QuanticaAlgebra IOpen in the courseChapter 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 called the whole thing a recipe, and 1.5 gave exponents their meaning. This lesson runs the recipe. Pick an input, substitute it for the variable, compute, and a single number comes out.
Problem
Take the expression \(7+2n^2\) 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 \(7+2(3)^2=7+2\cdot 9=7+18=\boxed{25}\). 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 \(5+9=\boxed{14}\). Substituting a bare \(9\) reads \(5-9=-4\), which is why the parentheses are worth writing.
Problem
Evaluate \(3x^2\) 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 \(3(-5)^2=3\cdot 25=\boxed{75}\). Squaring the product would give \((-15)^2=225\), and reading the square as \(-5^2=-(5^2)\) 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 \(w\)s become \(8\), so the expression reads \(8(8-5)\). The parentheses finish first, and \(8\cdot 3=\boxed{24}\). Reading it as \(8\cdot 8-5=59\) 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 \(-2+12=\boxed{10}\). Losing the second sign gives \(-2-12=-14\).
Problem
Evaluate \(\dfrac{m+14}{m-2}\) 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 \(20\div 4=\boxed{5}\). 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 \(8g^2+g\) at \(g=\frac{1}{2}\). The exponent question comes first, what \(\left(\frac{1}{2}\right)^2\) is, and 1.5 already answered it. Enter the value as a fraction in lowest terms.
Show a hint
- Squaring \(\frac{1}{2}\) makes it smaller, not the same and not doubled.
Show the full solution
Squaring gives \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\), so \(8\cdot\frac{1}{4}=2\), and \(2+\frac{1}{2}=\boxed{\frac{5}{2}}\). Squaring a fraction below \(1\) makes it smaller. Treating the square as \(\frac{1}{2}\) would give \(\frac{9}{2}\).
Problem
Formulas are expressions somebody already built for you. A dropped object falls \(16t^2\) 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, \(3^2=9\), then \(16\cdot 9=\boxed{144}\) feet. The exponent binds only to \(t\). Applying it to the whole product gives \((16\cdot 3)^2=2304\), which misreads the formula.
Problem
Everything in this lesson at once. Evaluate \(\dfrac{x^2-5x}{x+1}\) 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 \(24\div(-2)=\boxed{-12}\). Both wrong routes are sign slips, reading \(x^2\) as \(-3^2=-9\), or losing the sign in \(-5(-3)\). Parentheses at every appearance shut both down.
That closes Chapter 1. You can read any expression as its writer meant it, name its variables, terms, coefficients, and factors, unwind its exponents, and now run the recipe on any input, whole, negative, or fractional. Chapter 2 begins rewriting expressions themselves into friendlier 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 \(4\cdot 14=56\), but the product finishes before the sum, so \(4\cdot 5=20\) and \(20+9=\boxed{29}\).
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 \(30-12=\boxed{18}\).
Practice
Evaluate \(y^2-y\) 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 \(9^2-9\). The tempting slip is to subtract first and then square, which would give \((9-9)^2=0\). The exponent finishes before the subtraction, so \(81-9=\boxed{72}\).
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 \(7+8=\boxed{15}\).
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 \(\boxed{12}\).
Practice
Evaluate \(5x^2\) 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 \(-2^2=-4\) to get \(-20\). The exponent binds only to \(x\), and \(x\) is \(-2\) with its sign, so \(5(-2)^2=5\cdot 4=\boxed{20}\).
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 \(5\cdot 2\) and gives \(12-10=2\). With parentheses the substitution is \(4(3)-5(-2)=12+10=\boxed{22}\).
Practice
Evaluate \(\dfrac{p+11}{p-4}\) 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 \(20\div 5=\boxed{4}\).
Practice
Evaluate \(6q^2\) at \(q=\frac{1}{2}\). Enter the value as a fraction in lowest terms.
Show the solution
Treating the square of \(\frac{1}{2}\) as \(\frac{1}{2}\) gives \(3\), wrong. Squaring makes it smaller, \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\), so the value is \(6\cdot\frac{1}{4}=\boxed{\frac{3}{2}}\).
Practice
The expression \(6s^2\) gives the surface area of a cube with side length \(s\), six square faces of area \(s^2\) each. Evaluate it at \(s=4\).
Show the solution
The tempting slip squares the product, \((6\cdot 4)^2=576\). The exponent binds only to \(s\), so \(6(4)^2=6\cdot 16=\boxed{96}\).
Practice
Evaluate \(\dfrac{u^2+3u}{u+9}\) at \(u=-4\). Parentheses at every substitution. Enter the value as a fraction in lowest terms.
Show the solution
Reading the square as \(-4^2=-16\) 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 \(\boxed{\frac{4}{5}}\).
Practice
At \(n=1,2,3,4\) the expression \(n^2-n+17\) 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 \(17^2-17+17=289\), and \(289=17\cdot 17\), 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 \(\boxed{289}\).