Free math worksheets

- Add or subtract the coefficients of like terms. - Keep the common variable part unchanged. - Equal opposite terms cancel to zero.

1. \(8x+5x=(8+5)x=13x\). 2. \(14y-8y=(14-8)y=6y\). 3. \(11a-11a=0\). 4. \(3c+7c=(3+7)c=10c\).

a) \(13x\) b) \(6y\) c) \(0\) d) \(10c\)

- Group the strips that have the same length. - Use multiplication to represent repeated equal lengths. - Substitute each value only after simplifying the expression.

1. The four variable-length strips have total length \(x+x+x+x=4x\). 2. The two fixed-length strips have total length \(8+8=16\). 3. Therefore, \(L=4x+16\). 4. When \(x=5\), \(L=4\times5+16=36\,\text{cm}\). 5. When \(x=2.4\), \(L=4\times2.4+16=25.6\,\text{cm}\).

a) \(L=4x+16\) b) \(36\,\text{cm}\) and \(25.6\,\text{cm}\)

- Group terms that contain the same variable. - Keep each coefficient’s sign. - Terms cancel when their coefficients add to zero.

1. Combine the \(m\)-terms: \(12m+3m=15m\). 2. Combine the \(n\)-terms: \(-5n-8n+13n=(-5-8+13)n=0n\). 3. Therefore, the expression simplifies to \(15m\).

\(15m\)

- Count how many copies of each variable are present. - Combine only terms with the same variable part. - Fractions with the same denominator are added by adding their numerators.

1. Four equal \(a\)-terms combine to \(4a\). 2. Combine each variable separately: \(10x-4x=6x\) and \(5y+y=6y\). The result is \(6x+6y\). 3. Add the fractions with the same denominator: \(\frac{z+z+z+z+z}{6}=\frac{5z}{6}\).

1) \(4a\) 2) \(6x+6y\) 3) \(\frac{5z}{6}\)

- Count the number of identical terms. - Keep each term’s sign. - Only terms with identical variable parts can be combined.

1. Four identical \(uv\)-terms combine to \(4uv\). 2. Three identical \(xyz\)-terms combine to \(3xyz\). 3. Two positive \(a\)-terms combine to \(2a\), and three negative \(b\)-terms combine to \(-3b\). The result is \(2a-3b\).

1) \(4uv\) 2) \(3xyz\) 3) \(2a-3b\)

- Combine the numerical coefficients of like terms. - Keep the common variable part unchanged. - A term with coefficient zero has value zero.

1. Combine the coefficients: \(1.5+3.2-0.7=4\). The result is \(4x\). 2. Combine the coefficients: \(7-4=3\). The result is \(3y\). 3. Combine the coefficients: \(12-15+3=0\). Therefore, the expression equals \(0\).

1) \(4x\) 2) \(3y\) 3) \(0\)

- Like terms have the same variable part. - Add the coefficients of like terms. - Unlike terms cannot be combined.

1. \(12a+5a=17a\). 2. \(4x+(-9x)=-5x\). 3. \(-3m+3m=0\). 4. The terms \(7y\) and \(2z\) are not like terms, so their sum remains \(7y+2z\).

a) \(17a\) b) \(-5x\) c) \(0\) d) \(7y+2z\)

- Treat each blank as a missing like term. - Use subtraction to find a missing addend or subtracted term. - Substitute each result to check the equation.

1. For part a, \(12b-5b=7b\). 2. For part b, the amount subtracted is \(12k-3k=9k\). 3. For part c, \(10m-8m=2m\).

a) \(7b\) b) \(9k\) c) \(2m\)

- Group terms that have exactly the same variable part. - Combine coefficients within each group and keep unlike groups separate.

1. Combine the \(a\)-terms: \(4a+2a=6a\). 2. Combine the \(b\)-terms: \(3b-b=2b\). 3. The simplified expression is \(6a+2b\).

\(6a+2b\)

- Keep each sign attached to its coefficient. - Simplify the two variable groups independently before choosing.

1. Combine \(x\)-terms: \(9x-4x=5x\). 2. Combine \(y\)-terms: \(-2y+7y=5y\). 3. The equivalent expression is \(5x+5y\), choice a).

a) \(5x+5y\)

- Separate variable terms from constant terms. - The constants form their own like-term group.

1. Combine variable terms: \(5c+3c-2c=6c\). 2. Combine constants: \(-8+11=3\). 3. The simplified expression is \(6c+3\).

\(6c+3\)

- Compare the complete variable parts, not just one letter. - Terms with \(uv\) are not like terms with terms containing only \(u\).

1. Combine \(uv\)-terms: \(6uv+5uv-uv=10uv\). 2. Combine \(u\)-terms: \(-2u+9u=7u\). 3. The simplified expression is \(10uv+7u\).

\(10uv+7u\)

- Look for a pair of like terms whose coefficients sum to zero. - After cancellation, continue simplifying the remaining groups.

1. The \(r\)-terms cancel: \(12r-12r=0\). 2. The \(s\)-terms combine: \(5s-3s=2s\). 3. The constant remains \(4\). 4. The simplified expression is \(2s+4\).

\(2s+4\). The \(r\)-terms have opposite coefficients, so they cancel.

- Use the same variable part as the signal for a like-term group. - In the last row, remember that constants combine with constants.

1. \(2a+a+5a=(2+1+5)a=8a\). 2. \(9b-4b-b=(9-4-1)b=4b\). 3. \(3c+2-7c+6=(3-7)c+(2+6)=-4c+8\).

First row: \(8a\) Second row: \(4b\) Third row: \(-4c+8\)

- Remember the implied coefficient on a variable written alone. - After simplifying, read the signed numerical factors from the result.

1. Combine \(a\)-terms: \(a+3a+6a=10a\). 2. Combine \(b\)-terms: \(2b-5b+b=-2b\). 3. The simplified expression is \(10a-2b\). 4. The coefficients are \(10\) for \(a\) and \(-2\) for \(b\).

Simplified expression: \(10a-2b\) Coefficients: \(10\) and \(-2\)

- Reduce each expression to one coefficient times the variable. - Compare the resulting coefficients rather than the original term counts.

1. A simplifies to \((7-2+1)k=6k\). 2. B simplifies to \((4+3-1)k=6k\). 3. C simplifies to \((9-4+2)k=7k\). 4. A and B simplify to the same expression.

A and B; both simplify to \(6k\).

- Express all coefficients in a common number form. - Only the numerical factors need to be combined because the variable parts match.

1. Convert \(\frac{1}{2}\) to \(0.5\). 2. Add coefficients: \(0.25+0.5+1.25=2\). 3. The simplified expression is \(2x\).

\(2x\)

- Work with one variable group while leaving the other terms unchanged. - Then repeat the same process for the remaining like terms.

1. The \(p\)-terms combine as \(8p-5p+2p=5p\). 2. After that stage, the expression is \(5p-3q+q\). 3. The \(q\)-terms combine as \(-3q+q=-2q\). 4. The final simplified expression is \(5p-2q\).

After the first stage: \(5p-3q+q\) Final result: \(5p-2q\)

- Group the variable terms and constant terms separately. - Simplify the expression before substituting. - Multiply before adding.

1. Combine the variable terms: \(15x+5x=20x\). 2. Combine the constant terms: \(-7+17=10\). 3. The simplified expression is \(20x+10\). 4. Substitute \(x=1.2\): \(20\times1.2+10\). 5. Evaluate: \(24+10=34\).

\(34\)

- Combine only terms with the same variable part. - A variable with no visible coefficient has coefficient \(1\). - Work with the numerical coefficients while keeping each sign.

1. For part a, combine the coefficients: \(1.5+2.7-0.8=3.4\), so the result is \(3.4a\). 2. For part b, \(-3-4+12-1=4\), so the result is \(4k\). 3. For part c, \(\frac{1}{2}+\frac{3}{4}-1=\frac{2}{4}+\frac{3}{4}-\frac{4}{4}=\frac{1}{4}\), so the result is \(\frac{1}{4}x\). 4. For part d, \(12-15+3=0\), so all variable terms cancel and the result is \(0\).

a) \(3.4a\) b) \(4k\) c) \(\frac{1}{4}x\) d) \(0\)

- Write the implied coefficient of the final \(x\). - Combine all three coefficients, not only the first two. - Check how adding one more \(x\) changes \(3x\).

1. The coefficients are \(7\), \(-4\), and \(1\) because \(x=1x\). 2. Combine them: \(7-4+1=4\). 3. Therefore, the expression simplifies to \(4x\), so Ben is correct. 4. Anna likely ignored the implied coefficient \(1\) on the final \(x\) and stopped after calculating \(7-4=3\).

Ben is correct: \(7x-4x+x=(7-4+1)x=4x\). Anna likely overlooked the coefficient \(1\) on the final \(x\).

- Treat each blank as a missing term in an addition or subtraction statement. - Use an inverse operation to find the missing term. - Keep careful track of negative signs.

1. For part a, subtract the known term from the result: \(12m-5m=7m\). 2. For part b, the missing term is \(8.4t-2.1t=6.3t\). 3. For part c, add the opposite of \(-3b\), which is \(3b\). 4. For part d, add \(6p\) to the result: \(-10p+6p=-4p\).

a) \(7m\) b) \(6.3t\) c) \(3b\) d) \(-4p\)

- Perform multiplication before addition or subtraction. - Combine variable terms and constants separately. - Substitute only after simplifying.

1. For part a, multiply first and combine like terms: \(6x\times2-4x+9=12x-4x+9=8x+9\). 2. Substitute \(x=4\): \(8\times4+9=32+9=41\). 3. For part b, combine the constants and variable terms: \(3.5-8y+1.2-y=4.7-9y\). 4. Substitute \(y=0.5\): \(4.7-9\times0.5=4.7-4.5=0.2\).

a) Simplified expression: \(8x+9\); value: \(41\) b) Simplified expression: \(4.7-9y\); value: \(0.2\)

- Perform each multiplication before combining like terms. - Variable terms can cancel completely. - Substitute the given value only after simplifying \(B\).

1. Simplify \(A\): \(15x-4\times2x-7x=15x-8x-7x=0\). 2. Simplify \(B\): \(3\times0.5z=1.5z\), so \(B=1.2z+0.8z-1.5z=0.5z\). 3. Evaluate \(B\) at \(z=4\): \(0.5\times4=2\).

a) \(A=0\) b) \(B=0.5z\) For \(z=4\), \(B=2\).

- Perimeter is the sum of all side lengths. - Write the sum before combining any terms. - Combine like terms before substituting the value of \(x\).

1. Add the four side lengths: \(P=x+(x+2)+2x+5\). 2. Combine like terms: \(P=4x+7\). 3. Substitute \(x=3.5\): \(P=4\times3.5+7=21\,\text{cm}\).

a) \(P=4x+7\) b) \(21\,\text{cm}\)

- Express the width in terms of \(L\). - Use the formulas for rectangle perimeter and area. - Keep the expression \(L-10\) grouped when multiplying.

1. The width is \(L-10\). 2. The perimeter is \(P=2L+2(L-10)=4L-20\). 3. The area is \(A=L(L-10)\). 4. When \(L=25\), \(P=4\times25-20=80\,\text{m}\), and \(A=25\times(25-10)=375\,\text{m}^2\).

a) \(P=4L-20\) b) \(A=L(L-10)\) c) \(P=80\,\text{m}\); \(A=375\,\text{m}^2\)

- Combine only terms with the same variable part. - Work with the numerical coefficients. - A variable with no visible coefficient has coefficient \(1\).

1. Combine each variable separately: \(6.4m-2.1m=4.3m\) and \(3.5n-n=2.5n\). The result is \(4.3m+2.5n\). 2. Three copies of \(\frac{c}{3}\) equal \(c\), and two copies of \(\frac{d}{2}\) equal \(d\). The result is \(c-d\). 3. \(\frac{2}{9}k+\frac{7}{9}k=k\), so \(k-k=0\).

1) \(4.3m+2.5n\) 2) \(c-d\) 3) \(0\)

- Count how many times the entire grouped expression appears. - Combine the terms in the numerator before simplifying the fraction. - Rewrite repeated equal terms using a coefficient.

1. Three identical grouped expressions combine to \(3(c-d)\), which can also be written as \(3c-3d\). 2. The numerator contains six \(pq\)-terms, so \(\frac{pq+pq+pq+pq+pq+pq}{9}=\frac{6pq}{9}=\frac{2}{3}pq\). 3. The sum of five \(ab\)-terms is \(5ab\). Then \(5ab-2ab=3ab\).

1) \(3(c-d)\), or \(3c-3d\) 2) \(\frac{2}{3}pq\) 3) \(3ab\)

- A coefficient tells how many copies are added; an exponent tells how many equal factors are multiplied. - Expand each coefficient into repeated addition. - Can terms with different variable parts be counted as copies of the same quantity? - Count the total number of \(a\)-terms in part b).

1. The coefficient \(4\) in \(4x\) counts four addends: \(x+x+x+x\). The exponent \(4\) in \(x^4\) counts four factors: \(x\times x\times x\times x\). 2. Write \(2a+3a\) as \((a+a)+(a+a+a)\). This is \(a+a+a+a+a=5a\). 3. Write \(3x+2y\) as \(x+x+x+y+y\). The \(x\)-terms and \(y\)-terms are unlike terms, so they cannot be combined into a single term.

a) \(4x=x+x+x+x\), while \(x^4=x\times x\times x\times x\). b) \(2a+3a=(a+a)+(a+a+a)=a+a+a+a+a=5a\). c) \(3x+2y=x+x+x+y+y\). Since \(x\) and \(y\) are different variables, the terms are not like terms and cannot be combined.

- Treat each box as an unknown term. - Use inverse operations to isolate the missing term. - To make a sum equal zero, add the opposite term.

1. The missing subtracted term is \(7b-(-3b)=10b\). Check: \(7b-10b=-3b\). 2. Subtract \(-5k\) from \(2k\): \(2k-(-5k)=7k\). Check: \(7k+(-5k)=2k\). 3. First combine \(4m-9m=-5m\). Its opposite, \(5m\), must be added to obtain \(0\).

1) \(10b\) 2) \(7k\) 3) \(5m\)

- Group terms containing the same variable. - Align decimal points when combining coefficients. - Keep the negative signs on the \(n\)-terms.

1. Combine the \(m\)-terms: \(4.7+1.3-5.2=0.8\). 2. Combine the \(n\)-terms: \(-2.15-3.85=-6\). 3. The simplified expression is \(0.8m-6n\).

\(0.8m-6n\)

- Use subtraction to find a missing addend. - Treat each variable term like a signed quantity. - Compare matching terms on both sides.

1. For part a, \(15x-8x=7x\). 2. For part b, \(-2a-5a=-7a\). 3. For part c, \(6m-(-4m)=10m\). 4. For part d, \(3b\) already appears on both sides, so the missing expression is \(-2c\).

a) \(7x\) b) \(-7a\) c) \(10m\) d) \(-2c\)

- Group terms with the same variable. - Combine constant terms separately. - Keep each term’s sign.

1. Combine the \(x\)-terms: \(14x-5x=9x\). 2. Combine the \(y\)-terms: \(-8y+2y=-6y\). 3. Combine the constants: \(3-11=-8\). 4. The simplified expression is \(9x-6y-8\).

\(9x-6y-8\)

- Group terms containing the same variable. - An \(x\) with no visible coefficient has coefficient \(1\). - Use common denominators when combining fractional coefficients.

1. Combine the \(x\)-coefficients: \(\frac{3}{8}+\frac{1}{4}+1=\frac{3}{8}+\frac{2}{8}+\frac{8}{8}=\frac{13}{8}\). 2. Combine the \(y\)-coefficients: \(-\frac{1}{3}-\frac{5}{6}=-\frac{2}{6}-\frac{5}{6}=-\frac{7}{6}\). 3. The simplified expression is \(\frac{13}{8}x-\frac{7}{6}y\).

\(\frac{13}{8}x-\frac{7}{6}y\)

- Combine all \(a\)-terms separately from the constants. - A variable term disappears when its coefficient is zero. - Check whether the constants also cancel.

1. Write the sum: \((3a-5)+(2-4a)+(a+3)\). 2. Combine the variable terms: \(3a-4a+a=0\). 3. Combine the constants: \(-5+2+3=0\). 4. The total is \(0\), so the claim is correct.

The claim is correct. The sum is \(0\).

- Treat each box as a missing like term. - Subtract the known term from the result when it is being added. - Rewrite an implied coefficient of \(1\) as a fraction when needed.

1. The missing term is \(15x-7x=8x\). 2. The missing term is \(10y-4.5y=5.5y\). 3. Rewrite \(z\) as \(\frac{3}{3}z\). The missing term is \(\frac{3}{3}z-\frac{1}{3}z=\frac{2}{3}z\). 4. Add \(8w\) to the result: \(2w+8w=10w\).

1) \(8x\) 2) \(5.5y\) 3) \(\frac{2}{3}z\) 4) \(10w\)

- Write every coefficient as a fraction when helpful. - Use common denominators before adding or subtracting coefficients. - Convert a mixed-number coefficient to an improper fraction.

1. Rewrite \(a\) as \(\frac{5}{5}a\): \(\frac{3}{5}a+\frac{5}{5}a=\frac{8}{5}a\). 2. Rewrite \(\frac{1}{2}\) as \(\frac{2}{4}\): \(\frac{3}{4}b-\frac{2}{4}b=\frac{1}{4}b\). 3. Convert \(1\frac{2}{3}\) to \(\frac{5}{3}\): \(\frac{5}{3}c+\frac{5}{6}c=\frac{10}{6}c+\frac{5}{6}c=\frac{5}{2}c\).

1) \(\frac{8}{5}a\) 2) \(\frac{1}{4}b\) 3) \(\frac{5}{2}c\)

- Use a common denominator for all fractional coefficients. - Treat the blank as a missing term. - Subtract the target coefficient from the starting coefficient in part 2.

1. Use denominator \(18\): \(\frac{8}{18}x+\frac{6}{18}x-\frac{15}{18}x=-\frac{1}{18}x\). 2. The missing expression is the difference between \(\frac{7}{10}y\) and \(\frac{1}{5}y\): \(\frac{7}{10}y-\frac{2}{10}y=\frac{1}{2}y\).

1) \(-\frac{1}{18}x\) 2) \(\frac{1}{2}y\)

- Rewrite the coefficients in their original order with signs attached. - Check whether every operation sign was preserved in the student’s rewrite.

1. The final term is \(+2m\), so it should be added, not combined with \(-3m\) as another negative term. 2. Add the signed coefficients: \(8-3+2=7\). 3. The correct result is \(7m\).

Luis changed the sign of \(+2m\). The correct result is \(7m\).

- First combine only the terms that are already known. - Compare the resulting coefficient with the target coefficient.

1. The known terms combine to \(7p-2p=5p\). 2. The missing term must raise \(5p\) to \(12p\). 3. The needed coefficient is \(12-5=7\), so the missing term is \(7p\).

\(7p\)

- Sort the four terms by whether they contain the variable. - Check whether a numerical total may be attached to a variable after the groups are combined.

1. The terms \(2.4t\) and \(-0.9t\) are like terms, so their combined coefficient is \(2.4-0.9=1.5\). 2. The constants \(1.7\) and \(3.3\) combine to \(5\). 3. Variable terms and constants are not like terms, so the simplified expression is \(1.5t+5\).

Kai combined unlike terms. The correct simplified expression is \(1.5t+5\).

- Put the fractional coefficients into a form that makes their sizes directly comparable. - Decide whether the final coefficient is below, equal to, or above the stated benchmark.

1. Rewrite the coefficients with denominator \(12\): \(\frac{9}{12}+\frac{2}{12}-\frac{5}{12}\). 2. Combine them: \(\frac{9+2-5}{12}=\frac{6}{12}=\frac{1}{2}\). 3. The expression equals \(\frac{1}{2}q\), so it is not greater than \(\frac{1}{2}q\).

Mina is not correct. The expression simplifies to \(\frac{1}{2}q\), exactly equal to \(\frac{1}{2}q\).

- Simplify each expression independently. - Equivalent expressions must give the same simplified form for every value of the variable.

1. The first expression simplifies to \(11x-3\). 2. The second expression simplifies to \(11x-3\). 3. Since both simplify to the same expression, they are equivalent.

Yes. Both expressions simplify to \(11x-3\).

- Write the sum of all four side expressions before simplifying. - Combine variable terms and constants in separate groups.

1. Add the four sides: \((3x+2)+(x+5)+(3x+2)+(x+5)\). 2. Combine variable terms: \(3x+x+3x+x=8x\). 3. Combine constants: \(2+5+2+5=14\). 4. The perimeter is \(8x+14\) centimeters.

\(8x+14\,\text{cm}\)

- Simplify the complete expression first to see each combined coefficient. - Compare the current simplified form with the target one variable group at a time.

1. The full expression simplifies to \(2x+6y\). 2. To leave \(2x+2y\), the removed term must account for \(4y\). 3. Removing the term \(4y\) leaves \(5x+2y-3x=2x+2y\).

Remove \(4y\).

- Begin with the terms that share the same variable part. - Compare the simplified sum with what the grouped product represents.

1. Combine the \(m\)-terms on the left: \(2m+3m=5m\). 2. The left side becomes \(5m+5n\). 3. The right side \(5(m+n)\) represents \(5m+5n\). 4. The statement is always true.

Yes. The left side simplifies to \(5m+5n\), which is equivalent to \(5(m+n)\).

- Keep the subtraction sign attached to the constant that follows it. - Check the variable group and constant group separately.

1. The variable terms combine to \(3x+2x=5x\). 2. The constants combine to \(4-9=-5\). 3. The correct result is \(5x-5\). 4. The student treated \(-9\) as \(+9\).

\(5x-5\). The student changed the sign of the \(-9\) term.

- Simplify both the variable terms and constants in every choice. - One choice differs only in its constant group, so check both parts carefully.

1. a) simplifies to \(4x+3\). 2. b) simplifies to \(4x+3\). 3. c) simplifies to \(4x+3\). 4. d) simplifies to \(4x+4\). 5. Choice d) does not match.

d) \(2x+2x+6-2\)

- First remove the parentheses using the property that applies to multiplication over addition. - After distributing, identify which terms are like terms. - Compare the simplified result with the student's expression.

1. Distribute the \(3\): \(3(a+5)=3a+15\). 2. The full expression becomes \(3a+15+2a\). 3. Combine the like terms \(3a\) and \(2a\) to get \(5a+15\). 4. The student's claim is correct.

The claim is correct: \(3(a+5)+2a=3a+15+2a=5a+15\).

- Remove the parentheses before trying to combine terms. - After distributing, identify the terms that have the same variable part. - Leave the constant term separate from the variable terms.

1. Distribute \(4\) across \(x+3\): \(4(x+3)=4x+12\). 2. The expression becomes \(4x+12+2x\). 3. Combine the like terms \(4x\) and \(2x\) to get \(6x+12\).

\(6x+12\)

- Compare the expressions when \(k=1\). - How can repeated addition of the same quantity be written as multiplication? - What algebraic form guarantees that a number is a multiple of \(3\)?

1. For \(k=1\), Expression A is \(3\times1+1=4\), and Expression B is \(2\times1+2=4\). Since the expressions are equal for this value, Expression A is not greater for every positive integer \(k\). 2. More generally, the difference is \((3k+1)-(2k+2)=k-1\), which equals \(0\) when \(k=1\) and is positive when \(k>1\). 3. Combine like terms: \(k+k+k=3k\). Because \(3k\) is \(3\) times an integer, it is always a multiple of \(3\).

a) No. When \(k=1\), both expressions equal \(4\); Expression A is greater only when \(k>1\). b) Since \(k+k+k=3k\), the value always has a factor of \(3\) and is therefore a multiple of \(3\).