets of parentheses.
  • Both terms on top come out positive.
Show the full solution
The top is (−3)2−5(−3)=9+15=24, the bottom is −3+1=−2, and 24÷(−2)=−12. Both wrong routes are sign slips, reading x2 as −32=−9, or losing the sign in −5(−3). Parentheses at every appearance shut both down.

That closes Chapter 1. You can read any expression the way its writer meant it, name its variables, terms, coefficients, and factors, and unwind its exponents. You can also evaluate it at any input, whole, negative, or fractional. Chapter 2 rewrites expressions into simpler forms, and it opens with the distributive property in 2.1.

Practice these ideas

Practice
Evaluate 4n+9 at n=5.
Show the solution
Substituting gives 4(5)+9. Adding first reads 5+9=14 and computes 4⋅14=56, but the product finishes before the sum, so 4⋅5=20 and 20+9=29.
Practice
Evaluate 30−2k at k=6. By the 1.2 order the product finishes before the subtraction.
Show the solution
Working left to right reads 30−2=28 and then multiplies by 6, which misorders the steps. The product goes first, so 30−12=18.
Practice
Evaluate y2−y at y=9. Both appearances of y take the same value, and the exponent goes first.
Show the solution
Both appearances become 9 at the same moment, so the expression reads 92−9. The tempting slip is to subtract first and then square, which would give (9−9)2=0. The exponent finishes before the subtraction, so 81−9=72.
Practice
Evaluate 7−w at w=−8. Write the value inside parentheses, sign included, before computing anything.
Show the solution
The careless route reads 7−8=−1, which substitutes 8 and drops the sign that belongs to the value. Written as 7−(−8) it is 7+8=15.
Practice
Evaluate −t at t=−12. With the habit the substitution reads −(−12), the opposite of a negative.
Show the solution
The tempting answer is −12, keeping the value itself. The minus written in −t is not part of the value, it asks for the opposite of t, and the opposite of −12 is 12.
Practice
Evaluate 5x2 at x=−2. Decide exactly what the exponent touches before you multiply.
Show the solution
There are two wrong turns, squaring the product to get (−10)2=100, and reading the square as −22=−4 to get −20. The exponent binds only to x, and x is −2 with its sign, so 5(−2)2=5⋅4=20.
Practice
Evaluate 4a−5b at a=3 and b=−2. Substitute both values before simplifying.
Show the solution
Losing the sign on b reads the second term as 5⋅2 and gives 12−10=2. With parentheses the substitution is 4(3)−5(−2)=12+10=22.
Practice
Evaluate p+11p−4 at p=9. Finish the whole top and the whole bottom, then divide.
Show the solution
Collapsed onto one line the bar's grouping is lost and the pieces combine in the wrong order. The top is 9+11=20 and the bottom is 9−4=5, each finished on its own before any dividing. Then 20÷5=4.
Practice
Evaluate 6q2 at q=12. Enter the value as a fraction in lowest terms.
Show the solution
Treating the square of 12 as 12 gives 3, wrong. Squaring makes it smaller, (12)2=14, so the value is 6⋅14=32.
Practice
The expression 6s2 gives the surface area of a cube with side length s, six square faces of area s2 each. Evaluate it at s=4.
Show the solution
The tempting slip squares the product, (6⋅4)2=576. The exponent binds only to s, so 6(4)2=6⋅16=96.
Practice
Evaluate u2+3uu+9 at u=−4. Parentheses at every substitution. Enter the value as a fraction in lowest terms.
Show the solution
Reading the square as −42=−16 makes the top −28 and everything after it wrong. The correct top is (−4)2+3(−4)=16−12=4 over a bottom of −4+9=5, so the value is 45.
Practice
At n=1,2,3,4 the expression n2−n+17 gives 17, 19, 23, 29, a prime every time, and the streak continues for many inputs after that. It is tempting to conclude it always gives a prime. Evaluate it at n=17, then look hard at the output.
Show the solution
Running the recipe gives 172−17+17=289, and 289=17⋅17, which is not prime, so the pattern breaks at the seventeenth input. Sixteen prime outputs in a row never proved the claim, and one counterexample settles it, the same point 1.4 made about checking values. The output is 289.