Free Math Worksheets

- 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) \(7y+2z\)

- Add the signed coefficients. - Opposite terms cancel. - Subtracting a negative term is the same as adding a positive term.

1. \(-8x+(-5x)=(-8-5)x=-13x\). 2. \(14y-20y=(14-20)y=-6y\). 3. \(-11a+11a=0\). 4. \(3c-(-7c)=3c+7c=10c\).

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

- 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) \(0\)

- 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) \(-4p\)

- 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) \(\frac{5z}{6}\)

- 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 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. - Subtracting a negative term changes to addition. - A term with coefficient zero has value zero.

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

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

- 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) \(-2c\)

- Use inverse operations to isolate the missing expression. - Keep track of negative signs. - Substitute each result to check the equation.

1. For part a, \(2b-(-5b)=7b\). 2. For part b, the amount subtracted is \(12k-(-3k)=15k\). 3. For part c, \(-10m-(-8m)=-2m\).

a) \(7b\) b) \(15k\) c) \(-2m\)

- 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 the box as an unknown term. - Use inverse operations to isolate it. - Substitute your result to check each equation.

1. Solve \(7x-M=15x\): \(M=7x-15x=-8x\). 2. Solve \(-4y-M=-10y\): \(M=-4y-(-10y)=6y\). 3. Solve \(M-(-3z)=-z\): \(M+3z=-z\), so \(M=-4z\). 4. Solve \(M+(-8w)=2w\): \(M=2w-(-8w)=10w\).

1) \(-8x\) 2) \(6y\) 3) \(-4z\) 4) \(10w\)

- Subtracting a negative term changes to addition. - Convert whole or mixed-number coefficients to fractions. - Use common denominators.

1. \(\frac{3}{5}a-(-a)=\frac{3}{5}a+a=\frac{8}{5}a\). 2. \(-\frac{1}{2}b-\left(-\frac{3}{4}b\right)=-\frac{2}{4}b+\frac{3}{4}b=\frac{1}{4}b\). 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) \(\frac{5}{2}c\)

- Group the variable terms and constant terms separately. - Simplify the expression before substituting. - Use parentheses around the negative value when substituting.

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 \(x=-1.2\): \(20\times(-1.2)+10\). 5. Evaluate: \(-24+10=-14\).

\(-14\)

- 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 \(3\).

- 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 \(x\).

- Perform multiplication before addition or subtraction. - Combine variable terms and constants separately. - A term written as \(-y\) has coefficient \(-1\).

1. For part a, multiply first and combine like terms: \(6x\times2-4x+9=12x-4x+9=8x+9\). 2. Substitute \(x=-4\): \(8\times(-4)+9=-32+9=-23\). 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: \(-23\) 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 \(z=-4\): \(0.5\times(-4)=-2\).

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

- 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.

- Use a common denominator for all fractional coefficients. - Treat the blank as an unknown expression. - Pay close attention to subtraction of a negative term.

1. Use denominator \(18\): \(\frac{8}{18}x-\frac{15}{18}x+\frac{6}{18}x=-\frac{1}{18}x\). 2. Let the missing expression be \(M\). Then \(M=\frac{7}{10}y-\left(-\frac{1}{5}y\right)=\frac{7}{10}y+\frac{2}{10}y=\frac{9}{10}y\).

1) \(-\frac{1}{18}x\) 2) \(\frac{9}{10}y\)