Free math worksheets

- Which property lets you remove parentheses when a factor is outside them? - The outside factor applies to each term inside the parentheses. - After removing the parentheses, check that each original term has been accounted for.

Distribute \(5\) to both terms inside the parentheses: \(5(x + 2) = 5x + 10\).

\(5x + 10\)

- Expand the product before combining any terms. - Group the quadratic terms, linear terms, and constants separately. - Check whether the quadratic terms cancel completely.

1. Expand the product: \((2x+1)(x-3)=2x^2-5x-3\). 2. Substitute this into the expression: \(f(x)=2x^2-5x-3-2x^2+4x\). 3. Combine like terms. The quadratic terms cancel, and \(-5x+4x=-x\), so \(f(x)=-x-3\).

\(f(x)=-x-3\)

- Expand both products carefully before combining terms. - Keep track of the subtraction in front of \(x(x-10)\). - After simplifying, compare the result with the general linear form \(mx+b\).

1. Expand the square and product: \((x-4)^2=x^2-8x+16\) and \(x(x-10)=x^2-10x\). 2. Substitute and combine like terms: \(f(x)=x^2-8x+16-x^2+10x-15=2x+1\). 3. The quadratic terms cancel, leaving the form \(mx+b\). Therefore, \(f\) is linear.

\(f(x)=2x+1\). It is linear because the simplified expression has the form \(mx+b\).

- Remove the parentheses in \(f\) using the distributive property. - Combine constant terms separately from variable terms. - Arrange each result with the \(x\)-term first and the constant second.

1. Distribute and combine constants in \(f\): \(f(x)=2x-6+4=2x-2\). 2. Combine the constants in \(g\): \(g(x)=-\frac{1}{2}x+5-2=-\frac{1}{2}x+3\).

\(f(x)=2x-2\) \(g(x)=-\frac{1}{2}x+3\)

- Write the rectangle's area as length times width. - Use the distributive property to remove the parentheses.

1. The area is \(4(x + 3)\). 2. Distribute \(4\) to both terms inside the parentheses: \(4x + 12\).

\(4x + 12\) square units

- Check whether the outside factor was applied to every term inside the parentheses. - Compare the original and rewritten expressions term by term.

1. The rewrite is not correct because the factor \(5\) must multiply both terms inside the parentheses. 2. Distributing gives \(5(2x) - 5(3) = 10x - 15\).

No. The correct equivalent expression is \(10x - 15\). The student multiplied only the first term by \(5\).

- Remove the parentheses before combining terms. - Combine variable terms with variable terms and constants with constants.

1. Distribute \(3\): \(6x + 15 - 4x + 7\). 2. Combine like terms: \(6x - 4x = 2x\) and \(15 + 7 = 22\). 3. The simplified expression is \(2x + 22\).

\(2x + 22\)

- Simplify each expression separately. - In \(f\), check what happens to the two \(x^2\)-terms. - Equivalent expressions have the same simplified form for every value of the variable.

1. Distribute in \(f\): \(\frac{1}{2}x(4-2x)=2x-x^2\), so \(f(x)=2x-x^2+x^2+3=2x+3\). 2. Distribute in \(g\): \(g(x)=2x+3\). 3. Both expressions simplify to \(2x+3\), so they are equivalent for every value of \(x\).

\(f(x)=2x+3\) and \(g(x)=2x+3\). Yes, they are equivalent for every value of \(x\).

- First rewrite the expression with \(k\) by distributing and combining like terms. - For two linear expressions to be equivalent for every \(x\), compare both the variable coefficient and the constant term. - Check that the same value of \(k\) satisfies both comparisons.

1. Rewrite the left side as \((k + 2)x + 4k\). 2. Equivalent linear expressions must have matching coefficients of \(x\) and matching constant terms. 3. Matching the coefficient gives \(k + 2 = 7\), so \(k = 5\). Matching the constant gives \(4k = 20\), which also gives \(k = 5\). 4. Both conditions agree, so \(k = 5\).

\(k = 5\)

- A rectangle has two lengths and two widths. - Write the perimeter expression before simplifying it. - Distribute first, then combine like terms.

1. Use the perimeter formula: \(P = 2(2x + 1) + 2(x - 2)\). 2. Distribute: \(P = 4x + 2 + 2x - 4\). 3. Combine like terms: \(P = 6x - 2\).

\(6x - 2\) units