QuanticaAlgebra IOpen in the courseNow 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 \(\frac{2}{c}+\frac{5}{c}\) into a single fraction over \(c\). Enter the numerator.
Show a hint
- The bottoms already match, so you are just counting pieces of size \(\frac{1}{c}\).
Show the full solution
The bottoms already match, so just count the pieces, \(2+5=\boxed{7}\), giving \(\frac{7}{c}\). The bottom stays \(c\). Gathering more pieces of one size does not change the size of a piece.
Problem
Write \(\frac{9m+8}{7}-\frac{4m-3}{7}\) 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 \(\boxed{11}\). Drop the flip on the \(-3\) and you get \(5\) instead, which is the usual slip.
Problem
Simplify \(\frac{6n+15}{3}\) 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, \(\frac{3(2n+5)}{3}=2n+5\), and at \(n=4\) that is \(8+5=\boxed{13}\). 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 \(\frac{r+10}{r}\) to \(10\) by crossing out the \(r\)'s. Test the original \(\frac{r+10}{r}\) 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 \(\frac{2+10}{2}=\frac{12}{2}=\boxed{6}\), 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 \(\frac{8b-40}{b-5}\). 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, \(\frac{8(b-5)}{b-5}\). The whole chunk \((b-5)\) is shared, so it divides out and leaves \(\boxed{8}\). Nothing cancels until the top is factored, since a loose piece of a sum is not a factor.
Problem
Add \(\frac{3}{p}+\frac{7}{4p}\) over the common denominator \(4p\). Enter the numerator.
Show a hint
- \(4p\) is a multiple of \(p\). Scale \(\frac{3}{p}\) up by \(\frac{4}{4}\), an equivalent fraction.
Show the full solution
Scale the first fraction, \(\frac{3}{p}=\frac{12}{4p}\), then add over \(4p\), \(12+7=\boxed{19}\). Adding \(3+7\) straight across skips the scaling, and the pieces are different sizes until you do it.
Problem
Write \(\frac{3v+2}{4}-\frac{v-7}{3}\) 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 \(\boxed{34}\). Miss the flip on the \(-28\) and you land at \(-22\).
Problem
Combine \(\frac{5f+15}{f+3}-\frac{2}{f}\) into a single fraction of the form \(\frac{5f+k}{f}\). Enter \(k\) (typed like -7).
Show a hint
- Cancel the \((f+3)\) chunk in the first fraction first.
- Then write \(5\) as \(\frac{5f}{f}\) to subtract the \(\frac{2}{f}\).
Show the full solution
Cancel the shared chunk, \(\frac{5(f+3)}{f+3}=5\), then \(5-\frac{2}{f}=\frac{5f-2}{f}\), so \(k=\boxed{-2}\). Writing \(5\) as \(\frac{5f}{f}\) is what makes the subtraction possible. \(5-\frac{2}{f}\) is not \(\frac{3}{f}\).
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 \(\frac{4}{g}+\frac{5}{g}\). Enter the numerator of the sum over \(g\).
Show the solution
The bottoms already match, so count the pieces, \(4+5=\boxed{9}\). 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 \(\frac{8k-3}{5}-\frac{8k-7}{5}\) over \(5\). Enter the numerator.
Show the solution
The minus hits every term of \(8k-7\), so \(8k-3-8k+7=\boxed{4}\). Half-flipping to \(8k-3-8k-7\) lands the constant at \(-10\), so flip both signs of the second top.
Practice
Simplify \(\frac{10e+35}{5}\) by factoring the top and cancelling, then evaluate the result at \(e=3\). Enter the value.
Show the solution
Factor the top, \(\frac{5(2e+7)}{5}=2e+7\), and at \(e=3\) that is \(6+7=\boxed{13}\). 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 \(\frac{h+15}{h}\) and writes \(15\). Evaluate the original \(\frac{h+15}{h}\) at \(h=3\), the 1.4 way, and enter the true value.
Show the solution
At \(h=3\) the untouched fraction is \(\frac{3+15}{3}=\frac{18}{3}=\boxed{6}\), 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 \(\frac{7c-63}{c-9}\) by factoring the top and cancelling \((c-9)\). Enter what is left.
Show the solution
Factor the top, \(\frac{7(c-9)}{c-9}\). The whole chunk \((c-9)\) is shared, so it divides out and leaves \(\boxed{7}\). Nothing cancels until the top is factored, since a loose piece of a sum is not a factor.
Practice
Simplify \(\frac{10(x-3)}{4(x-3)}\) 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 \(\frac{10}{4}=\boxed{\frac{5}{2}}\). Stopping at \(\frac{10}{4}\) is the easy miss. The number part still reduces.
Practice
Add \(\frac{5}{y}+\frac{3}{2y}\). Scale the first to halves-of-\(y\), add over \(2y\), and enter the numerator.
Show the solution
Scale the first fraction, \(\frac{5}{y}=\frac{10}{2y}\), then add over \(2y\), \(10+3=\boxed{13}\). Adding \(5+3\) straight across skips the scaling, and the pieces are different sizes until you do it.
Practice
The denominators of \(\frac{4}{j}+\frac{7}{k}\) 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 \(\boxed{jk}\). Neither \(j\) nor \(k\) divides the other, so neither one alone can serve as a common bottom.
Practice
Combine \(\frac{r+3}{2}-\frac{r-8}{5}\) 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 \(\boxed{31}\). Miss the flip on the \(-16\) and you land at \(-1\).
Practice
In \(\frac{4x+9}{x}\) 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 \(\frac{4(3)+9}{3}=\frac{21}{3}=\boxed{7}\), 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 \(\frac{4s-3}{6}-\frac{2s-6}{3}\) into one fraction and reduce. The \(s\) terms cancel. Enter the result in lowest terms.
Show the solution
Scale the second fraction to sixths, \(\frac{2s-6}{3}=\frac{4s-12}{6}\). The minus takes the whole top, \((4s-3)-(4s-12)=9\), so the result is \(\frac{9}{6}=\boxed{\frac{3}{2}}\). Half-flipping to \(4s-3-4s-12\) gives \(-15\) instead.
Practice
Simplify \(\frac{12w}{4w}\) and enter that number.
Show the solution
Divide top and bottom by the whole shared factor \(4w\), leaving \(\frac{12}{4}=\boxed{3}\). Cancelling only the \(w\) and stopping at \(\frac{12}{4}\) is the easy miss, since the numbers still reduce.