QuanticaPrealgebraOpen in the courseA letter can stand for a number you do not know yet, or for one that is free to change. That lets one short line of math cover a whole family of situations instead of a single fixed case. Numbers and letters stitched together with the usual operations form an expression, like \(3x + 5\). Nothing you already know gets thrown out, since a letter is just a stand-in for a number. Two conventions to keep for good. \(3x\) means \(3 \times x\), and \(x^2\) means \(x \times x\).
Problem
A bead bracelet holds \(3\) colored beads per charm plus \(4\) fixed clasp beads, so a bracelet with \(n\) charms has \(3n + 4\) beads. A customer orders \(6\) charms. Evaluate \(3n + 4\) at \(n = 6\) by replacing \(n\) with \(6\). How many beads are on that bracelet? Type the single number.
Show a hint
- The expression is \(3n + 4\). To evaluate at \(n = 6\), swap the letter \(n\) for the number \(6\) everywhere it appears. You should be looking at \(3(6) + 4\).
- Do the multiplication first. \(3(6)\) means \(3 \times 6\), which is \(18\). Then add the \(4\) clasp beads.
Show the full solution
Replace \(n\) with \(6\) in \(3n + 4\). $$3(6) + 4 = 18 + 4 = 22.$$ The bracelet has \(\boxed{22}\) beads. Build the expression once and it handles any order, since a new charm count is just a new value to drop in.
Problem
A diver's height in meters is \(11 - 4t\), where \(t\) counts minutes of descent. An expression splits into terms separated by signs, here \(11\) and \(-4t\). The number multiplying \(t\), carried with its sign, is the coefficient of \(t\). Read off the coefficient of \(t\) and type that single value, sign included.
Show a hint
- A subtraction is really an addition of a negative. Rewrite \(11 - 4t\) as \(11 + (-4t)\) so the sign on the \(t\) term sits right out in the open.
- The coefficient of \(t\) is the whole number multiplying \(t\), the minus included. In \(-4t\) that number is \(-4\).
Show the full solution
Read \(11 - 4t\) as \(11 + (-4t)\). The term built from \(t\) is \(-4t\), so the coefficient of \(t\) is \(\boxed{-4}\). The minus belongs to the coefficient, not to the term next door. The lone \(11\) is the constant term, the diver's starting height.
Problem
Different letters stand for different kinds, so \(x\) and \(y\) never combine. In \(3x + 5y + 2x + y\), gather the \(x\) terms with each other and the \(y\) terms with each other, remembering the lone \(y\) means \(1y\). After you simplify, what is the coefficient of \(x\)? Type that single value.
Show a hint
- Sort the terms into two piles before you add anything. One pile holds the \(x\) terms, the other holds the \(y\) terms. A term with \(x\) can never land in the \(y\) pile, so the two letters get counted separately.
- Add the \(x\) terms by themselves, \(3x + 2x = 5x\), and leave the \(y\) terms alone. The number sitting in front of \(x\) is its coefficient, so read that off.
Show the full solution
Gather the \(x\) terms, \(3x + 2x = 5x\), and the \(y\) terms, remembering the lone \(y\) means \(1y\), so \(5y + y = 6y\). That gives $$3x + 5y + 2x + y = 5x + 6y.$$ The coefficient of \(x\) is \(\boxed{5}\). Different letters stand for different things, so an \(x\) term and a \(y\) term never combine.
Problem
Sort by shape before combining. In \(x^2 + 5x + 2x^2 + 3\), the squares belong with squares and the strips with strips. A thing squared and the thing itself are different kinds of object, so \(x^2\) and \(5x\) do not merge, which is exactly why counting terms is a real question. Write \(x^2 + 5x + 2x^2 + 3\) in simplest form. How many terms does it have once fully simplified? Type the single count.
Show a hint
- Sort the four pieces by their shape before you add anything. Which ones are square tiles \(x^2\), which is a strip \(5x\), and which is a plain number? Only pieces of the same shape can be combined.
- Combine just the squares, \(x^2 + 2x^2 = 3x^2\), and leave \(5x\) and \(3\) exactly where they are since nothing matches them. That gives \(3x^2 + 5x + 3\). Now count the separate pieces joined by plus signs.
Show the full solution
Combine the squares, \(x^2 + 2x^2 = 3x^2\). The \(5x\) and the \(3\) have nothing matching them, so $$x^2 + 5x + 2x^2 + 3 = 3x^2 + 5x + 3.$$ The pieces are \(3x^2\), \(5x\), and \(3\), which is \(\boxed{3}\) terms. Squaring \(x\) makes a different kind of object than \(x\) itself, so those two never fold together.
Problem
A school orders supply kits in two batches, \(3(x+5) + 4(x+8)\), where each kit holds \(x\) notebooks plus some pens. Distribute each group on its own, then combine the like terms so the order collapses to (notebooks)\(\,x\,\) plus (pens). How many pens does the order come to? Type that single number.
Show a hint
- Open each group separately. The first group \(3(x+5)\) becomes \(3x + 15\), and the second group \(4(x+8)\) becomes \(4x + 32\). The pens are the plain numbers with no \(x\) attached.
- Add the two plain-number pieces together. You have \(15\) pens from the first batch and \(32\) pens from the second, so the total number of pens is \(15 + 32\).
Show the full solution
Distribute each group. \(3(x+5) = 3x + 15\) and \(4(x+8) = 4x + 32\), so the order reads \(3x + 15 + 4x + 32\). Combine like terms, \(3x + 4x = 7x\) and \(15 + 32 = 47\), which gives $$7x + 47.$$ The pens are the plain number with no \(x\) on it, so the order comes to \(\boxed{47}\) pens.
Problem
This one leans on signs, so go slowly. Expand \(5(z - 3) + 3(7 - 2z)\), letting each subtraction ride along with its term and remembering \(3 \times (-2z) = -6z\). Combine the like terms all the way to one cleanest form, then read off the coefficient of \(z\). Type that single value, sign included.
Show a hint
- Line up your four pieces, \(5z - 15 + 21 - 6z\), then sort them into z terms and plain numbers. The coefficient of \(z\) is whatever number ends up sitting in front of \(z\) once you have gathered \(5z\) and \(-6z\) together.
- Add the z terms by adding their coefficients, \(5 + (-6) = -1\), so the z part becomes \(-z\). Writing \(-z\) is the same as writing \(-1z\), so the coefficient is that \(-1\).
Show the full solution
Distribute both groups, letting the subtractions travel with their terms. \(5(z - 3) = 5z - 15\), and \(3(7 - 2z) = 21 - 6z\) since \(3 \times (-2z) = -6z\). That gives \(5z - 15 + 21 - 6z\). The z terms combine as \(5z - 6z = -z\), and the constants as \(-15 + 21 = 6\), so the simplest form is \(-z + 6\). Since \(-z\) means \(-1z\), the coefficient of \(z\) is \(\boxed{-1}\).
Problem
A subtraction in front of a group flips the sign of everything inside, since the multiplier carries its minus into every piece. Take \(8 - 2(x - 5)\). The \(-2\) lands on both the \(x\) and the \(-5\), so the \(-5\) becomes \(+10\), not \(-10\). Distribute, combine the constants, and the expression collapses to (something)\(\,x\,\) plus a constant. What is the constant term once fully simplified? Type that single value.
Show a hint
- The number multiplying the group is the whole \(-2\), not just \(2\). Send it onto both pieces inside, so \(-2 \times x\) and \(-2 \times (-5)\). Watch what happens to that second sign.
- A negative times a negative is positive, so \(-2 \times (-5) = +10\). That gives \(8 - 2x + 10\). Add the two loose numbers \(8 + 10\) to get the constant, and read it off.
Show the full solution
The \(-2\) multiplies both pieces inside, so \(-2 \times x = -2x\) and \(-2 \times (-5) = +10\). $$8 - 2(x - 5) = 8 - 2x + 10 = -2x + 18.$$ The constant term is \(\boxed{18}\). The usual slip is sending only the \(2\) inside and landing on \(-10\), but the minus rides along with it.
Problem
Two expressions are equivalent when they agree for every value of the variable. Staring at them is not proof, so use this grading check. Build the difference \(A - B\) and plug in a value. If they truly are equivalent, that difference must come out \(0\). Let \(A = 2(3x+4) + x\) and \(B = 7x + 8\). Compute \(A - B\) at \(x = 5\) and type the single number you get.
Show a hint
- Evaluating at \(x = 5\) means replacing every \(x\) with \(5\). Find the value of \(A = 2(3x + 4) + x\) at \(x = 5\), then the value of \(B = 7x + 8\) at \(x = 5\), and finally subtract the second from the first.
- At \(x = 5\), \(A = 2(3 \cdot 5 + 4) + 5 = 2 \cdot 19 + 5 = 43\) and \(B = 7 \cdot 5 + 8 = 43\). The difference is \(43 - 43\).
Show the full solution
At \(x = 5\), \(A = 2(3 \cdot 5 + 4) + 5 = 2 \cdot 19 + 5 = 43\) and \(B = 7 \cdot 5 + 8 = 43\), so \(A - B = 43 - 43 = \boxed{0}\). Simplifying \(A\) shows this was never luck. \(2(3x + 4) + x = 6x + 8 + x = 7x + 8\), which is exactly \(B\), so the difference is \(0\) at every value of \(x\).
Problem
The target is \(6n + 10\), and four suspects line up beside it: \(2(3n+5)\), \(3(2n+4)\), \(5 + 6n + 5\), and \(10 + 6n\). Simplify each one until you can compare it cleanly with \(6n + 10\). How many of the four are equivalent to \(6n + 10\)? Type the single count.
Show a hint
- To test a suspect, rewrite it in the same plain form as the target, which is some number of \(n\) plus a constant. For the ones with parentheses, multiply the outside number by each piece inside. For the ones that are already a string of terms, gather the loose numbers together.
- Suspect one becomes \(6n + 10\), a match. Suspect two becomes \(6n + 12\), so its constant is off by 2 and it fails. Suspect three has \(5 + 5 = 10\) so it is \(6n + 10\), a match. Suspect four is \(10 + 6n\), the same two pieces just written in the other order, which is still a match. Now count the matches.
Show the full solution
Rewrite each suspect and line it up against \(6n + 10\). \(2(3n + 5) = 6n + 10\), a match. \(3(2n + 4) = 6n + 12\), and \(12 \ne 10\), so it fails. \(5 + 6n + 5 = 6n + 10\), a match. \(10 + 6n\) is the same two pieces in the other order, a match. That is \(\boxed{3}\) of the four. Reordering a sum never changes its value, so writing the constant first does not make a new expression.
Problem
Coefficients do not have to be whole numbers. A fraction can sit in front of a variable, and combining such terms is the same common-denominator move from Chapter 4 with an \(x\) riding along. The terms \(\tfrac{2x}{3}\) and \(\tfrac{5x}{7}\) are like terms, since both carry a single \(x\). Combine \(\tfrac{2x}{3} + \tfrac{5x}{7}\) into one term, then state the coefficient of \(x\) as a fraction in lowest terms, typed as a over b.
Show a hint
- Both terms are like terms, so think of this as adding the coefficients \(\tfrac{2}{3}\) and \(\tfrac{5}{7}\) while the \(x\) just rides along. To add those fractions you need a common denominator for \(3\) and \(7\).
- A common denominator of \(3\) and \(7\) is \(21\). Rewrite \(\tfrac{2x}{3}\) as \(\tfrac{14x}{21}\) and \(\tfrac{5x}{7}\) as \(\tfrac{15x}{21}\), then add the numerators over \(21\). The coefficient of \(x\) is whatever fraction lands in front.
Show the full solution
A common denominator of \(3\) and \(7\) is \(21\). Rewrite each term with denominator \(21\), so \(\tfrac{2x}{3} = \tfrac{14x}{21}\) and \(\tfrac{5x}{7} = \tfrac{15x}{21}\). Now add the numerators over the shared denominator. $$\frac{2x}{3} + \frac{5x}{7} = \frac{14x}{21} + \frac{15x}{21} = \frac{29x}{21}$$ That single term is \(\tfrac{29x}{21}\), and its coefficient of \(x\) is \(\tfrac{29}{21}\). Since \(29\) is prime and does not divide \(21\), the fraction is already in lowest terms. The coefficient of \(x\) is \(\boxed{29/21}\).
Problem
Sometimes the pieces you are adding live over different denominators, and one piece can be an expression. Combine \(\tfrac{a}{2} + \tfrac{6a - 5}{4}\) onto a common bottom, then write the result as a coefficient of \(a\) plus a constant, like \(2a - \tfrac{5}{4}\). In that clean form, type the constant term as a fraction in lowest terms, with its sign.
Show a hint
- The two denominators are \(2\) and \(4\), and \(4\) is a multiple of \(2\), so \(4\) is the common bottom. Rewrite \(\tfrac{a}{2}\) over \(4\) by doubling top and bottom, and leave \(\tfrac{6a-5}{4}\) exactly as it is. Then add the numerators over that one shared \(4\).
- You have \(\tfrac{2a}{4} + \tfrac{6a-5}{4} = \tfrac{2a + 6a - 5}{4} = \tfrac{8a-5}{4}\). Split that into two fractions, \(\tfrac{8a}{4} - \tfrac{5}{4} = 2a - \tfrac{5}{4}\). The constant term is the piece with no \(a\) attached.
Show the full solution
The common denominator is \(4\), since \(4\) is a multiple of \(2\). Double the top and bottom of the first piece, leave the second alone, and add over the shared bottom. $$\tfrac{a}{2} + \tfrac{6a - 5}{4} = \tfrac{2a}{4} + \tfrac{6a - 5}{4} = \tfrac{2a + 6a - 5}{4} = \tfrac{8a - 5}{4}$$ Split that back into an \(a\) piece and a number piece, \(\tfrac{8a}{4} - \tfrac{5}{4} = 2a - \tfrac{5}{4}\). The constant term is \(\boxed{-5/4}\). The minus stays with the constant because the \(5\) was being subtracted inside the numerator.
Problem
This one expression gathers the whole lesson, a piece scaled by a fraction, a piece scaled by a whole number, and a piece over a denominator. Simplify \(\frac{1}{2}(4x + 6) + 3(x - 1) + \frac{x - 5}{3}\) all the way down to the form (something)\(\,x\,\) plus (something), then report the coefficient of \(x\) as a fraction in lowest terms, typed as a over b.
Show a hint
- Knock out one group at a time before you mix anything. Half of \(4x + 6\) gives \(2x + 3\). Three times \(x - 1\) gives \(3x - 3\), and watch that the \(3\) hits the \(-1\) too. The last group \(\frac{x - 5}{3}\) is one quantity divided by \(3\), so it splits into \(\frac{x}{3} - \frac{5}{3}\).
- Now sweep up only the \(x\) terms and ignore the constants for this question. You have \(2x\), \(3x\), and \(\frac{x}{3}\), so the coefficient of \(x\) is \(2 + 3 + \frac{1}{3}\). Put \(2 + 3 = 5\) over a denominator of \(3\) to add the last piece, which makes \(\frac{15}{3} + \frac{1}{3}\).
Show the full solution
Take the three groups one at a time. \(\frac{1}{2}(4x + 6) = 2x + 3\), then \(3(x - 1) = 3x - 3\), then \(\frac{x - 5}{3} = \frac{x}{3} - \frac{5}{3}\). Lining them up gives \(2x + 3 + 3x - 3 + \frac{x}{3} - \frac{5}{3}\). The \(x\) coefficients add as \(2 + 3 + \frac{1}{3} = \frac{15}{3} + \frac{1}{3} = \frac{16}{3}\), and the constants give \(3 - 3 - \frac{5}{3} = -\frac{5}{3}\), so the expression is \(\frac{16}{3}x - \frac{5}{3}\) and the coefficient of \(x\) is \(\boxed{16/3}\). It is already in lowest terms, since \(16\) and \(3\) share no factor.
Practice these ideas
Practice
A weather balloon's altitude in meters is \(13 - 7p\), where \(p\) is the number of minutes a valve has been venting. The number multiplying a variable, carried with its sign, is its coefficient. State the coefficient of \(p\) and type that single value, sign included.
Show the solution
Reading the subtraction as adding a negative, \(13 - 7p\) is \(13 + (-7p)\). The term built from \(p\) is \(-7p\), so the coefficient of \(p\) is \(\boxed{-7}\). The minus belongs to the coefficient. The other term, \(13\), is the constant.
Practice
A print run is modeled by \(19 - 5q\), where \(q\) counts paper jams. The piece \(5q\) shifts as jams shift, but one piece does not change at all. That unchanging number is the constant term. State the constant term of \(19 - 5q\) and type that single number.
Show the solution
The two terms of \(19 - 5q\) are \(5q\), whose value shifts whenever \(q\) shifts, and \(19\), a plain number with no variable attached. The constant term is the one with no variable, so it is \(\boxed{19}\). Setting \(q = 0\) makes it visible, since \(19 - 5(0) = 19\).
Practice
A lone variable carries a hidden \(1\) in front, so \(m\) means \(1m\). Combine the like terms in \(3m + m + 5m\) into a single term, then type the coefficient of \(m\).
Show the solution
Every term here is a number of copies of \(m\), so they are like terms and can be merged. Reading the lone \(m\) as \(1m\), the expression becomes \(3m + 1m + 5m\). Add the numbers in front, \(3 + 1 + 5 = 9\), and keep the \(m\). The simplified expression is \(9m\). The coefficient of \(m\) is \(\boxed{9}\).
Practice
Tally each kind of term on its own, the plain numbers in one pile and the \(x\) terms in another. Simplify \(8 + 2x + 5 + 4x\), then read off the constant term and type that single value.
Show the solution
Group the like terms first. The plain numbers are \(8\) and \(5\), and the \(x\) terms are \(2x\) and \(4x\). Combine each group on its own. The numbers give \(8 + 5 = 13\), and the \(x\) terms give \(2x + 4x = 6x\). Putting both groups back together, the expression in simplest form is \(6x + 13\). The constant term is the piece with no \(x\) attached to it, which is \(13\). So the value you type is \(\boxed{13}\).
Practice
Square tiles each cover \(x^2\) and thin tiles each cover \(x\), and the two kinds never combine because \(x^2\) and \(x\) are different objects. Gather each kind in \(5x^2 + 2x + 3x^2 + x\), write it in simplest form, then count how many terms remain.
Show the solution
Keep the two kinds of tiles separate while you combine. The square tiles give \(5x^2 + 3x^2 = 8x^2\), and the thin x-tiles give \(2x + x = 3x\). Putting the piles back together, the simplest form is \(8x^2 + 3x\). A term is one chunk separated by a plus or minus sign, so \(8x^2\) is one term and \(3x\) is another. They cannot merge into a single piece because a square tile is not the same kind of tile as a thin one. That leaves two terms. \(\boxed{2}\)
Practice
Since \(a\) and \(b\) stand for different things, an \(a\) term and a \(b\) term can never merge. Combine only the matching kinds in \(6a + 2b + a + 3b\), write the result in simplest form, then type how many terms are left.
Show the solution
Collect the pieces of the same kind. The \(a\) terms give \(6a + a = 7a\), and the \(b\) terms give \(2b + 3b = 5b\), so $$6a + 2b + a + 3b = 7a + 5b.$$ That is two separate pieces joined by addition, so the count is \(\boxed{2}\). An \(a\) bundle and a \(b\) bundle measure different things, so they never merge into one term.
Practice
Expand \(8(3x + 2)\) by handing the \(8\) to every piece inside, so it lands on both the \(3x\) and the \(2\). In the result, the number multiplying \(x\) is its coefficient. What is the coefficient of \(x\)? Type that single value.
Show the solution
Hand the \(8\) to every piece inside the group. \(8 \times 3x = 24x\) and \(8 \times 2 = 16\). Putting those together, the expanded expression is \(24x + 16\). No parentheses are left, so it is in simplest form. The term holding the \(x\) is \(24x\), and the number multiplying \(x\) is \(24\). That number is the coefficient of \(x\). So the coefficient is \(\boxed{24}\).
Practice
Distribute across each group in \(2(x+6) + 5(x+1)\), then combine the like terms into simplest form. Read off the constant term, the lone number with no \(x\) attached, and type that single value.
Show the solution
Distribute each group on its own. The first group gives \(2(x+6) = 2x + 12\), and the second gives \(5(x+1) = 5x + 5\). Putting those together, the expression is \(2x + 12 + 5x + 5\). Now combine like terms. The \(x\) terms are \(2x\) and \(5x\), which add to \(7x\). The plain numbers are \(12\) and \(5\), which add to \(17\). So in simplest form the expression is \(7x + 17\). The constant term is the number standing alone with no \(x\), which is \(17\). So the answer is \(\boxed{17}\).
Practice
Distribute each number across its own parentheses in \(6(n-2) + 4(5-3n)\), letting every sign ride along, then collect the like terms. In simplest form, read off the coefficient of \(n\) and type that single value, sign included.
Show the solution
Distribute each factor across its parentheses, keeping every sign attached to its term. The first chunk gives \(6(n - 2) = 6n - 12\). The second chunk gives \(4(5 - 3n) = 20 - 12n\), where the \(4\) multiplies both the \(5\) and the \(-3n\). So far the expression reads \(6n - 12 + 20 - 12n\). Now collect like terms. The \(n\) terms are \(6n\) and \(-12n\), which combine to \(-6n\). The plain numbers are \(-12\) and \(20\), which combine to \(8\). The simplest form is \(-6n + 8\). The coefficient of \(n\) is the number multiplying \(n\), which is \(-6\). \(\boxed{-6}\)
Practice
In \(20 - 3(2x - 4)\), the \(-3\) distributes across the group and flips the sign of every piece inside because it is negative. Simplify into a number plus a multiple of \(x\), then find the constant term, the plain number with no \(x\), and type that single value.
Show the solution
Distribute the \(-3\) into \(2x - 4\). You get \(-3 \times 2x = -6x\) and \(-3 \times (-4) = +12\), so the expression becomes \(20 - 6x + 12\). The plain numbers combine, \(20 + 12 = 32\), while \(-6x\) has no partner, leaving \(32 - 6x\). The constant term is \(\boxed{32}\). A negative multiplier flips the sign of every piece inside, which is where most slips happen.
Practice
Two expressions are equivalent when their difference \(A - B\) is \(0\) for every value of \(x\). Let \(A = 3(2x + 5)\) and \(B = 6x + 15\). Evaluate \(A - B\) at \(x = 4\) by replacing every \(x\) with \(4\), and type the single number you get.
Show the solution
At \(x = 4\), \(A = 3(2 \cdot 4 + 5) = 3 \cdot 13 = 39\) and \(B = 6 \cdot 4 + 15 = 39\), so \(A - B = 39 - 39 = \boxed{0}\). Distributing shows it could not have come out any other way, since \(3(2x + 5) = 6x + 15\) is exactly \(B\), so the difference is \(0\) for every value of \(x\).
Practice
The target is \(8x + 12\). Simplify each of \(4(2x + 3)\), \(2(4x + 6)\), \(8x + 6\), and \(12 + 8x\), then compare with the target. How many of the four are equivalent to \(8x + 12\)? Type the single count.
Show the solution
Simplify each expression and compare it to the target \(8x + 12\). For the first, \(4(2x + 3) = 8x + 12\), which matches. For the second, \(2(4x + 6) = 8x + 12\), which also matches. For the third, \(8x + 6\) is already simplified, and its constant is \(6\) instead of \(12\), so it does not match. For the fourth, \(12 + 8x\) is the same two terms as \(8x + 12\) written in the other order, and order does not change the value, so it matches. Three of the four are equivalent to the target. \(\boxed{3}\)
Practice
Two dials add \(\tfrac{5x}{6}\) and \(\tfrac{x}{4}\) to a score, where \(x\) is the number of presses. Combine \(\tfrac{5x}{6} + \tfrac{x}{4}\) into one term using a common denominator of \(12\), keeping the \(x\). State the coefficient of \(x\) as a fraction in lowest terms, typed as a over b.
Show the solution
Rewrite each fraction with denominator \(12\). Since \(\tfrac{5x}{6} = \tfrac{10x}{12}\) and \(\tfrac{x}{4} = \tfrac{3x}{12}\), the sum becomes $$\frac{10x}{12} + \frac{3x}{12} = \frac{13x}{12}.$$ The combined term is \(\tfrac{13x}{12}\). The coefficient of \(x\) is the number multiplying it, which is \(\tfrac{13}{12}\). Since \(13\) is prime and does not divide \(12\), this fraction is already in lowest terms. The coefficient is \(\boxed{13/12}\).
Practice
A tiled border uses \(3(2s + 3)\) tiles in one row, \(2(s - 1)\) in another, plus \(5\) corner tiles, where \(s\) is a size setting. Build one expression for the total, simplify it, then evaluate at \(s = 4\) by replacing every \(s\) with \(4\). Type the single number you get.
Show the solution
Start by expanding each grouped piece. The first row is \(3(2s + 3) = 6s + 9\), and the second row is \(2(s - 1) = 2s - 2\). Adding everything, including the \(5\) corner tiles, gives \(6s + 9 + 2s - 2 + 5\). Combine the like terms. The \(s\) terms give \(6s + 2s = 8s\), and the plain numbers give \(9 - 2 + 5 = 12\), so the simplified total is \(8s + 12\). Now evaluate at \(s = 4\) by replacing \(s\) with \(4\), which gives \(8 \cdot 4 + 12 = 32 + 12 = 44\). \(\boxed{44}\)