QuanticaAlgebra IOpen in the courseEverything from 2.5 still works when more than one letter is in view. Take \(8ab+9ab\). The coefficients add the 2.4 way, \(8+9=17\), so \(8ab+9ab=17ab\), and the \(ab\) stays as it is. The one new question is what counts as the same kind of term, now that a single term can carry several letters.
Problem
Combine \(7st+6st\) into one term. Type its coefficient.
Show a hint
- \(st\) is one variable part, so this is a single family.
- Add the coefficients, the 2.4 way.
Show the full solution
\(st\) is one variable part, so the coefficients add, \(7+6=\boxed{13}\). Two letters in a term does not make it two families. The whole variable part is what has to match.
Problem
A classmate rewrote \(5cd+5c\) as \(10cd\), adding the coefficients as if the terms matched. Evaluate the original \(5cd+5c\) at \(c=2,\ d=4\). Type the value.
Show a hint
- Are \(5cd\) and \(5c\) really like terms? Compare their variable parts.
- Put \(c=2,\ d=4\) into the original \(5cd+5c\), one letter at a time.
Show the full solution
Substitute into the original. \(5cd=5\cdot2\cdot4=40\) and \(5c=5\cdot2=10\), so the value is \(40+10=\boxed{50}\). The classmate's \(10cd\) gives \(80\) at the same input, and one input with two different outputs proves \(5cd\) and \(5c\) are unlike.
Problem
Simplify \(11mn+9m-5nm\) by combining like terms. Type the coefficient of the \(mn\) term.
Show a hint
- \(nm\) is the same variable part as \(mn\), so the order does not matter.
- The minus travels with its term into the \(mn\) family; the \(9m\) is a different family.
Show the full solution
Read \(nm\) as \(mn\), so \(11mn-5mn=6mn\), and the \(9m\) is a separate family. The coefficient is \(\boxed{6}\). Order inside a term is free, so a backward-written term still joins its family, and its minus sign comes with it.
Problem
Distribute \(8r(2s+t)\). Type the coefficient of the \(rs\) term.
Show a hint
- The \(8r\) multiplies every term inside the parentheses.
- You only need the term that has \(r\) and \(s\).
Show the full solution
\(8r\cdot2s=16rs\) and \(8r\cdot t=8rt\), so the \(rs\) coefficient is \(\boxed{16}\). The \(8r\) multiplies both terms inside, not just the first.
Problem
Simplify \(2(4uv+5v)+7uv\) by distributing then combining like terms. Type the coefficient of the \(uv\) term.
Show a hint
- Distribute the \(2\) across both terms inside first.
- Only the \(uv\) terms combine; the plain \(v\) term is a different family.
Show the full solution
Distribute first, \(2(4uv+5v)=8uv+10v\). Only the \(uv\) terms combine, \(8uv+7uv=15uv\), so the coefficient is \(\boxed{15}\). The \(10v\) has a different variable part, so it never joins the \(uv\) count.
Problem
Find the greatest common factor of the terms in \(8pqr+12pqs\), typed like \(2xy\), the number first and the letters in alphabetical order.
Show a hint
- Take the GCD of \(8\) and \(12\) first.
- Which letters appear in BOTH terms? Only those come out front.
Show the full solution
The GCD of \(8\) and \(12\) is \(4\), and \(p\) and \(q\) are the letters sitting in both terms, so the GCF is \(\boxed{4pq}\). The \(r\) and \(s\) each appear in only one term, so neither can come out front.
Problem
Evaluate \(4gh-2g+h\) at \(g=5,\ h=2\). Type the value.
Show a hint
- \(4gh\) needs both letters, so \(4\cdot g\cdot h\).
- Handle the \(-2g\) and \(+h\) after the \(4gh\) term.
Show the full solution
\(4\cdot5\cdot2=40\), then \(-2\cdot5=-10\) and \(+2\), so \(40-10+2=\boxed{32}\). A term like \(4gh\) needs both values, and reading it as \(4g\) drops the \(h\).
Problem
Simplify \(2d(4e+f)+9de\) by distributing and combining, then evaluate the result at \(d=2,\ e=1,\ f=5\). Type the value.
Show a hint
- Distribute \(2d\) first, keeping the \(d\) on both new terms.
- Combine the \(de\) terms, then substitute \(d=2,\ e=1,\ f=5\).
Show the full solution
Distribute, \(2d(4e+f)=8de+2df\), then combine, \(8de+9de=17de\). Now substitute, \(17\cdot2\cdot1=34\) and \(2\cdot2\cdot5=20\), so \(34+20=\boxed{54}\). The \(d\) has to stay on both new terms, and \(de\) and \(df\) are different families, so they never merge.
That caps Chapter 2. Distribute, combine, factor, and evaluate are the whole toolkit, and none of them changed when several letters came into view. Only the like-term rule widened, since the variable part now has to match in full. Chapter 3 turns to exponents, the shorthand for a letter multiplied by itself over and over, along with the rules that come with them.
Practice these ideas
Practice
Combine \(5jk+9jk\). Type the coefficient.
Show the solution
\(jk\) is one variable part, so the coefficients add, \(5+9=\boxed{14}\).
Practice
Combine \(7gh+8hg\). Type the coefficient.
Show the solution
\(hg\) and \(gh\) are the same variable part, so both terms sit in one family and \(7+8=\boxed{15}\). Flipping the letters inside a term does not make a new family.
Practice
After combining like terms in \(5mp+2m+4mp\), how many terms remain?
Show the solution
The \(mp\) family gives \(5+4=9mp\), and \(2m\) stands on its own, so \(\boxed{2}\) terms remain. The \(2m\) has a different variable part, so it never joins the \(mp\) count.
Practice
Are \(6ab\) and \(6a\) like terms? Type yes or no.
Show the solution
The variable parts are \(ab\) and \(a\), and those differ, so \(\boxed{\text{no}}\). A shared coefficient and one shared letter are not enough, since the whole variable part has to match.
Practice
Evaluate \(4pq+3p\) at \(p=2,\ q=5\). Type the value.
Show the solution
\(4\cdot2\cdot5=40\) and \(3\cdot2=6\), so \(40+6=\boxed{46}\). The \(4pq\) term needs both values, and reading it as \(4p\) drops the \(q\).
Practice
Evaluate \(2mn+4m-n\) at \(m=3,\ n=1\). Type the value.
Show the solution
\(2\cdot3\cdot1=6\), \(4\cdot3=12\), and the last term is \(-1\), so \(6+12-1=\boxed{17}\). The \(2mn\) term needs both values, not just the \(m\).
Practice
Distribute \(2a(4b+3c)\). Type the coefficient of the \(ab\) term.
Show the solution
\(2a\cdot4b=8ab\), so the \(ab\) coefficient is \(\boxed{8}\). The \(2a\) multiplies both terms inside, not just the first.
Practice
Distribute \(5m(2n+4p)\). Type the coefficient of the \(mp\) term.
Show the solution
The \(mp\) term comes from \(5m\cdot4p=20mp\), so the coefficient is \(\boxed{20}\). The other product, \(5m\cdot2n=10mn\), is a different family.
Practice
Simplify \(3r(s+2)+4rs\). Type the coefficient of the \(rs\) term.
Show the solution
Distribute, \(3r(s+2)=3rs+6r\). Only the \(rs\) terms combine, \(3rs+4rs=7rs\), so the coefficient is \(\boxed{7}\). The \(6r\) has no \(s\), so it stays out of the \(rs\) count.
Practice
Factor \(8ab+12ac\) fully. Type the greatest common factor pulled out front, written like \(5x\) (number first, letters alphabetical).
Show the solution
The GCD of \(8\) and \(12\) is \(4\), and \(a\) is the only letter in both terms, so the GCF is \(\boxed{4a}\). The \(b\) and \(c\) each appear once, so neither comes out front.
Practice
Factor \(6cdh+15cdk\) fully. Type the greatest common factor, written like \(2xy\) (number first, letters alphabetical).
Show the solution
The GCD of \(6\) and \(15\) is \(3\), and both \(c\) and \(d\) sit in every term, so the GCF is \(\boxed{3cd}\). The \(h\) and \(k\) each appear in only one term, so they stay inside.
Practice
Evaluate \(5st-2s+t\) at \(s=2,\ t=3\). Type the value.
Show the solution
\(5\cdot2\cdot3=30\), \(-2\cdot2=-4\), and \(+3\), so \(30-4+3=\boxed{29}\). The \(5st\) term needs both values, and reading it as \(5s\) drops the \(t\).