MiloQuanticaAlgebra IOpen in the course

Algebra I · Lesson 2.4

Adding and Subtracting Expressions

Solve this lesson →All lessons

2.3 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=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 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.
Collect each family on its own3x52x35x + 8A minus in front multiplies every term inside by -1-(4h - 5)-4h + 5Correct-4h - 5Wrong
On top, each family collects on its own, so the x terms give 5x and the plain numbers give 8, and nothing crosses between them. Below, a minus in front of parentheses multiplies every term inside by −1, so both signs flip. Flipping only the first sign is the error to watch for.
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 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 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=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 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 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 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)−k2 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 k2 and k are different families.
Show the full solution
Expanding gives k2+9k−k2. The only like pair is k2−k2, which vanishes, leaving 9k, so the coefficient of k is 9. Combining k2 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=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=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 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=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=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=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 −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=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 (3x2+8x)+(9x2−2x) and enter the coefficient of x2 in the simplified result. Keep each family separate as you collect.
Show the solution
The x2 family gives 3+9=12 and the x family gives 8−2=6, so the result is 12x2+6x and the coefficient of x2 is 12. Adding all four coefficients to 18 treats x2 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=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 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 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.