QuanticaAlgebra IOpen in the course2.3 said this lesson would add and subtract whole expressions, with the sign flip from distributing \(-1\) as the main skill. That is where it ends up. It starts smaller, with plain counting, because adding expressions begins by counting how many copies of a letter you have.
Problem
Start by counting. Nine copies of \(k\) plus four more copies of \(k\) is some number of copies of \(k\), so \(9k+4k=ck\) for a single number \(c\). Enter \(c\).
Show a hint
- Count copies and ignore what one copy is worth. Nine of something plus four more of the same thing is how many of it?
Show the full solution
Nine copies plus four more copies is thirteen copies, whatever a copy is worth, so \(9k+4k=13k\) and \(c=\boxed{13}\). Copies add here, they never multiply, so the \(9\) and the \(4\) are added rather than multiplied.
Problem
Jae simplifies \(7w+3\) to \(10w\). Test the claim the 1.4 way, evaluate both expressions at \(w=2\), each on its own, and enter how far apart the two values are.
Show a hint
- One input, two values. Work out \(7w+3\) at \(w=2\), then \(10w\) at \(w=2\), and compare.
Show the full solution
At \(w=2\) the original gives \(14+3=17\) while \(10w\) gives \(20\), so the two values sit \(\boxed{3}\) apart. By 1.4, one input with two different outputs settles it, the claim is false. The merge \(7+3=10\) treated \(7w\) and \(3\) as the same kind of term.
Problem
Add two expressions. \((5d+9)+(2d+8)\) simplifies to one \(d\) term plus one constant. Enter the constant term of the result.
Show a hint
- In a pure sum the parentheses change nothing by 1.3, so drop them and collect each family on its own.
Show the full solution
Drop the parentheses and the sum is \(5d+9+2d+8\). The \(d\) family collects to \(7d\) and the constants to \(9+8=17\), so the result is \(7d+17\) and the constant term is \(\boxed{17}\). Piling all four numbers into one total would be wrong, since \(5d\) and \(9\) are unlike.
Problem
Now subtract. \((9p+7)-(4p+3)\) simplifies to one \(p\) term plus one constant. Enter the coefficient of \(p\) in the result.
Show a hint
- The minus applies to the whole second expression, so both \(4p\) and \(3\) get subtracted.
Show the full solution
The \(p\) terms give \(9p-4p=5p\) and the constants give \(7-3=4\), so the result is \(5p+4\) and the coefficient of \(p\) is \(\boxed{5}\). The minus applies to the whole second expression. Flipping only the first sign gives \(9p+7-4p+3=5p+10\), which adds the \(3\) instead of subtracting it.
Problem
Simplify \((11h+9)-(7h-5)\). The expression being subtracted carries a minus of its own this time. Enter the constant term of the simplified result.
Show a hint
- Rewrite the subtraction first. Distribute \(-1\) over the whole second expression, the 2.1 move.
- After the rewrite, check what happened to the \(-5\). What sign does it carry when the constants collect?
Show the full solution
The minus multiplies every term of \(7h-5\) by \(-1\), so the expression becomes \(11h+9-7h+5\). The \(h\) family gives \(4h\) and the constants give \(9+5=\boxed{14}\). Flipping only the first sign leaves \(11h+9-7h-5\) and a constant of \(4\), the classic error.
Problem
Simplify \((7u+12)-(7u+4)\) completely, flipping both signs of the second expression first. The result is shorter than usual. Enter the single number that remains.
Show a hint
- Run the sweep and watch what the \(u\) family adds up to.
Show the full solution
Flipping both signs gives \(7u+12-7u-4\). The \(u\) family totals \(7u-7u=0u\) and drops out entirely, and the constants give \(12-4=8\), so the whole expression is \(\boxed{8}\). Flipping only the first sign would leave \(+4\) and a total of \(16\).
Problem
Two letters now. Simplify \((4a+9b)+(6a-2b)\). The \(a\) terms and the \(b\) terms collect separately. Enter the coefficient of \(b\) in the result.
Show a hint
- Each letter is its own family, one sweep for the \(a\) terms and a separate sweep for the \(b\) terms.
Show the full solution
One sweep per letter. The \(a\) family gives \(4a+6a=10a\), the \(b\) family gives \(9b-2b=7b\), and the two totals never mix, so the result is \(10a+7b\). The \(10\) belongs to \(a\), and the coefficient of \(b\) is \(\boxed{7}\).
Problem
Three expressions chained together. Simplify \((2t-5)+(6-3t)+(4t+1)\). The constants sit in a different spot inside each pair of parentheses. Enter the constant term of the result.
Show a hint
- One sweep per family, and 1.3 says the order the terms arrive in does not matter.
Show the full solution
The \(t\) family gives \(2t-3t+4t=3t\) and the constants give \(-5+6+1=2\), so the result is \(3t+2\) and the constant term is \(\boxed{2}\). Where a constant sits inside its parentheses changes nothing, and the \(3\) is the coefficient, not the ask.
Problem
The whole chapter in one line. Expand \(k(k+9)\) with 2.1, then simplify \(k(k+9)-k^2\) completely. One family remains. Enter the coefficient of \(k\).
Show a hint
- Expand first. 2.1 turns \(k(k+9)\) into two terms.
- After expanding, decide which terms are actually like. 1.5 says \(k^2\) and \(k\) are different families.
Show the full solution
Expanding gives \(k^2+9k-k^2\). The only like pair is \(k^2-k^2\), which vanishes, leaving \(9k\), so the coefficient of \(k\) is \(\boxed{9}\). Combining \(k^2\) with \(9k\) is tempting, but 1.5 makes them different families.
Expressions now add, subtract, expand, and factor, four moves resting on the same handful of chapter 1 rules, and all of it reduces to counting copies and flipping signs. Next, 2.5 turns to fractions whose tops and bottoms hold variables, and the same care with rewriting carries straight over.
Practice these ideas
Practice
Twelve copies of \(g\) plus six more copies is some number of copies of \(g\). Write \(12g+6g\) as a single term \(cg\) and enter \(c\).
Show the solution
Twelve copies of \(g\) plus six more is eighteen copies, so \(12g+6g=18g\) and \(c=\boxed{18}\). Addition counts copies rather than multiplying them, so the \(72\) from multiplying the coefficients is not it.
Practice
A student simplifies \(8y+1\) to \(9y\). Settle it the 1.4 way. Evaluate both expressions at \(y=5\), each on its own, and enter how far apart their values are.
Show the solution
At \(y=5\) the original gives \(40+1=41\) while \(9y\) gives \(45\), so the gap is \(45-41=\boxed{4}\). One mismatch proves the claim false. The merge \(8+1=9\) treated the unlike terms \(8y\) and \(1\) as like.
Practice
Two expressions added. Simplify \((3c+10)+(8c+5)\) down to one \(c\) term plus one constant, and enter the constant term of the simplified result.
Show the solution
The parentheses drop and each family collects, \(3c+8c=11c\) and \(10+5=15\), so the result is \(11c+15\) and the constant term is \(\boxed{15}\). Entering \(11\) names the coefficient instead of the constant.
Practice
Now a subtraction. Simplify \((12v+11)-(8v+1)\) down to one \(v\) term plus one constant, and enter the constant term of the simplified result.
Show the solution
The minus multiplies both terms of \((8v+1)\), so the expression becomes \(12v+11-8v-1\) and the constants give \(11-1=\boxed{10}\). Flipping only the first sign leaves \(+1\) and a constant of \(12\).
Practice
Simplify \((9s+4)-(3s-7)\) down to one \(s\) term plus one constant, and enter the constant term of the simplified result.
Show the solution
The minus multiplies every term of \((3s-7)\), so the \(-7\) becomes \(+7\) and the constants give \(4+7=\boxed{11}\). Flipping only the first sign gives \(9s+4-3s-7\) and a constant of \(-3\).
Practice
Simplify \((5x+20)-(5x+14)\) completely and enter the single number that remains. Distribute the minus first, then total each family.
Show the solution
Distributing the minus gives \(5x+20-5x-14\). The \(x\) family totals \(5-5=0\) copies and vanishes, so all that remains is \(20-14=\boxed{6}\). Zero copies of \(x\) is nothing at all. Flipping only the first sign would land on \(34\).
Practice
Two letters and a subtraction. Simplify \((5a+6b)-(3a+13b)\) and enter the coefficient of \(b\) in the simplified result, sign included.
Show the solution
Distributing the minus makes the \(b\) family \(6b-13b=-7b\), so the coefficient of \(b\) is \(\boxed{-7}\). The minus stays glued to the coefficient. Flipping only the first sign leaves \(+13b\) and a total of \(19\).
Practice
Three expressions chained. Simplify \((6n+11)+(n-15)+(2n+4)\) with one sweep per family, then enter the constant term of the simplified result.
Show the solution
Every parenthesis drops in a pure sum, and the constants collect to \(11-15+4=\boxed{0}\). The bare \(n\) counts as one copy, which makes the \(n\) family \(6+1+2=9\), but \(9\) is the coefficient, not the constant.
Practice
Simplify \((3x^2+8x)+(9x^2-2x)\) and enter the coefficient of \(x^2\) in the simplified result. Keep each family separate as you collect.
Show the solution
The \(x^2\) family gives \(3+9=12\) and the \(x\) family gives \(8-2=6\), so the result is \(12x^2+6x\) and the coefficient of \(x^2\) is \(\boxed{12}\). Adding all four coefficients to \(18\) treats \(x^2\) and \(x\) as one family, and their factor structures differ.
Practice
Simplify first, then evaluate. Reduce \((6q+10)-(4q-3)\) to one \(q\) term plus one constant, then evaluate the simplified result at \(q=4\) and enter the value.
Show the solution
Subtracting the whole quantity flips the \(-3\) to \(+3\), so the simplified form is \(2q+13\), and at \(q=4\) the value is \(8+13=\boxed{21}\). Flipping only the first sign gives \(2q+7\) and a value of \(15\).
Practice
Expand, then combine. Rewrite \(4(r+8)-r\) with the parentheses gone, collect the \(r\) family, and enter the constant term of the simplified result.
Show the solution
The \(4\) multiplies both inside terms, so \(4(r+8)-r\) becomes \(4r+32-r=3r+32\) and the constant term is \(\boxed{32}\). Distributing onto the \(r\) alone gives \(4r+8-r\) and a constant of \(8\), the 2.1 partial-expansion error.
Practice
Three expressions, one subtraction. Simplify \((5z+16)+(7z-6)-(12z+9)\) completely and enter the single number that remains.
Show the solution
Distributing the minus gives \(5z+16+7z-6-12z-9\). The \(z\) family totals \(5+7-12=0\) and vanishes, and the constants total \(16-6-9=1\), so what remains is \(\boxed{1}\). A check at \(z=2\) gives \(26+8-33=1\). Flipping only the first sign of the last expression would keep the \(+9\) and give \(19\).