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Algebra I · Lesson 2.5

Fractions with Variables

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Now the tops and bottoms of fractions hold variables. A fraction bar is division, and a letter on the top or bottom is just an unknown number, so none of the fraction arithmetic from prealgebra changes. The care with signs from 2.4 carries straight over too.

Problem
Combine 2c+5c into a single fraction over c. Enter the numerator.
Show a hint
  • The bottoms already match, so you are just counting pieces of size 1c.
Show the full solution
The bottoms already match, so just count the pieces, 2+5=7, giving 7c. The bottom stays c. Gathering more pieces of one size does not change the size of a piece.
Problem
Write 9m+87−4m−37 as one fraction over 7. Enter the constant term of the numerator.
Show a hint
  • Subtracting the whole top subtracts both of its terms, the sign-flip from 2.4.
Show the full solution
The minus hits every term of 4m−3, so the top is 9m+8−4m+3=5m+11 and the constant is 11. Drop the flip on the −3 and you get 5 instead, which is the usual slip.
A factor multiplies the whole top and the whole bottom, so it cancels20 (u + 14)20 (u - 14)u + 14u - 14LegalA term is only one piece of a sum, so it does notu + 14u - 1414-14Wrong
Both routes start from a fraction sharing a piece top and bottom. Up top the 20 is a factor, it multiplies the whole numerator and the whole denominator, so dividing both by 20 is legal. Below, the u is only a term inside u+14, so crossing it out changes the value. A factor is a whole multiplied chunk, a term is one piece of a sum, and only factors cancel.
Problem
Simplify 6n+153 by factoring the top and cancelling, then evaluate the result at n=4. Enter the value.
Show a hint
  • Factor the top before cancelling. What divides both 6n and 15?
Show the full solution
Factor the top, 3(2n+5)3=2n+5, and at n=4 that is 8+5=13. Cancelling the 3 against only the 15 is the slip. The 3 has to divide the whole top before it can go.
Problem
A classmate simplifies r+10r to 10 by crossing out the r's. Test the original r+10r at r=2, the 1.4 way. Enter its value.
Show a hint
  • Evaluate the untouched fraction and compare with the 10 the cancel claims.
Show the full solution
At r=2 the untouched fraction is 2+102=122=6, not the 10 the cancel claims. Here r is a term of r+10, not a factor of the whole top, so crossing it out is illegal.
Problem
Simplify 8b−40b−5. Factor the top, cancel the shared (b−5), and enter what is left.
Show a hint
  • Nothing cancels until the top is factored. Pull the common factor out of 8b−40.
Show the full solution
Factor the top, 8(b−5)b−5. The whole chunk (b−5) is shared, so it divides out and leaves 8. Nothing cancels until the top is factored, since a loose piece of a sum is not a factor.
Problem
Add 3p+74p over the common denominator 4p. Enter the numerator.
Show a hint
  • 4p is a multiple of p. Scale 3p up by 44, an equivalent fraction.
Show the full solution
Scale the first fraction, 3p=124p, then add over 4p, 12+7=19. Adding 3+7 straight across skips the scaling, and the pieces are different sizes until you do it.
Problem
Write 3v+24−v−73 as one fraction over 12. Enter the constant term of the numerator.
Show a hint
  • The bottoms 4 and 3 share no factor, so 12 is their product.
  • Scale each fraction to twelfths, then subtract the whole second top.
Show the full solution
Over 12 the tops are 3(3v+2)=9v+6 and 4(v−7)=4v−28. The minus takes the whole second top, 9v+6−4v+28=5v+34, so the constant is 34. Miss the flip on the −28 and you land at −22.
Problem
Combine 5f+15f+3−2f into a single fraction of the form 5f+kf. Enter k (typed like -7).
Show a hint
  • Cancel the (f+3) chunk in the first fraction first.
  • Then write 5 as 5ff to subtract the 2f.
Show the full solution
Cancel the shared chunk, 5(f+3)f+3=5, then 5−2f=5f−2f, so k=−2. Writing 5 as 5ff is what makes the subtraction possible. 5−2f is not 3f.

Every fraction here had a single letter in it. 2.6 puts several letters in one expression at once. The counting of like pieces and the care with signs work exactly the same way with more letters to track, so the rewrites you just practiced are already the whole toolkit.

Practice these ideas

Practice
Add 4g+5g. Enter the numerator of the sum over g.
Show the solution
The bottoms already match, so count the pieces, 4+5=9. The bottom stays g. Adding the bottoms too is the common slip, but gathering more pieces of one size does not change the size of a piece.
Practice
Combine 8k−35−8k−75 over 5. Enter the numerator.
Show the solution
The minus hits every term of 8k−7, so 8k−3−8k+7=4. Half-flipping to 8k−3−8k−7 lands the constant at −10, so flip both signs of the second top.
Practice
Simplify 10e+355 by factoring the top and cancelling, then evaluate the result at e=3. Enter the value.
Show the solution
Factor the top, 5(2e+7)5=2e+7, and at e=3 that is 6+7=13. Cancelling the 5 against only the 35 is the slip. The 5 has to divide the whole top before it can go.
Practice
A student cancels the h's in h+15h and writes 15. Evaluate the original h+15h at h=3, the 1.4 way, and enter the true value.
Show the solution
At h=3 the untouched fraction is 3+153=183=6, not the 15 the cancel claims. Here h is a term of h+15, not a factor of the whole top, so crossing it out is illegal.
Practice
Simplify 7c−63c−9 by factoring the top and cancelling (c−9). Enter what is left.
Show the solution
Factor the top, 7(c−9)c−9. The whole chunk (c−9) is shared, so it divides out and leaves 7. Nothing cancels until the top is factored, since a loose piece of a sum is not a factor.
Practice
Simplify 10(x−3)4(x−3) by cancelling the shared (x−3). Enter the number left, in lowest terms.
Show the solution
The (x−3) is a whole factor of both, so it divides out and leaves 104=52. Stopping at 104 is the easy miss. The number part still reduces.
Practice
Add 5y+32y. Scale the first to halves-of-y, add over 2y, and enter the numerator.
Show the solution
Scale the first fraction, 5y=102y, then add over 2y, 10+3=13. Adding 5+3 straight across skips the scaling, and the pieces are different sizes until you do it.
Practice
The denominators of 4j+7k share no factor. Enter their least common denominator, typed like jk.
Show the solution
With no shared factor, the product of the bottoms works, so the least common denominator is jk. Neither j nor k divides the other, so neither one alone can serve as a common bottom.
Practice
Combine r+32−r−85 over 10. The subtraction flips every sign of the second top. Enter the constant term of the combined numerator.
Show the solution
Over 10 the tops are 5(r+3)=5r+15 and 2(r−8)=2r−16. The minus takes the whole second top, 5r+15−2r+16=3r+31, so the constant is 31. Miss the flip on the −16 and you land at −1.
Practice
In 4x+9x a student cancels the x from 4x against the bottom and writes 13. Evaluate the original at x=3 and enter the true value.
Show the solution
Keep the fraction whole and substitute. At x=3 it is 4(3)+93=213=7, not the 13 the cancel claims. The x in 4x is tied into one term of 4x+9, so it is not a factor of the whole top.
Practice
Combine 4s−36−2s−63 into one fraction and reduce. The s terms cancel. Enter the result in lowest terms.
Show the solution
Scale the second fraction to sixths, 2s−63=4s−126. The minus takes the whole top, (4s−3)−(4s−12)=9, so the result is 96=32. Half-flipping to 4s−3−4s−12 gives −15 instead.
Practice
Simplify 12w4w and enter that number.
Show the solution
Divide top and bottom by the whole shared factor 4w, leaving 124=3. Cancelling only the w and stopping at 124 is the easy miss, since the numbers still reduce.