Variables and Expressions — Free Pre-Algebra Review Games.

3(4) + 2 = 12 + 2 = 14.

Combine like terms: 5x + 3x = 8x.

The coefficient is the number multiplied by the variable, which is 7.

Distribute 3 to both terms: 3*x + 3*4 = 3x + 12.

2(5) - 3 = 10 - 3 = 7.

Combine like terms: (4x - x) + (2y + 5y) = 3x + 7y.

Distribute -2: -2(3x) + (-2)(-5) = -6x + 10.

\((-2)^2 + 3(-2) - 1 = 4 - 6 - 1 = -3\).

2x + 6 + 3x - 3 = 5x + 3.

6x - 2x - 8 = 4x - 8.

(2*2+1)(2-3) = (5)(-1) = -5.

6x - 3 - 2x - 8 + 5 = 4x - 6.

\(f(-1) = (-1)^2 - 4(-1) = 1 + 4 = 5\).

Twice a number is 2n, five less than that is 2n - 5.

\(x^2 + 3x - 2x^2 + 2 = -x^2 + 3x + 2\).

Substituting \(x = 3\) gives \(4(3) - 5 = 12 - 5 = 7\), so the correct value is \(7\).\(12\) is wrong because it stops after multiplying and forgets to subtract \(5\).Always substitute the value first, then follow order of operations to finish the arithmetic.

Since \(8y\) and \(-3y\) are like terms, you subtract the coefficients to get \(5y\).\(11y\) is wrong because it adds the coefficients instead of subtracting them.When combining like terms, only the numerical coefficients change while the variable part stays the same.

The coefficient is the number multiplying the variable, and here that number is \(-9\), including its sign.\(9\) is wrong because it drops the negative sign, changing the value's meaning.Always include the sign directly in front of a term when identifying its coefficient.

Distributing means multiplying \(5\) by each term inside the parentheses: \(5 \cdot x - 5 \cdot 2 = 5x - 10\).\(5x - 2\) is wrong because it only multiplies the first term and leaves the second term unchanged.Every term inside the parentheses must be multiplied by the outside factor, not just the first one.

Substituting gives \(3(4) + 2(1) = 12 + 2 = 14\), matching the correct choice.\(13\) is wrong because it likely comes from computing \(3(4) + 1\) and forgetting to double \(n\).When two variables appear, substitute both values carefully before performing any addition.

Combining the like terms \(6a\) and \(2a\) gives \(8a\), while the constant \(4\) has no like term and stays separate, so the answer is \(8a + 4\).\(12a\) is wrong because it incorrectly treats the constant \(4\) as if it were another \(a\)-term to combine.Only terms with the exact same variable part can be combined; constants stay separate from variable terms.

Distributing the \(2\) across both terms gives \(2 \cdot y + 2 \cdot 3 = 2y + 6\).\(2y + 3\) is wrong because it fails to multiply the \(3\) by \(2\), leaving that term undistributed.The distributive property requires multiplying the outside number by every single term inside the parentheses.

\(x^2\) means \(x\) multiplied by itself, so \(5^2 = 5 \times 5 = 25\).\(10\) is wrong because it comes from doubling \(x\) instead of squaring it.Exponents indicate repeated multiplication, not multiplication by the exponent value.

Adding opposite terms \(-4x\) and \(4x\) results in \(0\) because they cancel each other out.\(8x\) is wrong because it adds the coefficients' absolute values instead of accounting for the opposite signs.When combining like terms with opposite signs and equal magnitude, the sum is always zero.

The constant term is the number without a variable attached, which is \(9\) in this expression.\(2\) is wrong because it is the simplified coefficient of \(x\) after combining \(7x\) and \(-5x\), not the constant.Constants and variable coefficients are different parts of an expression and should not be confused.

Multiplying \(-3\) by each term gives \(-3 \cdot x = -3x\) and \(-3 \cdot (-4) = 12\), so the result is \(-3x + 12\).\(-3x - 12\) is wrong because it mishandles the sign when multiplying \(-3\) by \(-4\), which should produce a positive result.When distributing a negative number, remember that multiplying two negatives produces a positive term.

Substituting \(x = 8\) gives \(\frac{8}{2} + 3 = 4 + 3 = 7\).\(11\) is wrong because it adds \(8\) and \(3\) directly without dividing \(x\) by \(2\) first.Order of operations requires performing division before addition unless parentheses say otherwise.

Distributing gives \(3x + 6\), and then combining \(3x\) with \(-x\) leaves \(2x + 6\).\(3x + 6\) is wrong because it stops right after distributing and forgets to subtract the extra \(x\) term outside the parentheses.After distributing, always check the whole expression for additional like terms to combine.

Substituting gives \(2(9) - 3(3) + 1 = 18 - 9 + 1 = 10\).\(28\) is wrong because it likely comes from squaring the whole expression incorrectly, such as computing \((2 \cdot 3)^2\) instead of \(2 \cdot 3^2\).Always evaluate exponents on the base variable before multiplying by any coefficient.

Combining the \(x\)-terms gives \(5x + 3x = 8x\), and combining the \(y\)-terms gives \(-2y - 4y = -6y\), resulting in \(8x - 6y\).\(8x + 6y\) is wrong because it ignores the negative sign on both \(y\)-terms and treats them as positive.Group and combine like terms separately, keeping careful track of each term's sign.

Distributing gives \(8x - 12\), and then adding the separate \(5\) gives \(8x - 12 + 5 = 8x - 7\).\(8x - 12\) is wrong because it stops after distributing and forgets to add the \(5\) outside the parentheses.After distributing, always finish the problem by combining any remaining constant terms.

Substituting first gives \(3(2 + (-5)) = 3(-3) = -9\).\(9\) is wrong because it likely results from treating \(b\) as positive \(5\) instead of \(-5\).When a variable has a negative value, keep the negative sign throughout the substitution and simplification.

Distributing gives \(-2x + 6\), and combining with \(4x\) gives \(-2x + 4x + 6 = 2x + 6\).\(2x - 6\) is wrong because it mishandles the sign when distributing \(-2\) across \(-3\), which should yield a positive \(6\).When distributing a negative coefficient, apply the sign rule to every term inside the parentheses individually.

Distributing both terms gives \(6x + 12 - 3x + 3\), and combining like terms results in \(3x + 15\).\(3x + 9\) is wrong because it incorrectly adds \(12\) and \(3\) as if the second distributed term had no sign change, missing that \(-3(-1) = 3\) correctly but perhaps mis-combining constants.Distribute each term completely and carefully combine only the true like terms afterward.

Substituting \(x = 5\) gives \(\frac{2(5) + 6}{2} = \frac{16}{2} = 8\).\(16\) is wrong because it correctly computes the numerator but forgets to complete the division by \(2\).When a fraction bar is present, treat the numerator as a complete grouped expression before dividing.

Distributing \(-2\) gives \(-2x - 2\), and adding the \(7\) outside gives \(7 - 2x - 2 = 5 - 2x\).\(6 - 2x\) is wrong because it likely results from subtracting only \(1\) instead of \(2\) from the \(7\).When a term is subtracted before parentheses, distribute the negative sign into every term inside first.

The phrase 'sum of a number and four' means \(x + 4\), and 'three times' that sum means multiplying the entire sum by \(3\), giving \(3(x + 4)\).\(3x + 4\) is wrong because it only multiplies the variable by \(3\) and leaves \(4\) unaffected, ignoring that the whole sum should be tripled.Phrases like 'times the sum' require grouping the addition inside parentheses before applying multiplication.

Substituting gives \(4^2 - 3^2 = 16 - 9 = 7\).\(1\) is wrong because it likely results from subtracting \(x - y\) first and squaring the difference instead of squaring each variable separately.Exponents apply only to the individual variable they're attached to, not to a combined expression unless parentheses indicate otherwise.

Distributing gives \(2x + 3x - 12 + 5\), and combining like terms results in \(5x - 7\).\(5x - 12\) is wrong because it correctly combines the \(x\)-terms but forgets to add the separate constant \(5\) at the end.After distributing, gather all constant terms together with any other constants in the expression.

A negative sign in front of parentheses distributes as \(-1\), so \(-1(3x) = -3x\) and \(-1(-5) = 5\), giving \(-3x + 5\).\(-3x - 5\) is wrong because it fails to flip the sign of the second term, treating \(-1 \times (-5)\) as if it stayed negative.A leading negative sign flips the sign of every term inside the parentheses, not just the first one.

Substituting gives \(2(6 - 2) + 3(2) = 2(4) + 6 = 8 + 6 = 14\).\(8\) is wrong because it stops after computing \(2(x-y)\) and forgets to add the separate \(3y\) term.After substituting into a multi-part expression, remember to complete every operation, including terms outside any grouped part.

Distributing \(\frac{1}{2}\) across both terms gives \(\frac{1}{2}(4x) + \frac{1}{2}(6) = 2x + 3\).\(4x + 3\) is wrong because it only multiplies the constant term by \(\frac{1}{2}\) and leaves the \(x\)-term unchanged.Fractional coefficients distribute the same way whole numbers do: multiply every term inside the parentheses.

'Twice a number' is \(2x\), and 'the sum of the number and three' is \(x + 3\), which must be subtracted as a whole group, giving \(2x - (x + 3)\).\(2x - x - 3\) is a common trap that gives the same simplified value but represents the phrase incorrectly before distributing, since the parentheses show the sum is subtracted as one quantity, matching the language before simplification.When a phrase involves subtracting a 'sum', that sum must be grouped in parentheses to preserve the intended order of operations.

Substituting first gives \((3 + 2)^2 = 5^2 = 25\).\(11\) is wrong because it squares \(x\) first and then adds \(2\), ignoring the grouping shown by the parentheses.When an expression is grouped in parentheses and raised to a power, evaluate inside the parentheses completely before applying the exponent.

Distributing gives \(6x - 6x + 3\), and since \(6x - 6x = 0\), the simplified result is just \(3\).\(0\) is wrong because it correctly cancels the \(x\)-terms but forgets that the constant \(+3\) remains in the expression.Even when variable terms cancel completely, remaining constants must still be included in the final simplified answer.

Substituting gives \(3(2(1) - 1)(1 + 1) = 3(1)(2) = 6\).\(3\) is wrong because it likely comes from multiplying only the first two factors and forgetting the last factor of \(2\).When an expression has multiple factors, multiply all of them together after substituting the given value.

Distributing gives \(4x - 8 - 6x - 3 + 10\), and combining like terms results in \(-2x + (-8 - 3 + 10) = -2x - 1\).\(-2x - 11\) is wrong because it miscombines the constants, likely treating \(10\) as negative or misadding \(-8 - 3 - 10\).When multiple grouped terms and constants appear together, combine all like terms carefully in one final step, tracking every sign.

Substituting \(x = -2\) gives \(2(-2)^2 - 3(-2) + 4 = 2(4) + 6 + 4 = 8 + 6 + 4 = 18\).\(14\) is wrong because it likely results from computing \((-2)^2\) as \(-4\) instead of the correct positive \(4\).Remember that squaring a negative number always produces a positive result, since a negative times a negative is positive.

'The difference of a number and two' is \(x - 2\), and 'three times' that difference means the whole quantity is multiplied by \(3\), giving \(3(x - 2)\), and 'four more' adds \(4\) at the end, giving \(3(x - 2) + 4\).\(3x - 2 + 4\) is wrong because it only multiplies the variable by \(3\) and leaves the \(-2\) outside the intended grouping, misrepresenting which part gets tripled.Phrases involving 'times the difference' require the subtraction to be grouped in parentheses before distributing.

Distributing gives \(2x^2 + 6x - x^2 + 4x\), and combining like terms results in \(x^2 + 10x\).\(x^2 + 2x\) is wrong because it mishandles the subtraction sign when distributing \(-x\) across \((x - 4)\), incorrectly producing \(-4x\) instead of \(+4x\).When distributing a negative term across a subtraction, remember that a negative times a negative produces a positive result.

Substituting gives \(\frac{3(16) - 8}{4} = \frac{48 - 8}{4} = \frac{40}{4} = 10\).\(46\) is wrong because it forgets to divide the final numerator by \(x\) and only completes the subtraction.The entire numerator must be simplified completely before dividing by the denominator.

Distributing both negative signs gives \(-3x - 6 - x + 5\), and combining like terms results in \(-4x - 1\).\(-4x - 11\) is wrong because it fails to flip the sign of \(-5\) when distributing the leading negative across the second parentheses, treating \(-(-5)\) as \(-5\) instead of \(+5\).A lone negative sign in front of parentheses acts like multiplying by \(-1\), flipping every sign inside.

Computing \(h(2) = 6 - 4 = 2\) and \(h(-1) = -3 - 1 = -4\), the sum is \(2 + (-4) = -2\).\(6\) is wrong because it likely only computes \(h(2)\) and forgets to add the value of \(h(-1)\).When a problem asks for the sum of two function values, evaluate each one separately and completely before adding them together.

Distributing gives \(10x - 15 - 6x + 2 + 4x\), and combining like terms yields \(8x - 13\).\(8x - 15\) is wrong because it correctly combines the \(x\)-terms but forgets to add the \(+2\) that results from \(-2(-1)\).When simplifying multi-term expressions, track and combine every constant produced by distribution, not just the largest one.

First distribute inside the brackets: \(3(x-1) + 4 = 3x - 3 + 4 = 3x + 1\), then multiply by \(2\) to get \(6x + 2\).\(6x - 2\) is wrong because it likely skips simplifying inside the bracket first and mishandles the constant sign during the outer distribution.When expressions have nested grouping symbols, simplify the innermost group completely before distributing the outer factor.

Substituting first gives \((2(2) - 1)^2 = (4 - 1)^2 = 3^2 = 9\).\(7\) is wrong because it likely squares \(x\) and \(y\) separately before combining instead of evaluating the whole grouped expression first.When a binomial expression is raised to a power, simplify inside the parentheses completely before squaring the result.

Simplifying inside the brackets first gives \(2x - 3x - 3 = -x - 3\), and then \(4x - (-x - 3) = 4x + x + 3 = 5x + 3\).\(5x - 3\) is wrong because it forgets to flip the sign of the constant when distributing the negative across the whole bracketed expression.When subtracting an entire bracketed expression, distribute the negative sign to every term inside, including constants.

Perimeter is \(2(\text{length} + \text{width}) = 2[(2x+3) + (x-1)] = 2(3x + 2) = 6x + 4\).\(3x + 2\) is wrong because it only adds the length and width once and forgets to multiply the sum by \(2\) for the full perimeter.Remember that the perimeter formula requires doubling the sum of length and width, not just adding them once.

Distributing the negative sign gives \(x^2 + 2x - x^2 + 3x - 5\), and the \(x^2\) terms cancel, leaving \(5x - 5\).\(x - 5\) is wrong because it fails to correctly combine \(2x\) and \(3x\), likely treating the sign on \(3x\) incorrectly during distribution.When subtracting a polynomial in parentheses, distribute the negative sign to every term before combining like terms.

Substituting \(x = 5\) gives \(-2(5) + 3(5 - 4) = -10 + 3(1) = -10 + 3 = -7\).\(5\) is wrong because it likely miscalculates \(3(5-4)\) or drops the negative sign on the first term during substitution.When a function combines multiplication and distribution, substitute the value everywhere it appears and then follow order of operations carefully.